Đề thi tuyển sinh vào lớp 10 môn Toán chuyên năm 2021-2022 sở GD&ĐT Lào Cai có đáp án

a) Với: \[\left\{ \begin{array}{l}a > 0\\a\, \ne \,\left\{ {1\,,\,2} \right\}\end{array} \right.\]

Ta có:

\[A = \left( {\frac{{a\sqrt a  - 1}}{{a - \sqrt a }} - \frac{{a\sqrt a  + 1}}{{a + \sqrt a }}} \right)\,\,:\,\,\left( {\frac{{a + 2}}{{a - 2}}} \right)\, = \left( {\frac{{\left( {\sqrt a  - 1} \right)\left( {a + \sqrt a  + 1} \right)}}{{\sqrt a \left( {\sqrt a  - 1} \right)}} - \frac{{\left( {\sqrt a  + 1} \right)\left( {a - \sqrt a  + 1} \right)}}{{\sqrt a \left( {\sqrt a  + 1} \right)}}} \right)\,\,:\,\,\left( {\frac{{a + 2}}{{a - 2}}} \right)\]

\[A = \left( {\frac{{a + \sqrt a  + 1}}{{\sqrt a }} - \frac{{a - \sqrt a  + 1}}{{\sqrt a }}} \right)\,\,:\,\,\left( {\frac{{a + 2}}{{a - 2}}} \right)\, = 2 \cdot \,\left( {\frac{{a - 2}}{{a + 2}}} \right)\, = \frac{{2a - 4}}{{a + 2}}\,\, = \,2 - \frac{8}{{a + 2}}\]

Để \[A \in \mathbb{Z}\,\, \Rightarrow \,\,2 - \frac{8}{{a + 2}}\, \in \,\mathbb{Z}\,\, \Rightarrow \,a + 2\, \in \,U\left( 8 \right)\, = \left\{ { \pm \,1\,;\, \pm \,2\,;\, \pm \,4\,;\, \pm \,8} \right\}\]

Do: \[\left\{ \begin{array}{l}a \in {\mathbb{Z}^ + }\\a\,\, \ne \,\,\left\{ {1\,;\,2} \right\}\end{array} \right.\,\, \Rightarrow \,\,a + 2\,\,\, \ge \,\,5\,\, \Rightarrow \,\,a + \,2\,\, \in \,\,\left\{ {8\,} \right\}\,\, \Rightarrow \,\,a = \,6\,\,\left( {TM} \right)\]

Vậy \[a = \,6\,\,\, \Rightarrow \,A\,\, \in \,\mathbb{Z}\]

b) Đặt :\[M = {x^5} - 2{x^4} - 2021{x^3} + 3{x^2} + 2018x - 2021\]

\[ = {x^5} - 2{x^4} - 2020{x^3} - {x^3} + 2{x^2} + 2020x + {x^2} - 2x - 2020 - 1.\]

\[\begin{array}{l} = {x^3}\left( {{x^2} - 2{x^{}} - 2020} \right) - x\left( {{x^2} - 2{x^{}} - 2020} \right) + \left( {{x^2} - 2x - 2020} \right) - 1\\ = \,\left( {{x^2} - 2x - 2020} \right)\left( {{x^3} - x + 1} \right) - 1\end{array}\]

Mà: \[x = 1 + \sqrt {2021} \,\, \Leftrightarrow \,x - 1 = \sqrt {2021} \,\, \Leftrightarrow \,{\left( {x - 1} \right)^2} = 2021\,\, \Leftrightarrow \,{x^2} - 2x - 2020 = 0.\]

\[ \Rightarrow M =  - 1\]

1) Gọi vận tốc dự định của xe đạp là: \[x\,\,\left( {km/h} \right)\,;\,\,x\,\, > \,0.\]

Vận tốc sau khi tăng tốc là: \[x + 3\,\,\left( {km/h} \right)\,.\]

Thời gian dự định là: \[\frac{{40}}{x}\,\left( h \right)\,.\]

Quãng đường từ lúc tăng tốc là: \[40 - 20 = 20\,\,\left( {km} \right)\,.\]

Thời gian lúc chưa tăng tốc là: \[\frac{{20}}{x}\,\,\left( h \right)\,.\]

Thời gian từ lúc tăng tốc là: \[\frac{{20}}{{x + 3}}\,\,\left( h \right)\,.\]

