Đề thi tuyển sinh vào lớp 10 môn Toán năm 2023-2024 Chuyên Ninh Bình có đáp án

a)*  P = \(\frac{{a\sqrt a  - b\sqrt b  - a\left( {\sqrt a  - \sqrt b } \right) + b\left( {\sqrt a  - \sqrt b } \right)}}{{a - b}}\)

       = \(\frac{{a\sqrt a  - b\sqrt b  - a\sqrt a  + a\sqrt b  + b\sqrt a  + b\sqrt b }}{{a - b}}\)

       = \(\frac{{\sqrt {ab} \left( {\sqrt a  + \sqrt b } \right)}}{{\left( {\sqrt a  + \sqrt b } \right)\left( {\sqrt a  - \sqrt b } \right)}}\) = \(\frac{{\sqrt a }}{{\sqrt a \; - \;\sqrt b }}\) .

*  (1 – a)(1 – b) + 2\(\sqrt {ab} \) = 1 \( \Leftrightarrow \) 1 – b – a + ab + 2\(\sqrt {ab} \) = 1

\( \Leftrightarrow \) a - 2\(\sqrt {ab} \) + b = ab \( \Leftrightarrow \) \({\left( {\sqrt a  - \sqrt b } \right)^2}\)= \({\left( {\sqrt {ab} } \right)^2}\)

\( \Leftrightarrow \) \(\left[ {\begin{array}{*{20}{c}}{\sqrt a  - \sqrt b  = \sqrt {ab} \;(khi\;a > b)}\\{\sqrt a  - \sqrt b  =  - \sqrt {ab} \;(khi\;a < b)}\end{array}} \right.\) \( \Leftrightarrow \;\left[ {\begin{array}{*{20}{c}}{\frac{{\sqrt {ab} }}{{\sqrt a \; - \;\sqrt b }} = 1}\\{\frac{{\sqrt {ab} }}{{\sqrt a \; - \;\sqrt b }} =  - 1}\end{array}} \right.\)

* Vậy P =1 (khi a > b) hoặc P = -1 (khi a < b)

b)ĐK: x\(\; \ne \;\)0 ; y\(\; \ne \;\)0

Đặt a = \(x + \frac{1}{y}\) , b = \(y + \frac{1}{x}\)

HPT đã cho trở thành \(\left\{ {\begin{array}{*{20}{c}}{a + b = \frac{9}{2}\;\;\;\;\;\;\;\;\left( 1 \right)}\\{\frac{9}{4} + \frac{3}{2}a = ab\;\;\;\;\left( 2 \right)}\end{array}} \right.\)

Từ (1): b = \(\frac{9}{2} - a\). Thay vào (2):

\(\frac{9}{4} + \frac{3}{2}a\) = a\(\left( {\frac{9}{2} - a} \right)\) \( \Leftrightarrow \) 9 + 6a = 2a (9 – 2a)

\( \Leftrightarrow \) 4a2 – 12a + 9 = 0 \( \Leftrightarrow \;{\left( {2a - 3} \right)^2}\) = 0

\( \Leftrightarrow \) 2a – 3 = 0 \( \Leftrightarrow \;\)a = \(\frac{3}{2}\) \( \Rightarrow \) b = 3

Vậy  \(\left\{ {\begin{array}{*{20}{c}}{x + \frac{1}{y} = \frac{3}{2}}\\{y + \frac{1}{x} = 3}\end{array}} \right.\) \( \Rightarrow \) \(\left\{ {\begin{array}{*{20}{c}}{2xy + 2 = 3y\;\;\;\;\left( 3 \right)}\\{xy + 1 = 3x\;\;\;\;\;\;\left( 4 \right)}\end{array}} \right.\)

\( \Rightarrow \) y = 2x. Thay vào (4):

2x2 – 3x + 1 = 0 \( \Leftrightarrow \) \(\left[ {\begin{array}{*{20}{c}}{x = 1}\\{x = \frac{1}{2}}\end{array}} \right.\)

x = 1 \( \to \) y = 2

x = \(\frac{1}{2}\) \( \to \) y = 1              (t/m đk)

