a, √x+2 + √x+7=5 b, √x² - 4x +6 = x + 4 c, √x-3= √x²-5x+6 Giúp mình vs ạ
1 câu trả lời
Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge - 2\\
\sqrt {x + 2} + \sqrt {x + 7} = 5\\
\Leftrightarrow x + 2 + 2\sqrt {x + 2} .\sqrt {x + 7} + x + 7 = 25\\
\Leftrightarrow 2\sqrt {x + 2} .\sqrt {x + 7} = 16 - 2x\\
\Leftrightarrow \sqrt {\left( {x + 2} \right)\left( {x + 7} \right)} = 8 - x\left( {dk:x \le 8} \right)\\
\Leftrightarrow {x^2} + 9x + 14 = {\left( {8 - x} \right)^2}\\
\Leftrightarrow {x^2} + 9x + 14 = 64 - 16x + {x^2}\\
\Leftrightarrow 25x = 50\\
\Leftrightarrow x = 2\left( {tmdk} \right)\\
Vậy\,x = 2\\
b)Dkxd:x \ge - 4\\
\sqrt {{x^2} - 4x + 6} = x + 4\\
\Leftrightarrow {x^2} - 4x + 6 = {x^2} + 8x + 16\\
\Leftrightarrow 12x = - 10\\
\Leftrightarrow x = - \frac{5}{6}\left( {tmdk} \right)\\
Vậy\,x = - \frac{5}{6}\\
c)Dkxd:\left\{ \begin{array}{l}
x - 3 \ge 0\\
{x^2} - 5x + 6 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 3\\
\left( {x - 2} \right)\left( {x - 3} \right) \ge 0
\end{array} \right. \Leftrightarrow x \ge 3\\
\sqrt {x - 3} = \sqrt {{x^2} - 5x + 6} \\
\Leftrightarrow x - 3 = {x^2} - 5x + 6\\
\Leftrightarrow x - 3 - \left( {x - 2} \right)\left( {x - 3} \right) = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {1 - x + 2} \right) = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {3 - x} \right) = 0\\
\Leftrightarrow x = 3\left( {tmdk} \right)\\
Vậy\,x = 3
\end{array}$