Question

tìm x 
a, 2√x+2=  √x3−8
 b,(√x-2 )( 5- √x ) = 4- x

 

Answer 1

a, Không rõ đề bạn ơi ;-;

b, ĐKXĐ : \(x\ge0\)

Ta có : \(\left(\sqrt{x}-2\right)\left(5-\sqrt{x}\right)=4-x=\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)=\left(\sqrt{x}-2\right)\left(2+\sqrt{x}\right)\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-5-\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\sqrt{x}-2=0\)

\(\Leftrightarrow x=4\) ( TM )

Vậy ...

Answer 2

`b)(sqrtx-2)(5-sqrtx)=4-x`

`đk:0<=x`

`pt<=>(sqrtx-2)(sqrtx-5)=x-4`

`<=>x-7sqrtx+10=x-4`

`<=>7sqrtx=14`

`<=>sqrtx=2`

`<=>x=4(tmđk).`

Answer 3

a) Ta có: \(2\sqrt{x+2}=\sqrt{x^3-8}\)

\(\Leftrightarrow2\sqrt{x+2}-\sqrt{x+2}\cdot\sqrt{x^2-2x+4}=0\)

\(\Leftrightarrow\sqrt{x+2}\left(2-\sqrt{x^2-2x+4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x^2-2x+4=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x^2-2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\\x=2\end{matrix}\right.\)

Vậy: S={0;-2;2}