Vậy (x;y)  \(\left\{ {\left( {1;2} \right);\left( {\frac{1}{2};1} \right)} \right\}\)

Áp dụng BĐT Cô-si ta có:

\(\sqrt {{b^3} + 1} \) = \(\sqrt {\left( {b + 1} \right)\left( {{b^2} - b + 1} \right)} \) \( \le \) \(\frac{{b + 1 + \;{b^2} - b + 1}}{2} = \) \(\frac{{{b^2} + \;2}}{2}\)

T2: \(\sqrt {{c^3} + 1} \) \( \le \) \(\frac{{{c^2} + \;2}}{2}\)  ;  \(\sqrt {{a^3} + 1} \) \( \le \) \(\frac{{{a^2} + \;2}}{2}\) 

Do đó VT \( \ge \) \(\frac{{2a}}{{{b^2} + \;2}}\) + \(\frac{{2b}}{{{c^2} + \;2}}\) + \(\frac{{2c}}{{{a^2} + \;2}}\)

Ta cần CM: S = \(\frac{{2a}}{{{b^2} + \;2}}\) + \(\frac{{2b}}{{{c^2} + \;2}}\) + \(\frac{{2c}}{{{a^2} + \;2}}\) \( \ge \) 2

Ta có: \(\frac{{2a}}{{{b^2} + \;2}}\) = \(\frac{{a\left( {{b^2} + 2} \right) - a{b^2}}}{{{b^2} + \;2}}\) = a - \(\frac{{a{b^2}}}{{{b^2} + \;2}}\)   

Lại có : \(\frac{{a{b^2}}}{{{b^2} + \;2}}\) = \(\frac{{2a{b^2}}}{{{b^2} + {b^2} + \;4}}\) \( \le \) \(\frac{{2a{b^2}}}{{{3^3}\sqrt {{b^4}.\;\;4} }}\) = \(\frac{{{a^3}\sqrt {2{b^2}} }}{3}\)  \( \le \) \(\frac{{a.\left( {2 + b + b} \right)}}{9}\) = \(\frac{{2a.\left( {b + 1} \right)}}{9}\)

T2 ta được S \( \ge \) a + b + c - \(\frac{{2.\left( {a + b + c} \right)}}{9}\) - \(\frac{{2\left( {ab + bc + ca} \right)}}{9}\) = \(\frac{{7.\left( {a + b + c} \right)}}{9}\) - \(\frac{{2\left( {ab + bc + ca} \right)}}{9}\)

Ta có ab + bc + ca \( \le \) \(\frac{{{{\left( {a + b + c} \right)}^2}}}{3}\)

Do đó S \( \ge \) \(\frac{{7\;.\;6}}{9}\) - \(\frac{2}{9}\) . \(\frac{{{6^2}}}{9}\) = 2

Dấu bằng xảy ra khi a = b = c = 2. Ta có đpcm.

a) Vì p là SNT lẻ nên p chỉ có 1 trong 2 dạng:

              4k + 1  hoặc  4k + 3

Vì (x2 + 1) \( \vdots \;\)p nên p có dạng 4x + 1, hay p – 1 = 4k \( \vdots \) 4.

b) Tồn tại STN x sao cho 2023p + 23p – 24 = x2

                             \( \Leftrightarrow \) x2 + 1 = 2023p + 23p – 23

Theo Fermat nhỏ, ta có 23p – 23 \( \equiv \) 0 (mod p)

=> 2023p + 23p – 23 \( \equiv \) 0 (mod p)

=> x2 + 1\( \equiv \) 0 (mod p) => p = 4k + 1

=> 2023p + 23p – 24 \( \equiv \;\)-p + (-1)p \( \equiv \) 2 (mod 4)

Mà x2 \( \equiv \) 0,1 (mod 4), mâu thuẫn

Vậy 2023p + 23p – 24 không là số chính phương.