\(x^3-2x^2+x-2=0\\ \Leftrightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ Vậy:x=2\\ ---\\ 2x\left(3x-5\right)=10-6x\\ \Leftrightarrow6x^2-10x-10+6x=0\\ \Leftrightarrow6x^2-4x-10=0\\ \Leftrightarrow6x^2+6x-10x-10=0\\ \Leftrightarrow6x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(6x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}6x-10=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)

\(4-x=2\left(x-4\right)^2\\ \Leftrightarrow4-x=2\left(x^2-8x+16\right)\\ \Leftrightarrow2x^2-16x+32+x-4=0\\ \Leftrightarrow2x^2-15x+28=0\\ \Leftrightarrow2x^2-8x-7x+28=0\\ \Leftrightarrow2x\left(x-4\right)-7\left(x-4\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\\ ---\\ 4-6x+x\left(3x-2\right)=0\\ \Leftrightarrow4-6x+3x^2-2x=0\\ \Leftrightarrow3x^2-8x+4=0\\ \Leftrightarrow3x^2-6x-2x+4=0\\ \Leftrightarrow3x\left(x-2\right)-2\left(x-2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

=> cái vế  trong ngoặc là 0.

rồi bạn làm tiếp nha

\(\dfrac{-12}{25}.\left(\dfrac{3}{4}-x+\dfrac{6}{-11}-\dfrac{5}{6}\right)=0\)

\(\dfrac{3}{4}-x+\dfrac{-6}{11}-\dfrac{5}{6}=0\)

\(\dfrac{3}{4}-x+\dfrac{-91}{66}=0\)

\(\dfrac{3}{4}-x=0-\left(\dfrac{-91}{66}\right)\)

\(\dfrac{3}{4}-x=\dfrac{91}{66}\)

\(x=\dfrac{-83}{132}\)

a: Ta có: \(-\dfrac{12}{25}\cdot\left(\dfrac{3}{4}-x+\dfrac{6}{-11}-\dfrac{5}{6}\right)=0\)

\(\Leftrightarrow x+\dfrac{83}{132}=0\)

hay \(x=-\dfrac{83}{132}\)

Bài 1: 

( x - 4 ) ( x + 4 ) < 0

=> xx + 4x - 4x + 16 < 0

x ( x + 4 - 4 )  + 16 < 0

x ( x + 0 ) + 16 < 0

xx + 16 < 0

x² + 16 < 0

x² < -16 

Ta có : x² > 0 với mọi x

-16 < 0

=> không tìm đc x t/m 

Đề bài thế này thì minh cũng chịu. Bài một thì cấp 2 mới học số âm. Bài 2 thì giữa 2 phần ngoặc kép lại ko có dấu gì cả. Nếu giữa 2 phần ngoặc kép là dấu cộng thì nó dễ hiểu hơn rồi.

\(\left(x-1\right)\left(x+y\right)=33\)

\(\Rightarrow x-1;x+y\in\text{Ư}\left(33\right)\)

\(\Rightarrow x-1;x+y\in\left\{\pm1;\pm33\right\}\)

Ta có bảng :

Vậy ...

x(x-1)-4(x-1)=0

(x-4)(x-1)=0 

=> x-4=0 hoặc x-1=0 

=> x=4 hoặc x= 1

Vậy x = { 4;1}

\(x\left(x-1\right)-4\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)

`x xx 6/7=5/14`

`=>x=5/14:6/7`

`=>x=5/14xx7/6`

`=>x=35/84`

`=>x=5/12`

Vậy `x=5/12`

__

`x:2/3=4/9`

`=>x=4/9xx2/3`

`=>x=8/27`

Vậy `x=8/27`

__

`x-1/4=3/2`

`=>x=3/2+1/4`

`=>x=6/4+1/4`

`=>x=7/4`

Vậy `x=7/4`

__

`x+4/5=8/9`

`=>x=8/9-4/5`

`=>x=40/45-36/45`

`=>x=4/45`

Vậy `x=4/45`

\(x\cdot\dfrac{6}{7}=\dfrac{5}{14}\)

\(x\)         \(=\dfrac{5}{14}:\dfrac{6}{7}\)

\(x\)           \(=\dfrac{5}{12}\)

\(x:\dfrac{2}{3}=\dfrac{4}{9}\)

\(x\)        \(=\dfrac{4}{9}\cdot\dfrac{2}{3}\)

\(x\)          \(=\dfrac{8}{27}\)

\(x-\dfrac{1}{4}=\dfrac{3}{2}\)

\(x\)          \(=\dfrac{3}{2}+\dfrac{1}{4}\)

\(x\)            \(=\dfrac{7}{4}\)

\(x+\dfrac{4}{5}=\dfrac{8}{9}\)

\(x\)          \(=\dfrac{8}{9}-\dfrac{4}{5}\)

\(x\)            \(=\dfrac{4}{45}\)