1)Rút gọn A={[(x-y) ÷ (x+y)] + [(x+y) ÷(x-y)] } ÷ [(x ÷y)+(y ÷x)]

Lời giải:

a) ĐK: $x\geq 0; y\geq 0; x\neq y$

\(A=\left[\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{x}-\sqrt{y}}-\frac{(\sqrt{x}-\sqrt{y})(x+\sqrt{xy}+y)}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}\right]:\frac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)

\(=\left(\sqrt{x}+\sqrt{y}-\frac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right).\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

\(=\frac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}.\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

b) \(1-A=\frac{(\sqrt{x}-\sqrt{y})^2}{x-\sqrt{xy}+y}>0\) với mọi $x\neq y; x,y\geq 0$

$\Rightarrow A< 1$

\(A=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4\left(y^2-1\right)\)

   \(=\left(x-y-x-y\right)^2-4\left(y^2-1\right)\)

   \(=\left(-2y\right)^2-4y^2+4=4\)

a, x(x-y)+y(x-y)

=x²-xy+xy-y²

=x²-y² 

b, x^n-1(x+y)-y(x^n-1+y^n-1)

=xⁿ+x^n-1y-x^n-1y-yⁿ

=xⁿ-yⁿ

Chúc bn học giỏi nhoa!!!

a, x(x-y)+y(x-y)

=x²-xy+xy-y²

=x²-y² 

A=\(\frac{x^2}{\left(x+y\right)\left(1-y\right)}-\frac{y^2}{\left(x+y\right)\left(1+x\right)}\)\(-\frac{x^2y^2}{\left(1+x\right)\left(1-y\right)}\)

A=\(\frac{x^2\left(1+x\right)-y^2\left(1-y\right)-x^2y^2\left(x+y\right)}{\left(1+x\right)\left(1-y\right)\left(x+y\right)}\)

A=\(\frac{x^2+x^3-y^2+y^3-x^2y^2\left(x+y\right)}{\left(1+x\right)\left(1-y\right)\left(x+y\right)}\)

A=\(\frac{\left(x+y\right)\left(x-y\right)+\left(x+y\right)\left(x^2-xy+y^2\right)-x^2y^2\left(x+y\right)}{\left(1+x\right)\left(1+y\right)\left(x+y\right)}\)

A=\(\frac{\left(x+y\right)\left(x-y+x^2-xy+y^2-x^2y^2\right)}{\left(x+y\right)\left(x+1\right)\left(1-y\right)}\)

A=\(\frac{x\left(x+1\right)-y\left(x+1\right)+y^2\left(1-x\right)\left(1+x\right)}{\left(x+1\right)\left(1-y\right)}\)

A=\(\frac{\left(x+1\right)\left(x-y+y^2-y^2x\right)}{\left(x+1\right)\left(1-y\right)}\)

A=\(\frac{-y\left(1-y\right)+x\left(1-y\right)\left(1+y\right)}{\left(1-y\right)}\)

A=\(\frac{\left(1-y\right)\left(-y+x+xy\right)}{1-y}\)=\(x-y+xy\)

a: Sửa đề: \(\left(x-z\right)^2+z\left(x-z\right)\left(x+z\right)+\left(x+z\right)^2\)

\(=x^2-2xz+z^2+x^2+2xz+z^2+z\left(x^2-z^2\right)\)

\(=2x^2+2z^2+x^2z-z^3\)

b: \(\left(x-y\right)^2+\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)

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

\(=3x^2+y^2\)

D  nhá!!

(-3*x-1)*y+3*x^2+x

đã tắt máy chưa để cho mình giải nha

Giúp mik nha mọi người :)

a) x (x - y) + y (x - y) = x² – xy+ yx – y²

                                = x² – xy+ xy – y²

                                = x² – y²

b) x^{n – 1} (x + y) – y(x^{n – 1} + y^{n – 1}) =xⁿ+ x^{n – 1}y – yx^{n – 1} - yⁿ

                                                    = xⁿ + x^{n – 1}y - x^{n – 1}y - yⁿ

                                                    = xⁿ – yⁿ.

a) x (x - y) + y (x - y) = x² – xy+ yx – y²

                                = x² – xy+ xy – y²

                                = x² – y²

b) x^{n – 1} (x + y) – y(x^{n – 1} + y^{n – 1}) =xⁿ+ x^{n – 1}y – yx^{n – 1} - yⁿ

                                                    = xⁿ + x^{n – 1}y - x^{n – 1}y - yⁿ

                                                    = xⁿ – yⁿ.

a: ĐKXĐ: x<>y; x<>1/2; x<>-2

b:

\(\left(\frac{x+y}{y}-\frac{2y}{y-x}\right):\frac{x^2+y^2}{y-x}\)

\(=\frac{\left(y+x\right)\left(y-x\right)-2y^2}{y\left(y-x\right)}\cdot\frac{y-x}{x^2+y^2}\)

\(=\frac{-x^2-y^2}{y\left(x^2+y^2\right)}=\frac{-1}{y}\)

\(\left(\frac{x^2+1}{2x-1}-\frac{x}{2}\right)\cdot\frac{1-2x}{x+2}\)

\(=\frac{2\left(x^2+1\right)-x\left(2x-1\right)}{2\left(2x-1\right)}\cdot\frac{-\left(2x-1\right)}{x+2}\)

\(=\frac{2x^2+2-2x^2+x}{2}\cdot\frac{-1}{x+2}=\frac{x+2}{-2\left(x+2\right)}=\frac{-1}{2}\)

Ta có: \(A=\left(\frac{x+y}{y}-\frac{2y}{y-x}\right):\frac{x^2+y^2}{y-x}+\left(\frac{x^2+1}{2x-1}-\frac{x}{2}\right)\cdot\frac{1-2x}{x+2}\)

\(=\frac{-1}{y}+\frac{-1}{2}=\frac{-y-2}{2y}\)