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a, (\(x\) + y).(\(x\) + y)2 - 3\(xy\).(\(x\) + y)
= (\(x+y\))3 - 3\(x^2\)y - 3\(xy^2\)
= \(x^3\) + 3\(x^2\).y + 3\(xy^2\) + y3 - 3\(x^2\).y - 3\(xy^2\)
= \(x^3\) + y3
b, (\(x-y\)).(\(x-y\))2 - 3\(xy\).(\(x-y\))
= (\(x\) - y)3 - 3\(x^2\).y + 3\(xy^2\)
= \(x^3\) - 3\(x^2\)y + 3\(xy^2\) - y3 - 3\(x^2\)y + 3\(xy^2\)
= \(x^3\) - 6\(x^2\)y + 6\(xy^2\) - y3
\(B=\frac{x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)}{x^2y-x^2z+y^2z-y^3}\)
\(=\frac{x^2y-x^2z+zy^2-xy^2+z^2x-z^2y}{x^2\left(y-z\right)-y^2\left(y-z\right)}\)
\(=\frac{\left(x^2y-z^2y\right)-\left(xy^2-zy^2\right)-\left(x^2z-z^2x\right)}{\left(x^2-y^2\right)\left(y-z\right)}\)
\(=\frac{\left[y\left(x+z\right)-y^2-xz\right]\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{\left(xy+zy-y^2-xz\right)\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{\left[\left(xy-y^2\right)-\left(xz-zy\right)\right]\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{\left[y\left(x-y\right)-z\left(x-y\right)\right]\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{\left(y-z\right)\left(x-y\right)\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{x-z}{x+y}\)
\(A=\frac{\left(x^2-y\right)\left(y+1\right)+x^2y^2-1}{\left(x^2+y\right)\left(y+1\right)+x^2y^2+1}\)
\(=\frac{x^2y-y^2+x^2-y+x^2y^2-1}{x^2y+y^2+x^2+y+x^2y^2+1}\)
\(=\frac{\left(x^2y+x^2\right)+\left(x^2y^2-y^2\right)-\left(y+1\right)}{\left(x^2y+x^2\right)+\left(x^2y^2+y^2\right)+\left(y+1\right)}\)
\(=\frac{x^2\left(y+1\right)+y^2\left(x^2-1\right)-\left(y+1\right)}{x^2\left(y+1\right)+y^2\left(x^2+1\right)+\left(y+1\right)}\)
\(=\frac{\left(x^2-1\right)\left(y+1\right)+y^2\left(x^2-1\right)}{\left(x^2+1\right)\left(y+1\right)+y^2\left(x^2+1\right)}\)
\(=\frac{\left(x^2-1\right)\left(y^2+y+1\right)}{\left(x^2+1\right)\left(y^2+y+1\right)}\)
\(=\frac{x^2-1}{x^2+1}\)
a) ĐKXĐ: \(x\ne2y,x\ne-y;x\ne-1\)
b) \(B=\left(\dfrac{x-y}{2y-x}-\dfrac{x^2+y^2+y-2}{x^2-xy-2y^2}\right):\dfrac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\)
\(B=\left[\dfrac{y-x}{x-2y}-\dfrac{x^2+y^2+y-2}{\left(x+y\right)\left(x-2y\right)}\right]:\dfrac{4x^4+4x^2y+y^2-4}{x\left(x+y\right)+\left(x+y\right)}\)
\(B=\left[\dfrac{\left(y-x\right)\left(x+y\right)}{\left(x-2y\right)\left(x+y\right)}-\dfrac{x^2+y^2+y-2}{\left(x+y\right)\left(x-2y\right)}\right]:\dfrac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+1\right)\left(x+y\right)}\)
\(B=\dfrac{y^2-x^2-x^2-y^2-y+2}{\left(x+y\right)\left(x-2y\right)}:\dfrac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+1\right)\left(x+y\right)}\)
\(B=\dfrac{-2x^2-y+2}{\left(x+y\right)\left(x-2y\right)}\cdot\dfrac{\left(x+1\right)\left(x+y\right)}{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}\)
\(B=\dfrac{-\left(2x^2+y-2\right)}{\left(x+y\right)\left(x-2y\right)}\cdot\dfrac{\left(x+1\right)\left(x+y\right)}{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}\)
\(B=\dfrac{-\left(x+1\right)}{\left(x-2y\right)\left(2x^2+y+2\right)}\)
\(1)A=2x\left(x-y\right)-y\left(y-2x\right)\)
\(=2x^2-2xy-y^2+2xy\)
\(=2x^2-y^2=2.\left(-\dfrac{2}{3}\right)^2-\left(-\dfrac{1}{3}\right)^2\)
\(=\dfrac{8}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)
\(2)B=5x\left(x-4y\right)-4y\left(y-5x\right)\)
\(=5x^2-20xy-4y^2+20xy\)
\(=5x^2-4y^2=5.\left(-\dfrac{1}{5}\right)^2-4.\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{5}-1=-\dfrac{4}{5}\)
\(3)C=\text{x.(x^2-y^2)-x^2(x+y)+y(x^2-x)}\)
\(=x^3-xy^2-x^3-x^2y+x^2y-xy\)
\(=-xy\left(x+1\right)\)
a/ \(\left(x-2y\right)^2+3\left(x-2y\right)\left(x+2y\right)\)
\(=\left(x-2y\right)\left(x-2y+3x-6y\right)=\left(x-2y\right)\left(4x-8y\right)\)
\(=4\left(x-2y\right)\left(x-2y\right)=4\left(x-2y\right)^2\)
b/ \(\left(y^2+1\right)\left(y+2\right)-\left(y+2\right)\left(y^2-2y+4\right)\)
\(=y^3+2y^2+y+2-y^3-8\)
\(=2y^2+y-6=2y^2+4y-3y-6\)
\(=\left(y+2\right)\left(2y-3\right)\)
riêng câu b mình có sửa đề lại, bn xem có đúng hong nha. Chúc bn hc tốt nhé ^^
\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\)
\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)
\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)
\(a.\left(x+y\right)^2+\left(x-y\right)^2\\ =x^2+2xy+y^2+x^2-2xy+y^2\\ =2x^2+2y^2\\ b.\left(x-y\right)^2-\left(x+y\right)^2\\ =x^2-2xy+y^2-x^2-2xy-y^2\\ =-4xy\\ c.2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\\ =2\left(x^2-y^2\right)+\left(x^2+2xy+y^2\right)+\left(x^2-2xy+y^2\right)\\ =2x^2-2y^2+x^2+2xy+y^2+x^2-2xy+y^2\\ =4x^2\\ d.\left(x+y\right)^2-4xy-\left(x-y\right)^2\\ =x^2+2xy+y^2-4xy-x^2+2xy-y^2\\ =0\\ e.\left(x-2y\right)\left(x+2y\right)+\left(x+2y\right)^2\\ =x^2-4y^2+x^2+4xy+4y^2\\ =2x^2+4xy\)