\(\left(x+y\right)^3-\left(x-y\right)^3=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3=6x^2y+2y^3\)

\(25\left(x+y\right)^2-16\left(x-y\right)^2\)

\(=\left(5x+5y\right)^2-\left(4x-4y\right)^2\)

\(=\left(5x+5y+4x-4y\right)\left(5x+5y-4x+4y\right)\)

\(=\left(9x+y\right)\left(x+9y\right)\)

Rep ib tui với

\(\dfrac{x+2}{x-3}+\dfrac{x-2}{x}=\dfrac{x^2+2x+6}{x\left(x-3\right)}\)  đkxđ: x khác 3 , x khác 0

\(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-3\right)}+\dfrac{\left(x-2\right)\left(x-3\right)}{x\left(x-3\right)}-\dfrac{x^2+2x+6}{x\left(x-3\right)}=0\)

\(\Leftrightarrow\dfrac{x^2+2x}{....}+\dfrac{x^2-3x-2x+6}{.....}-\dfrac{x^2+2x+6}{...}=0\)

\(\Leftrightarrow x^2+2x+x^2-3x-2x+6-x^2-2x-6=0\)

\(\Leftrightarrow x^2-5x=0\)

\(\Leftrightarrow x\left(x-5\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

giúp mình câu này nx bn : x+3/x-3 - 4x^2/x^2-9=x-3/x+3

Ta có: |x+1|>=0 với mọi x

           |y+2|>=0 với mọi y

           |x-y+z|>=0 với mọi x,y,z

=>|x+1|+|y+2|+|x-y+z|>=0+0+0 với mọi x,y,z

Mà |x+1|+|y+2|+|x-y+z|=0

=>|x+1|=|y+2|=|x-y+z|=0

=>x+1=y+2=x-y+z=0

=>x=-1 và y=-2 và -1-(-2)+z=0

=>x=-1,y=-2 và z=-1

\(2,=\left(x-y\right)^2-2\left(x-y\right)=\left(x-y\right)\left(x-y-2\right)\\ 3,=\left(3x-5\right)\left(x+1\right)\\ 4,sai.đề\\ 5,=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\\ 6,=\left(x+3\right)\left(x+5\right)\)

\(\Rightarrow x^2+2x+1-y^2-4y-4-7=0\\ \Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\\ \Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=16\\\left(y+2\right)^2=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x+1=4\\y+2=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=-4\\y+2=-3\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

Bạn làm như thế này là sai rồi nhé bạn dùng HDT số 3 rồi xét các ước của pt=> nghiệm nha

m) \(\dfrac{1}{4}x^2-4x^2=\left(\dfrac{1}{2}x-2x\right)\left(\dfrac{1}{2}x+2x\right)\)

n) \(\dfrac{4}{49}-4x^2=\left(\dfrac{2}{7}-2x\right)\left(\dfrac{2}{7}+2x\right)\)

o) \(\left(x-3\right)\left(x+3\right)=x^2-9\)

Mik cần gấp 

a: \(\dfrac{3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2}{\left(y-x\right)^2}\)

\(=\dfrac{3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2}{\left(x-y\right)^2}\)

\(=3\left(x-y\right)^2+2\left(x-y\right)-5\)

b: \(\dfrac{\left(x-2y\right)^3}{x^2-4xy+4y^2}\)

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

=x-2y

c: \(\dfrac{x^3+y^3}{x+y}\)

\(=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x+y}\)

\(=x^2-xy+y^2\)