Tìm m , n , p để
\(\frac{1}{\left(y-1\right)^2\left(y-2\right)}=\frac{m}{y-1}+\frac{n}{\left(y-1\right)^2}+\frac{p}{y-2}\)
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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!
Theo bất đẳng thức Cô-Si, ta có \(1=x+y\ge2\sqrt{xy}\to xy\le\frac{1}{4}.\) Do vậy áp dụng bất đẳng thức Cô-Si
\(xy+\frac{1}{xy}=xy+\frac{1}{16xy}+\frac{15}{16xy}\ge2\sqrt{xy\cdot\frac{1}{16xy}}+\frac{15}{16\cdot\frac{1}{4}}=\frac{17}{4}.\)
a. Ta có \(M=\left(xy\right)^2+\frac{1}{\left(xy\right)^2}+2=\left(xy+\frac{1}{xy}\right)^2\ge\left(\frac{17}{4}\right)^2=\frac{289}{16}.\) Dấu bằng xảy ra khi \(x=y=\frac{1}{2}.\) Vây giá trị bé nhất của M là \(\frac{289}{16}.\)
b. Theo bất đẳng thức Cô-Si
\(N\ge2\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)=2\left(xy+\frac{1}{xy}\right)+2\left(\frac{x}{y}+\frac{y}{x}\right)\ge2\cdot\frac{17}{4}+4\sqrt{\frac{x}{y}\cdot\frac{y}{x}}=\frac{25}{2}.\)
Dấu bằng xảy ra khi và chỉ \(x=y=\frac{1}{2}.\)
Cho BT trên là S
Ta có: \(1+x^2=x^2+xy+yz+zx=\left(x+y\right)\left(x+z\right)\\ 1+y^2=\left(y+x\right)\left(y+z\right);1+z^2=\left(z+x\right)\left(z+y\right)\\ \Rightarrow S=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+xz+yz\right)=2\)
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