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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) 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!
ĐKXĐ: ...
Đặt \(\left(x^2;x^2+y^2;x^2+y^2+z^2\right)=\left(a;b;c\right)\Rightarrow1\le a\le b\le c\)
\(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\Leftrightarrow a+b+c=abc\)
\(\Rightarrow abc\le3c\Rightarrow ab\le3\Rightarrow ab=\left\{1;2;3\right\}\)
- TH1: \(ab=1\Rightarrow a=b=1\Rightarrow2+c=c\) (vô nghiệm)
- TH2: \(ab=2\Rightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\) \(\Rightarrow c+3=2c\Rightarrow c=3\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\x^2+y^2=2\\x^2+y^2+z^2=3\end{matrix}\right.\) \(\Rightarrow x^2=y^2=z^2=1\Rightarrow...\)
- TH3: \(ab=3\Rightarrow\left\{{}\begin{matrix}a=1\\b=3\end{matrix}\right.\) \(\Rightarrow c+4=3c\Rightarrow c=2< b\) (loại)
Vậy...
\(B=\left[\left(\frac{x}{y}-\frac{y}{x}\right):\left(x-y\right)-2.\left(\frac{1}{y}-\frac{1}{x}\right)\right]:\frac{x-y}{y}\)
\(=\left[\frac{x^2-y^2}{xy}.\frac{1}{x-y}-2.\frac{x-y}{xy}\right].\frac{y}{x-y}\)
\(=\left(\frac{\left(x-y\right)\left(x+y\right)}{xy.\left(x-y\right)}-\frac{2.\left(x-y\right)}{xy}\right).\frac{y}{x-y}\)
\(=\left(\frac{x+y}{xy}-\frac{2x-2y}{xy}\right).\frac{y}{x-y}=\frac{x+y-2x+2y}{xy}.\frac{y}{x-y}=\frac{y.\left(3y-x\right)}{xy.\left(x-y\right)}=\frac{3y-x}{x.\left(x-y\right)}\)
\(C=\left(\frac{x+y}{2x-2y}-\frac{x-y}{2x+2y}-\frac{2y^2}{y-x}\right):\frac{2y}{x-y}\)
\(=\left(\frac{x+y}{2.\left(x-y\right)}-\frac{x-y}{2.\left(x+y\right)}+\frac{2y^2}{x-y}\right).\frac{x-y}{2y}\)
\(=\frac{\left(x+y\right)^2-\left(x-y\right)^2+2.2y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{\left(x+y+x-y\right)\left(x+y-x+y\right)+4y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{4xy+4xy^2+4y^3}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}=\frac{4y.\left(x+xy+y^2\right).\left(x-y\right)}{4y.\left(x-y\right)\left(x+y\right)}=\frac{x+xy+y^2}{x+y}\)
\(D=3x:\left\{\frac{x^2-y^2}{x^3+y^3}.\left[\left(x-\frac{x^2+y^2}{y}\right):\left(\frac{1}{x}-\frac{1}{y}\right)\right]\right\}\)
\(=3x:\left\{\frac{\left(x+y\right)\left(x-y\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}.\left[\frac{xy-x^2-y^2}{y}:\frac{y-x}{xy}\right]\right\}\)
\(=3x:\left[\frac{x-y}{x^2-xy+y^2}.\left(\frac{xy-x^2-y^2}{y}.\frac{xy}{y-x}\right)\right]\)
\(=3x:\left(\frac{x-y}{x^2-xy+y^2}.\frac{xy.\left(x^2-xy+y^2\right)}{y.\left(x-y\right)}\right)\)
\(=3x:\frac{xy.\left(x-y\right)\left(x^2-xy+y^2\right)}{y.\left(x-y\right)\left(x^2-xy+y^2\right)}=3x:x=3\)
\(E=\frac{2}{x.\left(x+1\right)}+\frac{2}{\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+3\right)}\)
\(=2.\left(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\right)\)
\(=2.\frac{\left(x+2\right)\left(x+3\right)+x.\left(x+3\right)+x.\left(x+1\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{x^2+2x+3x+6+x^2+3x+x^2+x}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{3x^2+9x+6}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=2.\frac{3.\left(x^2+3x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x^2+x+2x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6.\left[x.\left(x+1\right)+2.\left(x+1\right)\right]}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x+1\right)\left(x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6}{x.\left(x+3\right)}\)
Bài làm
\(\frac{1}{\left(y-1\right)^2\left(y-2\right)}=\frac{m}{y-1}+\frac{n}{\left(y-1\right)^2}+\frac{p}{y-2}\)
ĐKXĐ : \(\hept{\begin{cases}y\ne1\\y\ne2\end{cases}}\)
MTC của VP : ( y - 1 )2( y - 2 )
<=> \(\frac{1}{\left(y-1\right)^2\left(y-2\right)}=\frac{m\left(y-1\right)\left(y-2\right)}{\left(y-1\right)^2\left(y-2\right)}+\frac{n\left(y-2\right)}{\left(y-1\right)^2\left(y-2\right)}+\frac{p\left(y-1\right)^2}{\left(y-1\right)^2\left(y-2\right)}\)
<=> \(\frac{1}{\left(y-1\right)^2\left(y-2\right)}=\frac{m\left(y^2-3y+2\right)}{\left(y-1\right)^2\left(y-2\right)}+\frac{ny-2n}{\left(y-1\right)^2\left(y-2\right)}+\frac{p\left(y^2-2y+1\right)}{\left(y-1\right)^2\left(y-2\right)}\)
<=> \(\frac{1}{\left(y-1\right)^2\left(y-2\right)}=\frac{my^2-3my+2m}{\left(y-1\right)^2\left(y-2\right)}+\frac{ny-2n}{\left(y-1\right)^2\left(y-2\right)}+\frac{py^2-2py+p}{\left(y-1\right)^2\left(y-2\right)}\)
<=> \(\frac{1}{\left(y-1\right)^2\left(y-2\right)}=\frac{my^2-3my+2m+ny-2n+py^2-2py+p}{\left(y-1\right)^2\left(y-2\right)}\)
<=> \(\frac{1}{\left(y-1\right)^2\left(y-2\right)}=\frac{\left(m+p\right)y^2+\left(-3m+n-2p\right)y+\left(2n-2n+p\right)}{\left(y-1\right)^2\left(y-2\right)}\)
Khử mẫu
<=> \(\left(m+p\right)y^2+\left(-3m+n-2p\right)y+\left(2m-2n+p\right)=1\)
Đồng nhất hệ số ta có :
\(\hept{\begin{cases}m+p=0\\-3m+n-2p=0\\2m-2n+p=1\end{cases}}\Rightarrow\hept{\begin{cases}m=n=-1\\p=1\end{cases}}\)< mình dùng máy 580VN X để giải hệ này >
Vậy m = n = -1 ; p = 1