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Đề bài sai, đề đúng thì phân thức đằng sau dấu chia phải là:
\(\dfrac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\)
a) \(\left(x-2y\right)^2+\left(x+2y\right)^2=x^2-4xy+4y^2+x^2+4xy+4y^2=2x^2+8y^2\)
b) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2=2\left(x^2-y^2\right)+x^2+2xy+y^2+x^2-2xy^2+y^2\)
\(=2x^2-2y^2+2x^2+2y^2=4x^2\)
\(a,\left(x-2y\right)^2+\left(x+2y\right)^2\)
\(=\left(x^2-4xy+4y^2\right)
+\left(x^2+4xy+4y^2\right)\)
\(=2x^2+8y^2\)
\(b,2\left(x-y\right).\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=2\left(x^2-y^2\right)+\left(x^2+2xy+y^2\right)+\left(x^2-2xy+y^2\right)\)
\(=2x^2-2y^2+2x^2+2y^2\)
\(=4x^2\)
\(=\left[\left(\dfrac{-\left(x-y\right)}{x-2y}-\dfrac{x^2+y^2+y-2}{\left(x-2y\right)\left(x+y\right)}\right):\dfrac{\left(2x^2+y\right)^2-4}{x\left(x+y\right)+\left(x+y\right)}\right]:\dfrac{x+1}{2x^2+y+2}\)
\(=\dfrac{-x^2+y^2-x^2-y^2-y+2}{\left(x-2y\right)\left(x+y\right)}\cdot\dfrac{\left(x+y\right)\left(x+1\right)}{\left(2x^2+y-2\right)\left(2x^2+y+2\right)}\cdot\dfrac{2x^2+y+2}{x+1}\)
\(=\dfrac{-2x^2-y+2}{\left(x-2y\right)}\cdot\dfrac{\left(x+1\right)}{\left(2x^2+y-2\right)\left(2x^2+y+2\right)}\cdot\dfrac{2x^2+y+2}{x+1}\)
\(=\dfrac{-1}{x-2y}\)
Thay $x=-1,76$ và $y=\dfrac{3}{25}$ vào $P=\dfrac{-1}{x-2y}$, ta được:
$P=\dfrac{-1}{-1,76-2.(\dfrac{3}{25})}=\dfrac{1}{2}$.
1) \(2\left(x-1\right)^3-\left(x-1\right)=\left(x-1\right)\left(2\left(x-1\right)^2-1\right)\)
2) \(y\left(x-2y\right)^2+xy^2\left(2y-x\right)=\left(2y-x\right)\left(2\left(2y-x\right)+1\right)=\left(2y-x\right)\left(4y-2x+1\right)\)
3) \(xy\left(x+y\right)-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\) (xem lại đề sửa -2x thành -x mới đúng)
4) \(xy\left(x-3y\right)-2x+6y=xy\left(x-3y\right)-2\left(x-3y\right)=\left(x-3y\right)\left(xy-2\right)\)
\(\left(2x+y\right)^2=4x^2+4xy+y^2\)
\(\left(x-\frac{y}{2}\right)^2=x^2-xy+\frac{y^2}{4}\)
\(\left(x^2+\frac{y}{2}\right)\left(x^2-\frac{y}{2}\right)=x^4-\frac{y^2}{4}\)
\(\left(x-2y\right)^2\left(x+2y\right)^2=\left(x^2-4y^2\right)^2\)
\(=x^4-8x^2y^2+16y^4\)
\(\left(x+y\right)^2=x^2+2xy+y^2\)
\(\left(x-2y\right)^2=x^2-4xy+4y^2\)
\(\left(xy^2+1\right)\left(xy^2-1\right)=x^2y^4-1\)
\(\left(x+y\right)^2-4\left(x-y\right)+4=x^2+2xy+y^2-4x+4y+4\)
\(\left(2x+y\right)^2=4x^2+4xy+y^2\)
\(\left(x-\frac{y}{2}\right)^2=x^2-xy+\frac{y^2}{4}\)
\(\left(x^2+\frac{y}{2}\right)\left(x^2-\frac{y}{2}\right)=x^4-\frac{x^2y}{2}+\frac{x^2y}{2}-\frac{y^2}{4}=x^4-\frac{y^2}{4}\)
\(\left(x-2y\right)^2\left(x+2y\right)^2=x^4-8x^2y^2+16y^4\)
\(\left(x+y\right)^2=x^2+2xy+y^2\)
\(\left(x-2y\right)^2=x^2-4xy+4y^2\)
\(\left(xy^2+1\right)\left(xy^2-1\right)=x^2y^4-xy^2+xy^2-1=x^2y^4-1\)
\(\left(x+y\right)^2-4\left(x-y\right)+4=x^2+2xy+y^2-4x+4y+4\)