Giải phương trình
a)3+2 √x = 5
b) √x²-10+25=5
c) √x²-12 = x+3
d)x-9 √x + 14=0
e) √x²+6x+9 =x-1
f) √x+√2x-1 + √x-√2x-1= √2
Giải phương trình a)3+2 √x = 5 b) √x²-10+25=5 c) √x²-12 = x+3 d)x-9 √x + 14=0 e) √x²+6x+9 =x-1 f) √x+√2x-1 + √x-√2x-1= √2
By Iris
Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0\\
3 + 2\sqrt x = 5\\
\Rightarrow 2\sqrt x = 2\\
\Rightarrow \sqrt x = 1\\
\Rightarrow x = 1\left( {tmdk} \right)\\
b)\sqrt {{x^2} – 10 + 25} = 5\\
\Rightarrow \sqrt {{{\left( {x – 5} \right)}^2}} = 5\\
\Rightarrow \left| {x – 5} \right| = 5\\
\Rightarrow \left[ \begin{array}{l}
x – 5 = 5\\
x – 5 = – 5
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 10\\
x = 0
\end{array} \right.\\
c)\sqrt {{x^2} – 12} = x + 3\left( {dkxd:x \ge – 3} \right)\\
\Rightarrow {x^2} – 12 = {\left( {x + 3} \right)^2}\\
\Rightarrow {x^2} – 12 = {x^2} + 6x + 9\\
\Rightarrow 6x = – 21\\
\Rightarrow x = \dfrac{{ – 7}}{2}\left( {tmdk} \right)\\
d)x – 9\sqrt x + 14 = 0\left( {dkxd:x \ge 0} \right)\\
\Rightarrow {\left( {\sqrt x } \right)^2} – 2\sqrt x – 7\sqrt x + 14 = 0\\
\Rightarrow \left( {\sqrt x – 2} \right)\left( {\sqrt x – 7} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = 2\\
\sqrt x = 7
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 4\left( {tmdk} \right)\\
x = 49\left( {tmdk} \right)
\end{array} \right.\\
e)\sqrt {{x^2} + 6x + 9} = x – 1\left( {dkxd:x \ge 1} \right)\\
\Rightarrow {x^2} + 6x + 9 = {x^2} – 2x + 1\\
\Rightarrow 8x = – 8\\
\Rightarrow x = – 1\left( {ktm} \right)\\
\Rightarrow pt\,vô\,nghiệm\\
f)\sqrt {x + \sqrt {2x – 1} } + \sqrt {x – \sqrt {2x – 1} } = \sqrt 2 \\
\left( {Dkxd:x \ge \dfrac{1}{2}} \right)\\
\Rightarrow x + \sqrt {2x – 1} + 2\sqrt {x + \sqrt {2x – 1} } .\sqrt {x – \sqrt {2x – 1} } \\
+ x – \sqrt {2x – 1} = 2\\
\Rightarrow 2x + 2\sqrt {{x^2} – \left( {2x – 1} \right)} = 2\\
\Rightarrow \sqrt {{x^2} – 2x + 1} = 1 – x\left( {dkxd:x \le 1} \right)\\
\Rightarrow \left| {x – 1} \right| = 1 – x\\
\Rightarrow x – 1 \le 0\\
\Rightarrow x \le 1\\
Vay\,\dfrac{1}{2} \le x \le 1
\end{array}$