Theo đề bài ta có: \[\frac{{20}}{x} + \frac{1}{3} + \frac{{20}}{{x + 3}}\, = \frac{{40}}{x}\,\, \Leftrightarrow \,\left[ \begin{array}{l}x = 12\,\,\,\,\left( {TM} \right)\\x =  - 15\,\,\left( {KTM} \right)\end{array} \right.\]

Vậy vận tốc dự định của xe đạp là: 12 (km/h)

2) a) Ta có: \[\Delta ' = {\left[ { - \left( {m - 1} \right)} \right]^2} - 2m + 5 = {m^2} - 4m + 6 = {\left( {m - 2} \right)^2} + 2\,\, > \,\,0\,\,\forall \,m\]

Suy ra phương trình luôn có 2 nghiệm phân biệt với mọi m.

b) Theo Vi-et ta có: \[\left\{ \begin{array}{l}x_1^{} + x_2^{} = 2\left( {m - 1} \right)\\x_1^{}x_2^{} = 2m - 5\end{array} \right.\]

Do: \[x_1^{};\,\,x_2^{}\] là nghiệm của phương trình nên ta có:

\[\begin{array}{l}\left\{ \begin{array}{l}x_{^1}^2 - 2\left( {m - 1} \right)x_1^{} + 2m - 5 = 0\\x_{^2}^2 - 2\left( {m - 1} \right)x_2^{} + 2m - 5 = 0\end{array} \right.\,\,\\ \Rightarrow \,\,\left\{ \begin{array}{l}x_{^1}^2 - 2mx_1^{} + 2x_1^{} + 2m - 1 - 4 = 0\\x_{^2}^2 - 2mx_2^{} + 2x_2^{} + 2m - 1 - 4 = 0\end{array} \right.\,\,\\ \Rightarrow \,\,\left\{ \begin{array}{l}x_{^1}^2 - 2mx_1^{} + 2m - 1 = 4 - 2x_1^{}\\x_{^2}^2 - 2mx_2^{} + 2m - 1 = 4 - 2x_2^{}\end{array} \right.\end{array}\]

Mà: \[\left( {x_1^2 - 2m{x_1} + 2m - 1} \right)\,\left( {x_2^2 - 2m{x_2} + 2m - 1} \right)\,\, < \,\,0\,\,\]

\[\begin{array}{l} \Leftrightarrow \,\,\left( {4 - 2{x_1}} \right)\,\left( {4 - 2{x_2}} \right)\,\, < \,0\,\\\, \Leftrightarrow \,\,16 - 8\left( {{x_1} + {x_2}} \right)\, + 4{x_1}{x_2}\,\, < \,0\,\end{array}\]

\[ \Leftrightarrow \,\,16 - 8.2\left( {m - 1} \right)\, + 4\left( {2m - 5} \right)\,\, < \,0\,\,\, \Leftrightarrow \,\,12 - 8m\,\, < \,0\,\, \Leftrightarrow \,\,m\,\, < \,\frac{3}{2}\,\].

a) Xét \[\Delta \,AFP\,\] và \[\Delta \,ADF\] có:

 (đpcm).

b) Vì: AF và AE là 2 tiếp tuyến của \[\left( I \right)\,\, \Rightarrow \,AI\,\] là trung trực của FE \[\, \Rightarrow \,AI\, \bot \,FE\] tại Q.

\[\, \Rightarrow \,A\,{F^2} = AQ.AI\] (hệ thức lượng) \[\, \Rightarrow \,AQ.AI = AP.AD\,\,\left( { = A\,{F^2}} \right)\,\, \Rightarrow \,\frac{{AP}}{{AQ}} = \frac{{AI}}{{AD}}\].

