11 bài tập Toán 9 Cánh diều Bài 1. Căn bậc hai và căn thức bậc hai có đáp án

a)  $x_1=\sqrt{5}\approx 2,236;x_2=-\sqrt{5}\approx -2,236$

b) $x_1=\sqrt{2,5}\approx 1,581;x_2=-\sqrt{2,5}\approx -1,581$

c)  $x_1=\sqrt{\sqrt{5}}\approx 1,495;x_2=-\sqrt{\sqrt{5}}\approx -1,495$

a) \(a - \sqrt a  = \sqrt a (\sqrt a  - 1) > 0\) nếu \(a > 1\).                                                            

b) \(a - \sqrt a  = \sqrt a \left( {\sqrt a  - 1} \right) < 0\) nếu \(0 < a < 1\).

a) ĐK: \[ - 3x + 2 \ge 0\] nên \[ - 3x \ge  - 2\] hay \[x \le \frac{2}{3}.\]

b) ĐK: \[\left\{ \begin{array}{l}\frac{4}{{2x + 3}} \ge 0\\2x + 3 \ne 0\end{array} \right.\] nên \[2x + 3 > 0\] hay \[2x >  - 3\], suy ra \[x > \frac{{ - 3}}{2}.\]

c) Đk: \[\left\{ \begin{array}{l}\frac{2}{{{x^2}}} \ge 0\\{x^2} \ne 0\end{array} \right.\] nên \[{x^2} > 0\] hay \[x \ne 0\]

d) ĐK: \[x\left( {x + 2} \right) \ge 0\]

TH1: \[\left\{ \begin{array}{l}x \ge 0\\x + 2 \ge 0\end{array} \right.\] nên \[\left\{ \begin{array}{l}x \ge 0\\x \ge  - 2\end{array} \right.\], suy ra \[x \ge 0\].

TH2: \[\left\{ \begin{array}{l}x \le 0\\x + 2 \le 0\end{array} \right.\] nên \[\left\{ \begin{array}{l}x \le 0\\x \le  - 2\end{array} \right.\], suy ra \[x \le  - 2\].

e) ĐK: \[9{x^2} - 6x + 1 \ge 0 \Leftrightarrow {\left( {3x - 1} \right)^2} \ge 0,\forall x.\]

f) ĐK: \[\frac{{2x - 1}}{{2 - x}} \ge 0\]

TH1: \[\left\{ \begin{array}{l}2x - 1 \ge 0\\2 - x > 0\end{array} \right.\] nên \[\left\{ \begin{array}{l}x \ge \frac{1}{2}\\x < 2\end{array} \right.\], suy ra \[\frac{1}{2} \le x < 2\]

TH2: \[\left\{ \begin{array}{l}2x - 1 \le 0\\2 - x < 0\end{array} \right.\] nên \[\left\{ \begin{array}{l}x \le \frac{1}{2}\\x > 2\end{array} \right.\]. Không có giá trị \(x\) thỏa mãn.

g) ĐK: \[5{x^2} - 3x + 8 \ge 0\]

\[5{x^2} - \left( {8 - 5} \right)x + 8 \ge 0\]

\[\left( {5{x^2} - 8x} \right) + \left( {5x - 8} \right) \ge 0\]

\[\left( {5x - 8} \right)\left( {x + 1} \right) \ge 0\]

TH1: \[\left\{ \begin{array}{l}5x - 8 \ge 0\\x + 1 \ge 0\end{array} \right.\] nên \[x \ge \frac{8}{5}\].

TH1: \[\left\{ \begin{array}{l}5x - 8 \le 0\\x + 1 \le 0\end{array} \right.\] nên \[x \le  - 1\].

h) ĐK: \[5{x^2} + 4x + 7 \ge 0\]

\[25{x^2} + 20x + 35 \ge 0\]

\[\left( {25{x^2} + 2.5x.2 + 4} \right) + 31 \ge 0\]

\[{\left( {5x + 2} \right)^2} + 31 \ge 0,\] với mọi \[x.\]

a) \[ - 0,8\sqrt {{{\left( { - 0,125} \right)}^2}}  =  - 0,8\left| { - 0,125} \right| =  - 0,8.0,125 =  - 0,1.\]

b) \[\sqrt {{{\left( { - 2} \right)}^6}}  = \sqrt {{{\left( { - 2} \right)}^{3.2}}}  = \sqrt {{{\left[ {{{\left( { - 2} \right)}^3}} \right]}^2}}  = \left| {{{\left( { - 2} \right)}^3}} \right| = {2^3} = 8.\]

c) \[\sqrt {{{\left( {\sqrt 3  - 2} \right)}^2}}  = \left| {\sqrt 3  - 2} \right| = 2 - \sqrt 3 .\]

d) \[\sqrt {{{\left( {2\sqrt 2  - 3} \right)}^2}}  = \left| {2\sqrt 2  - 3} \right| = 3 - 2\sqrt 2 .\]

e) \[\sqrt {{{\left( {\frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 3 }}} \right)}^2}}  = \left| {\frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 3 }}} \right| = \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 3 }}.\]

f) \[\sqrt {{{\left( {0,1 - \sqrt {0,1} } \right)}^2}}  = \left| {0,1 - \sqrt {0,1} } \right| = \left| {\sqrt {0,1} \left( {\sqrt {0,1}  - 1} \right)} \right| = \sqrt {0,1} \left( {1 - \sqrt {0,1} } \right) = \sqrt {0,1}  - 1.\]

g) \[\sqrt {4 - 2\sqrt 3 }  = \sqrt {3 - 2\sqrt 3 .1 + 1}  = \sqrt {{{\left( {\sqrt 3  - 1} \right)}^2}}  = \left| {\sqrt 3  - 1} \right| = \sqrt 3  - 1.\]

h) \[\sqrt {3 + 2\sqrt 2 }  = \sqrt {2 + 2.\sqrt 2 .1 + 1}  = \sqrt {{{\left( {\sqrt 2  + 1} \right)}^2}}  = \left| {\sqrt 2  + 1} \right| = \sqrt 2  + 1.\]

i) \[\sqrt {9 - 4\sqrt 5 }  = \sqrt {5 - 2.\sqrt 5 .2 + 4}  = \sqrt {{{\left( {\sqrt 5  - 2} \right)}^2}}  = \left| {\sqrt 5  - 2} \right| = \sqrt 5  - 2.\]

j) \[\sqrt {16 - 6\sqrt 7 }  = \sqrt {\left( {7 - 2.\sqrt 7 .3 + 9} \right)}  = \sqrt {{{\left( {\sqrt 7  - 3} \right)}^2}}  = \left| {\sqrt 7  - 3} \right| = 3 - \sqrt 7 .\]