Question

1, tìm x

a,\(\sqrt{2x}\)=12

b,\(\sqrt{9x^2-6x}\)+1=10

c,\(\sqrt{5}\times x^2-\sqrt{125}\)=0

Answer 1

a) \(\sqrt{2x}=12\left(đk:x\ge0\right)\)

\(2x=144\)

\(x=72\)

b) \(\sqrt{9x^2-6x}+1=10\)\(\left(Đk:x\le0;x\ge\dfrac{2}{3}\right)\)

\(\sqrt{9x^2-6x}=9\)

\(9x^2-6x=81\)

\(\left(3x-1\right)^2=82\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{82}+1}{3}\\x=\dfrac{1-\sqrt{82}}{3}\end{matrix}\right.\)

c) \(x^2\sqrt{5}-\sqrt{125}=0\)

\(x^2\sqrt{5}=5\sqrt{5}\)

\(x^2=5\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)

Answer 2

các thầy cô giúp e vs ạ