Xét \[\Delta \,APQ\] và \[\Delta \,AID\] có: \[\,\frac{{AP}}{{AQ}} = \frac{{AI}}{{AD}}\,\,\left( {cmt} \right)\,;\,\,\widehat {\,A\,\,\,}Chung\]

 nội tiếp (vì: \[\widehat {\,AQP\,}\]là góc ngoài tại đỉnh Q)

Ta có: \[\widehat {\,{A_1}\,}\, = \,\,\widehat {\,{A_2}\,}\] (vì: AI là tia phân giác)

Xét \[\Delta \,ABN\] và \[\Delta \,BMN\] có: \[\,\widehat {\,{B_1}\,}\, = \,\,\widehat {\,{A_2}\,}\,\,\left( {cmt} \right)\,;\,\,\widehat {\,N\,\,\,}Chung\]

 (đpcm)

c) Ta có:

Mà: \[\,\left\{ \begin{array}{l}\,\,\widehat {\,IDP\,} = \,\,\widehat {\,AQP\,}\,\left( {cmt} \right)\\\widehat {\,AQT\,}\, = \,\,\widehat {\,IQD\,}\,\left( {doi\,\,dinh} \right)\end{array} \right.\,\, \Rightarrow \,\,\widehat {\,AQP\,}\, = \,\widehat {\,AQT\,}\,\, \Rightarrow \,\] đpcm

d) Gọi \[K\] là giao điểm của \[AI\,\] với

Mà:  đpcm.

a) Dự đoán điểm rơi:

\[x = y\,\, = \,\,\frac{1}{3}\,\, \Rightarrow \,\frac{1}{{{x^2}}} + ax\, + ax\,\,\mathop \ge \limits^{Co - Si} \,\,3.\sqrt[3]{{\frac{1}{{{x^2}}} \cdot ax \cdot ax}} = 3.\sqrt[3]{{{a^2}}}\,\, \Rightarrow \,\,\frac{1}{{{x^2}}} = ax\,\, \Rightarrow a = 27\]

Ta có: \[A = 53x + 53y\,\, + \,\,\frac{1}{{{x^2}}}\, + \,\,\frac{1}{{{y^2}}}\, = \left( {\,27x + 27x\,\, + \,\,\frac{1}{{{x^2}}}} \right)\, + \,\,\left( {\,27y + 27y\,\, + \,\,\frac{1}{{{y^2}}}} \right) - \left( {x + y} \right)\]

\[ \Rightarrow A\,\,\,\mathop \ge \limits^{Co - Si} \,\,3.\sqrt[3]{{27x \cdot 27x\,\, \cdot \,\,\frac{1}{{{x^2}}}}}\, + \,\,3.\sqrt[3]{{27y \cdot 27y\,\, \cdot \,\,\frac{1}{{{y^2}}}}} - \left( {x + y} \right)\,\, = 27 + 27 - \,\left( {x + y} \right)\,\, \ge \,\,54 - \frac{2}{3} = \frac{{160}}{3}\]Dấu “=” xảy ra khi \[x = y\,\, = \,\,\frac{1}{3}\]

Vậy \[ \Rightarrow Min\,A\,\,\, = \frac{{160}}{3}\,\, \Leftrightarrow \,\,x = y = \frac{1}{3}\]

b) Ta có: \[{x^4}\, + \,1\,\, \ge \,2.\sqrt {{x^4}\,.\,1} = 2{x^2}\,\,;\,\,{y^4}\, + \,1\,\, \ge \,2.\sqrt {{y^4}\,.\,1} = 2{y^2}\,\,;\,\,{z^4}\, + \,1\,\, \ge \,2.\sqrt {{z^4}\,.\,1} = 2{z^2}\]

\[ \Rightarrow {x^4}\, + \,{y^4} + {z^4}\,\, \ge \,\,2\left( {{x^2} + {y^2} + {z^2}} \right) - 3\,\, \Rightarrow \,\,VT\,\, \ge \,\,2\left( {{x^2} + {y^2} + {z^2}} \right) - 3\,\, + \,{x^3} + {y^3} + {z^3}\]

Tương tự: \[{x^3}\, + \,x\,\, \ge \,2.\sqrt {{x^3}\,.\,x} = 2{x^2}\,\,;\,\,{y^3}\, + \,y\,\, \ge \,2.\sqrt {{y^3}\,.\,y} = 2{y^2}\,\,;\,\,{z^3}\, + \,z\,\, \ge \,2.\sqrt {{z^3}\,.\,z} = 2{z^2}\]

\[\begin{array}{l} \Rightarrow {x^3}\, + \,{y^3}\, + {z^3}\,\, \ge \,\,2\left( {{x^2} + {y^2} + {z^2}} \right)\, - \left( {x + y + z} \right)\,\,\,\\ \Rightarrow \,\,\,VT\,\, \ge \,\,2\left( {{x^2} + {y^2} + {z^2}} \right)\, - \left( {x + y + z} \right)\,\, + \,\,2\left( {{x^2} + {y^2} + {z^2}} \right)\, - 3\end{array}\]

