Tìm \(x\) không âm, biết

a) \(\sqrt x = 5\);              b) \(\sqrt x = \sqrt 2 \);  c) \(\sqrt x = - 2\).

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

\[VT = 5 + 2.\sqrt 5 .2 + 4 = {\sqrt 5 ^2} + 2.\sqrt 5 .2 + {2^2} = {\left( {\sqrt 5  + 2} \right)^2} = VT\]

b). Ta có biến đổi:

\[\begin{array}{l}VT = \sqrt {5 + 2.\sqrt 5 .2 + 4}  - \sqrt 5  = \sqrt {{{\sqrt 5 }^2} + 2.\sqrt 5 .2 + {2^2}}  - \sqrt 5 \\\,\,\,\,\,\,\, = \sqrt {{{\left( {\sqrt 5  + 2} \right)}^2}}  - \sqrt 5  = \left| {\sqrt 5  + 2} \right| - \sqrt 5  = \sqrt 5  + 2 - \sqrt 5  = 2 = VP\end{array}\]

c). \[VT = \sqrt {16 + 2.4.\sqrt 7  + 7}  - \sqrt 7  = \sqrt {{4^2} + 2.4.\sqrt 7  + {{\sqrt 7 }^2}}  - \sqrt 7 \]

\[ = \sqrt {{{\left( {4 + \sqrt 7 } \right)}^2}}  - \sqrt 7  = \left| {4 + \sqrt 7 } \right| - \sqrt 7  = 4 + \sqrt 7  - \sqrt 7  = 4 = VP\]

d). \[VT = \sqrt {a - 2 + 2.\sqrt {a - 2} 2 + 4}  + \sqrt {a - 2 - 2.\sqrt {a - 2} .2 + 4} \]

\[ = \left| {\sqrt {a - 2}  + 2} \right| + \left| {\sqrt {a - 2}  - 2} \right| = \sqrt {a - 2}  + 2 + 2 - \sqrt {a - 2}  = 4 = VP\] (vì \[2 \le a \le 6\])

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 \[x.\]