\(P=\left[\left(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-xy-2y^2}\right):\frac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\right]:\frac{x+1}{2x^2+y+2}\)

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

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

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

\(P=\frac{1}{2y-x}\)

Tại \(x=-1,76\) và \(y=\frac{3}{25}\) thì giá trị của \(Q=\frac{1}{2}\)

Đặt \(A=\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-xy-2y^2}\)

      \(B=\frac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\)

    \(C=\frac{x+1}{2x^2+y+2}\)

Ta có: 

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

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

B=\(\frac{\left(2x^2\right)^2+2.2x^2.y+y^2-4}{x^2+xy+x+y}=\frac{\left(2x^2+y\right)^2-4}{x\left(x+y\right)+\left(x+y\right)}=\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+1\right)\left(x+y\right)}\)

=>\(P=\left(A:B\right):C\)

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

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

        \(=\frac{1}{2y-x}\)

=>\(P=\frac{1}{2y-x}\)

Thế x=-1,76 và y=3/25 vào P

=>\(P=\frac{1}{2.\frac{3}{25}-1,76}=\frac{1}{2}\)

(Không ghi đề bài)

= (x - 5)²/ [x.(x² - 25)]

=  (x - 5)² / [x.(x - 5).(x+5)]

= (x-5) / x.(x+5)

Thay x = -3/5

=> (-3/5-5) / [-3/5.(-3/5+5)]

= -28/5 : (-3/5 . 22/5)

= -28/5 : (-66/25)

=  -28/5 . -25/66

= 70/33

Đây nhé!! Chúc bạn học tốt!!✨

đề sai 

cho M: \(\left(\frac{x^2-25}{x^3-10x^2+25}\right):\left(\frac{y-2}{y^2-y-2}\right)\)

Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)

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

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

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

Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)

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

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

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

\(x^2+9y^2-4xy-2xy+\left|x-3\right|=0\)

\(\Leftrightarrow\left(x-3y\right)^2+\left|x-3\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=3y\\x=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\) Thay vào M rồi tính nha bạn dễ ẹc

Bài 1:

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

Bài 2:

a)

ĐKXĐ: \(\left\{\begin{matrix} x-5\neq 0\\ x^2-25\neq 0\\ x+5\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x-5\neq 0\\ (x-5)(x+5)\neq 0\\ x+5\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+5\neq 0\\ x-5\neq 0\end{matrix}\right.\Leftrightarrow x\neq \pm 5\)

b)

\(A=\frac{x(x+5)}{(x+5)(x-5)}-\frac{10x}{(x-5)(x+5)}-\frac{5(x-5)}{(x-5)(x+5)}=\frac{x(x+5)-10x-5(x-5)}{(x-5)(x+5)}\)

\(=\frac{x^2-10x+25}{(x-5)(x+5)}=\frac{(x-5)^2}{(x-5)(x+5)}=\frac{x-5}{x+5}\)

c)

Khi $x=9$ thì $A=\frac{9-5}{9+5}=\frac{2}{7}$

Bài 3:

a)

ĐKXĐ: \(\left\{\begin{matrix} x^3-4x\neq 0\\ x+2\neq 0\\ 2x-4-x^2\neq 0\\ 2x^2+4x\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x(x-2)(x+2)\neq 0\\ x+2\neq 0\\ -3-(x-1)^2\neq 0\\ 2x(x+2)\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+2\neq 0\\ x-2\neq 0\\ x\neq 0\end{matrix}\right.\)

\(\Leftrightarrow x\neq \pm 2; x\neq 0\)

\(B=\left[\frac{4}{x(x-2)(x+2)}+\frac{x(x-2)}{x(x-2)(x+2)}\right]:\frac{2x-4-x^2}{2x(x+2)}\)

\(=\frac{4+x^2-2x}{x(x-2)(x+2)}.\frac{2x(x+2)}{-(x^2+4-2x)}=\frac{2}{x-2}\)

b)

Khi $x=1$ thì $B=\frac{2}{1-2}=-2$

c)

Để $B$ nguyên thì $\frac{2}{x-2}$ nguyên

$\Rightarrow 2\vdots x-2$

$\Rightarrow x-2\in\left\{\pm 1;\pm 2\right\}$

$\Rightarrow x\in\left\{3; 1; 0; 4\right\}$

Mà $x\neq \pm 2; x\neq 0$ nên $x\in\left\{3;1;4\right\}$

d: \(=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)