\[\begin{array}{l} \Rightarrow \,\,\,VT\,\, \ge \,\,\left( {{x^2} + {y^2} + {z^2}} \right)\, - \left( {x + y + z} \right)\,\, + \,\,3\left( {{x^2} + {y^2} + {z^2}} \right)\, - 3\,\\ \Rightarrow VT\, \ge \,\,\left( {{x^2} + {y^2} + {z^2}} \right)\, - \left( {x + y + z} \right)\,\, + \,\,3.3\, - 3\end{array}\]\[ \Rightarrow \,\,\,VT\,\, \ge \,\,\,\left( {{x^2} + {y^2} + {z^2}} \right)\, - \left( {x + y + z} \right)\,\, + \,\,6\]

Mà: \[{x^2}\, + \,1\,\, \ge \,2.\sqrt {{x^2}\,.\,1} = 2x\,\,;\,\,{y^2}\, + \,1\,\, \ge \,2.\sqrt {{y^2}\,.\,1} = 2y\,\,;\,\,{z^2}\, + \,1\,\, \ge \,2.\sqrt {{z^2}\,.\,1} = 2z\]

\[ \Rightarrow {x^2}\, + \,{y^2} + {z^2}\,\, \ge \,\,2\left( {x + y + z} \right) - 3\,\,\,\]

\[ \Rightarrow \,\,VT\,\, \ge \,\,2\left( {x + y + z} \right) - 3\, - \,\left( {x + y + z} \right) + 6\, = \,\left( {x + y + z} \right) + 3\] (đpcm)

a) Ta có:

\[\begin{array}{l}{x^2} - 2x + 2{y^2} = 2\left( {xy + 1} \right)\,\,\\ \Leftrightarrow \,\,{x^2} - 2x + 2{y^2} = 2xy + 2\,\,\\ \Leftrightarrow \,\,{x^2} - 2xy + {y^2} + {y^2} - 2x = 2\,\,\end{array}\]

\[\begin{array}{l} \Leftrightarrow {\left( {x - y} \right)^2} - 2x + {y^2} - 2y + 1 + 2y = 3\,\,\\ \Leftrightarrow \,\,{\left( {x - y} \right)^2} - 2\left( {x - y} \right) + {\left( {y - 1} \right)^2} = 3\,\,\\ \Leftrightarrow \,\,{\left( {x - y} \right)^2} - 2\left( {x - y} \right) + 1 + {\left( {y - 1} \right)^2} = 4\end{array}\]

\[ \Leftrightarrow {\left( {x - y - 1} \right)^2} + {\left( {y - 1} \right)^2} = 4\,\,\left( { = {0^2} + {2^2}} \right)\]

\[ \Leftrightarrow \,\,\left\{ \begin{array}{l}x - y - 1 = 0\\y - 1 = 2\end{array} \right.\,\,\, \vee \,\,\,\left\{ \begin{array}{l}x - y - 1 = 0\\y - 1 = - 2\end{array} \right.\,\,\, \vee \,\,\,\left\{ \begin{array}{l}y - 1 = 0\\x - y - 1 = - 2\end{array} \right.\,\, \vee \,\,\,\left\{ \begin{array}{l}y - 1 = 0\\x - y - 1 = 2\end{array} \right.\]

\[ \Leftrightarrow \,\,\left\{ \begin{array}{l}x = 4\\y = 3\end{array} \right.\,\,\, \vee \,\,\,\left\{ \begin{array}{l}x = 0\\y = - 1\end{array} \right.\,\,\, \vee \,\,\,\left\{ \begin{array}{l}y = 1\\x = 0\end{array} \right.\,\, \vee \,\,\,\left\{ \begin{array}{l}y = 1\\x = 4\end{array} \right.\]

