ĐK:\(x\ge0\)

\(\left(x^2-1\right)\sqrt{x}=0\Leftrightarrow\left(x-1\right)\left(x+1\right)\sqrt{x}=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\\sqrt{x}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(ktm\right)\\x=0\left(tm\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

Ủa lớp 7 sao học căn r nè

đánh đề bằng latex cho rõ đi bạn, không biết nào dấu nào biến:v

\(\Leftrightarrow\left[{}\begin{matrix}x=20\\x=-\dfrac{10}{7}\end{matrix}\right.\)

\(\left(x-20\right).\left(7x+10\right)=0\)
\(=>\left[{}\begin{matrix}x-20=0\\7x+10=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=20\\x=-\dfrac{10}{7}\end{matrix}\right.\)

7x(x-20)+10(x-20)=0

(x-20)x(7x+10)=0

(x-20)=0 hoặc (7x+10)=0

x=20 hoặc 7x=-10

x=20 hoặc x=-10/7

=41/10

x - 1/2 = 18/5

x = 18/5 + 1/2

x = 41/10

\(x-\dfrac{1}{2}=\dfrac{3}{5}\times6\)

\(x-\dfrac{1}{2}=\dfrac{18}{5}\)

\(x=\dfrac{18}{5}+\dfrac{1}{2}\)

\(x=\dfrac{41}{10}\)

x(x+1)-(x-2)(x+1)=0

\(\left(x+1\right)\left(x-x+2\right)=0\\ \left(x+1\right)\cdot2=0\\ =>x+1=0\\ x=0-1\\ x=-1\)

=>(x+1)(x-x+2)=0

=>x+1=0

=>x=-1

Đề trước đó: 

(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x

<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x

<=>x^2-11x-6=0

<=>x^2-2x. 11/2 + 121/4-145/4=0

<=>(x-11/2)^2=145/4

<=>|x-11/2|=căn(145)/2

<=>x=[11+-căn(145)]/2

`(x+2)/(x-2)-2/(x^2-2x)=1/x`

ĐK:`x ne 0,x ne +-2`

Nhân 2 vế với `x^2-2x ne 0` ta có pt

`x(x+2)-2=x(x-2)`

`<=>x^2+2x-2=x^2-2x`

`<=>4x=2`

`<=>x=1/2.(tm)`

Vậy `S={1/2}`

Đề bài yêu cầu gì?