Ta có: \(3x-6=\left|x-1\right|+\left|x-2\right|\ge0\left(\forall x\right)\)

\(\Rightarrow3x-6\ge0\Leftrightarrow x-2\ge0\Rightarrow x\ge2\)

Khi đó: \(\hept{\begin{cases}\left|x-1\right|=x-1\\\left|x-2\right|=x-2\end{cases}}\)

PT trở thành: \(x-1+x-2=3x-6\)

\(\Rightarrow x=3\)

Vậy x = 3

6x(x-4)+2x(2-3x)=-25

<=> 6x²-24x+4x-6x²=-25

<=> -20x=-25

<=> x=\(\frac{5}{4}\)

6x(x - 4) + 2x(2 - 3x) = -25

6x² - 24 + 4x - 6x² = -25

4x - 24 = -25

4x = -25 + 24

4x = -1

x = -1 : 4 = -0,25

\(6x\left(x-4\right)+2x\left(2-3x\right)=-25\)

\(\Rightarrow6x^2-24x+4x-6x^2=-25\)

\(\Rightarrow\left(6x^2-6x^2\right)+\left(-24x+4x\right)=-25\)

\(\Rightarrow-20x=-25\)

\(\Rightarrow x=-25:-20\)

\(\Rightarrow x=\frac{5}{4}\)

Vậy  \(x=\frac{5}{4}\)

_Chúc bạn học tốt_

x=2 nhaaaaaaaa

( x+4)=16

x       = 16-4

x       = 12

x=2

\(chuc^{bn}hoc^{gioi}\)

\(nha^{bn}bn^{nha}nha^{bn}bn^{nha}\)

\(3x^{n-2}\left(x^{n+2}-y^{n+2}\right)+y^{n+2}\left(3x^{n-2}-y^{n-2}\right)\)

\(=3x^{2n}-3x^{n-2}y^{n+2}+y^{n+2}\left(3x^{n-2}-y^{n-2}\right)\)

\(=3x^{2n}-3x^{n-2}y^{n+2}+3x^{n-2}y^{n+2}-y^{2n}\)

\(=3x^{2n}-y^{2n}\)

\(3x^{n-2}\left(x^{n+2}-y^{n+2}\right)+y^{n+2}\left(3x^{n-2}-y^{n-2}\right)\)

\(=3x^{2n}-3x^{n-2}y^{n+2}+y^{n+2}\left(3x^{n-2}-y^{n-2}\right)\)

\(=3x^{2n}-3x^{n-2}y^{n+2}+3x^{n-2}y^{n+2}-y^{2n}\)

\(=3x^{2n}-y^{2n}\)

Cho mk xin cái li ke

a=8;b=0 nha bn!

                                                        Bài giải

\(a+b=a-b\)

\(a+b+b=a\)

\(a+2b=a\)

\(2b=a-a\)

\(2b=0\)

\(b=0\text{ : }2\)

\(b=0\)

Mà \(a+b=8\text{ nên }a+0=8\)

                                 \(a=8-0=8\)

Vậy \(a=8\text{ ; }b=0\)

a+b=a-b=8 

=>a=8,b=0

a) (x-y)²-(x²-2xy)

=y²-2xy+x²-x²+2xy

=y²-(-2xy+2xy)+(x²-x²)

=y²

b)(x-y)²+x²+2xy-(x+y)²

=y²-2xy+x²+x²+2xy-y²-2xy-x²

=(y²-y²)-(2xy+2xy-2xy)+(x²+x²-x²)

=x²-2xy