Vậy \[\left( {x\,\,;\,\,y} \right)\,\, = \,\,\left( {4\,\,;\,\,3} \right)\,\,;\,\,\left( {0; - 1} \right)\,\,;\,\,\left( {0\,\,;\,\,1} \right)\,\,;\,\,\left( {4\,\,;\,\,1} \right).\]

b) Ta có:

\[\begin{array}{l}{x^3} + {y^3} - p = 6xy - 8\,\,\\ \Leftrightarrow \,\,p\, = \,{x^3} + {y^3} - 6xy + 8\,\,\\ \Leftrightarrow \,\,p\, = \,{\left( {x + y} \right)^3} - 3xy\left( {x + y} \right) - 6xy + 8\end{array}\]

\[\begin{array}{l} \Leftrightarrow \,\,p\, = \,\left[ {{{\left( {x + y} \right)}^3} + 8} \right] - 3xy\left( {x + y + 2} \right)\,\\\, \Leftrightarrow \,p\, = \,\left( {x + y + 2} \right)\,\left[ {{{\left( {x + y} \right)}^2} - 2\left( {x + y} \right) + 4 - 3xy} \right]\end{array}\]

Do \[p\] là số nguyên tố nên:

\[\left[ \begin{array}{l}\left( {x + y + 2} \right) = 1\,\,\\{\left( {x + y} \right)^2} - 2\left( {x + y} \right) + 4 - 3xy\, = 1\end{array} \right.\,\,\, \Rightarrow \,\,{\left( {x + y} \right)^2} - 2\left( {x + y} \right) + 4 - 3xy\, = 1\,\,\,\]

(Vì: \[x\,;\,y\,\, \in \,\,{\mathbb{Z}^ + } \Rightarrow x + y + 2\,\, \ge \,\,4\])

\[\begin{array}{l} \Rightarrow \,\,{\left( {x + y} \right)^2} - 2\left( {x + y} \right) + 4 - 3xy\, = 1\,\,\,\\ \Leftrightarrow {x^2}\, + 2xy + {y^2} - 2x - 2y - 3xy = - 3\,\,\\ \Leftrightarrow \,{x^2}\, - xy + {y^2} - 2x - 2y = - 3\end{array}\]

\[\begin{array}{l}\, \Leftrightarrow \,\,4{x^2}\, - 4xy + 4{y^2} - 8x - 8y = - 12\,\,\\ \Leftrightarrow \,\,{\left( {2x - y} \right)^2}\, + 3{y^2} - 4\left( {2x - y} \right) + 4 - 12y + 12 = 4\end{array}\]

\[\, \Leftrightarrow \,\,{\left( {2x - y - 2} \right)^2}\, + 3{\left( {y - 2} \right)^2} = 4\,\left( { = {1^2} + {{3.1}^2}} \right)\]

\[ \Leftrightarrow \,\,\left\{ \begin{array}{l}2x - y - 2 = 1\\y - 2 = 1\end{array} \right.\,\,\, \vee \,\,\,\left\{ \begin{array}{l}2x - y - 2 = 1\\y - 2 = - 1\end{array} \right.\,\,\, \vee \,\,\,\left\{ \begin{array}{l}2x - y - 2 = - 1\\y - 2 = 1\end{array} \right.\,\, \vee \,\,\,\left\{ \begin{array}{l}2x - y - 2 = - 1\\y - 2 = - 1\end{array} \right.\]

\[ \Leftrightarrow \,\,\left\{ \begin{array}{l}x = 3\\y = 3\end{array} \right.\,\,\, \vee \,\,\,\left\{ \begin{array}{l}x = 2\\y = 1\end{array} \right.\,\,\, \vee \,\,\,\left\{ \begin{array}{l}x = 2\\y = 3\end{array} \right.\,\, \vee \,\,\,\left\{ \begin{array}{l}x = 1\\y = 1\end{array} \right.\]

TH1: \[\left\{ \begin{array}{l}x = 3\\y = 3\end{array} \right.\,\,\, \Rightarrow \,\,p = 8\,\,\left( {KTM} \right)\]

TH2: \[\left\{ \begin{array}{l}x = 2\\y = 1\end{array} \right.\,\,\, \Rightarrow \,\,p = 5\,\,\left( {TM} \right)\,\]

TH3: \[\left\{ \begin{array}{l}x = 2\\y = 3\end{array} \right.\,\,\, \Rightarrow \,\,p = 7\,\,\left( {TM} \right)\]

TH4: \[\left\{ \begin{array}{l}x = 1\\y = 1\end{array} \right.\,\,\, \Rightarrow \,\,p = 4\,\,\left( {KTM} \right)\]

Vì: \[p\] là số nguyên tố lớn nhất \[ \Rightarrow \,\,p = 7\]

Vậy \[p = 7\] thỏa mãn yêu cầu bài toán.