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Lời giải:
a)
$H=\frac{(x^2+y^2)(x+y)-x^2(x+1)-y^2(y-1)}{(x+1)(y-1)(x+y)}$
$=\frac{x^2y+xy^2-x^2+y^2}{(x+1)(y-1)(x+y)}$
$=\frac{xy(x+y)-(x-y)(x+y)}{(x+1)(y-1)(x+y)}=\frac{(x+y)(xy-x+y)}{(x+1)(y-1)(x+y)}$
$=\frac{xy-x+y}{(x+1)(y-1)}=\frac{xy-x+y}{xy-x+y-1}=1+\frac{1}{(x+1)(y-1)}$
b)
$H=6\Leftrightarrow \frac{1}{(x+1)(y-1)}=5$
$\Leftrightarrow (x+1)(y-1)=\frac{1}{5}$ (vô lý với mọi $x,y$ nguyên.
a) Rút gọn:
\(M=\frac{x^2}{\left(x+y\right).\left(1-y\right)}-\frac{y^2}{\left(x+y\right).\left(x+1\right)}-\frac{x^2y^2}{\left(1+x\right).\left(1-y\right)}\)
\(M=\frac{x^2}{\left(x+y\right).\left(1-y\right)}-\frac{y^2}{\left(x+y\right).\left(x+1\right)}-\frac{x^2y^2}{\left(x+1\right).\left(1-y\right)}\)
\(M=\frac{x^2.\left(x+1\right)}{\left(x+y\right).\left(1-y\right).\left(x+1\right)}-\frac{y^2.\left(1-y\right)}{\left(x+y\right).\left(1-y\right).\left(x+1\right)}-\frac{x^2y^2.\left(x+y\right)}{\left(x+y\right).\left(1-y\right).\left(x+1\right)}\)
\(M=\frac{x^2.\left(x+1\right)}{\left(x+y\right).\left(1-y\right).\left(x+1\right)}+\frac{-y^2.\left(1-y\right)}{\left(x+y\right).\left(1-y\right).\left(x+1\right)}+\frac{-x^2y^2.\left(x+y\right)}{\left(x+y\right).\left(1-y\right).\left(x+1\right)}\)
\(M=\frac{x^2.\left(x+1\right)-y^2.\left(1-y\right)-x^2y^2.\left(x+y\right)}{\left(x+y\right).\left(1-y\right).\left(x+1\right)}\)
\(M=x^2-y^2-x^2y^2.\)
Chúc bạn học tốt!
a)\(M=\frac{x^2}{\left(x+y\right)\left(1-y\right)}-\frac{y^2}{\left(x+y\right)\left(1+x\right)}-\frac{x^2y^2}{\left(1+x\right)\left(1-y\right)}\left(ĐKXĐ:x\ne-1;y\ne1\right)\)
\(M=\frac{x^2\left(1+x\right)-y^2\left(1-y\right)-x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)
\(M=\frac{x^2+x^3-y^2+y^3-x^3y^2-x^2y^3}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)
\(M=\frac{\left(x-y\right)\left(x+y\right)-x^2y^2\left(x+y\right)+x^3+y^3}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)
\(M=\frac{\left(x-y\right)\left(x+y\right)-x^2y^2\left(x+y\right)+\left(x+y\right)\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)
\(M=\frac{\left(x+y\right)\left(x-y-x^2y^2+x^2-xy+y^2\right)}{\left(x+y\right)\left(1-y\right)\left(1+x\right)}\)
\(M=\frac{x-y-x^2y^2+x^2-xy+y^2}{\left(1-y\right)\left(1+x\right)}\)
\(M=\frac{x-xy+x^2-x^2y^2+y^2-y}{\left(1-y\right)\left(1+x\right)}\)
\(M=\frac{x\left(1-y\right)+x^2\left(1-y\right)\left(1+y\right)-y\left(1-y\right)}{\left(1-y\right)\left(1+x\right)}\)
\(M=\frac{\left(1-y\right)\left(x+x^2\left(1+y\right)-y\right)}{\left(1-y\right)\left(1+x\right)}\)
\(M=\frac{x\left(x+1\right)+y\left(x-1\right)\left(x+1\right)}{1+x}\)
\(M=x+xy-y\)
b)Ta có:\(x+xy-y=-7\)
\(x\left(y+1\right)-y-1+8=0\)
\(\left(x-1\right)\left(y+1\right)=-8\)
Ta có : -8 = 8 . -1 = -8 . 1 = -2.4=-4.2
Rồi chỗ đó tự thay nha
Đây là bài dài nhất trong olm của mk
\(A\)xác định \(\Leftrightarrow x^2y^2+1+\left(x^2-y\right)\left(1-y\right)\ne0\)
\(\Leftrightarrow x^2y^2+1+x^2-x^2y-y+y^2\ne0\)
\(\Leftrightarrow\left(x^2y^2+y^2\right)+\left(x^2+1\right)-\left(x^2y+y\right)\ne0\)
\(\Leftrightarrow y^2\left(x^2+1\right)+\left(x^2+1\right)-y\left(x^2+1\right)\ne0\)
\(\Leftrightarrow\left(x^2+1\right)\left(y^2-y+1\right)\ne0\)
\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\)
Ta có: \(\hept{\begin{cases}x^2+1>0\forall x\\\left(y-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall y\end{cases}}\)\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]>0\forall x;y\)
\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\forall x;y\)
\(\Leftrightarrow A\ne0\forall x;y\)