Toán Lớp 9: a, √x+2 + √x+7=5
b, √x² – 4x +6 = x + 4
c, √x-3= √x²-5x+6
Giúp mình vs ạ
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Toán Lớp 9: a, √x+2 + √x+7=5
b, √x² – 4x +6 = x + 4
c, √x-3= √x²-5x+6
Giúp mình vs ạ
Comments ( 1 )
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}$