A) Giải Phương Trình Sau:căn 9x^2 - 12x + 4 = 2x + 3 B) Rút Gọn B

Lời giải của Tự Học 365

Giải chi tiết:

a) Giải phương trình sau: \(\sqrt {9{x^2} - 12x + 4}  = 2x + 3\)

Cách 1:

Điều kiện bài toán: \(2x + 3 \ge 0 \Leftrightarrow 2x \ge  - 3 \Leftrightarrow x \ge  - \frac{2}{3}\) Khi đó, ta có:

\(\begin{array}{l} \,\sqrt {9{x^2} - 12x + 4}  = 2x + 3\\\Leftrightarrow \sqrt {{{\left( {3x - 2} \right)}^2}}  = 2x + 3\\ \Leftrightarrow \left| {3x - 2} \right| = 2x + 3\\ \Leftrightarrow \left[ \begin{array}{l}3x - 2 = 2x + 3\\3x - 2 =  - 2x - 3\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}3x - 2x = 3 + 2\\3x + 2x =  - 3 + 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 5\\5x =  - 1\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = {5^{}}\left( n \right)\\x =  - {\frac{1}{5}^{}}\left( n \right)\end{array} \right.\end{array}\)

Vậy:  \(S = \left\{ { - \frac{1}{5},5} \right\}\)

Cách 2:

\(\begin{array}{l}\,\,\,\,\,\sqrt {9{x^2} - 12x + 4}  = 2x + 3\\ \Leftrightarrow \left\{ \begin{array}{l}2x + 3 \ge 0\\{\left({\sqrt {9{x^2} - 12x + 4} } \right)^2} = {\left( {2x + 3}\right)^2}\end{array} \right.\\\Leftrightarrow \left\{ \begin{array}{l}2x \ge  - 3\\9{x^2} - 12x + 4 = 4{x^2} + 12x + 9\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge  - \frac{3}{2}\\5{x^2} - 24x - 5 = 0\end{array} \right.\Leftrightarrow \left\{ \begin{array}{l}x \ge  - \frac{3}{2}\\5{x^2} - 25x + x - 5= 0\end{array} \right.\\\Leftrightarrow \left\{ \begin{array}{l}x \ge  - \frac{3}{2}\\5x\left( {x - 5} \right) + \left( {x - 5} \right) = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge  - \frac{3}{2}\\\left( {x - 5} \right).\left( {5x + 1} \right) = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge  -\frac{3}{2}\\\left[ \begin{array}{l}x - 5 = 0\\5x + 1 = 0\end{array} \right.\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge  - \frac{3}{2}\\\left[ \begin{array}{l}x = 5\left( {tm} \right)\\x =  - \frac{1}{5}\left( {tm} \right)\end{array} \right.\end{array} \right.\end{array}\)

Vậy: \(S = \left\{ { - \frac{1}{5},5} \right\}\)

b) Rút gọn biểu thức:

\(\begin{array}{l}A = \left( {\sqrt 5  - 2} \right)\left( {\sqrt 5  + 2} \right) - \frac{{\sqrt {7 - 4\sqrt 3 } }}{{\sqrt 3  - 2}}\\A = {\left( {\sqrt 5 } \right)^2} - {2^2} + \frac{{\sqrt {{2^2} - 2.2.\sqrt 3  + {{\left( {\sqrt 3 } \right)}^2}} }}{{2 - \sqrt 3 }}\\A = 5 - 4 + \frac{{\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} }}{{2 - \sqrt 3 }}\\A = 1 + \frac{{\left| {2 - \sqrt 3 } \right|}}{{2 - \sqrt 3 }}\\A = 1 + \frac{{2 - \sqrt 3 }}{{2 - \sqrt 3 }}\,\,\,\left( {2 - \sqrt 3  > 0} \right)\\A = 1 + 1\\A = 2\end{array}\)