Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)
\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)
\(=\frac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}\)
\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)
Bài 1a/
\(\frac{1}{1+x+xy}=\frac{xyz}{xyz+x+xy}=\frac{yz}{1+y+yz}\)
\(\frac{1}{1+z+xz}=\frac{y}{y+yz+xyz}=\frac{y}{1+y+yz}\)
Vậy \(M=\frac{1}{1+y+yz}+\frac{y}{1+y+yz}+\frac{yz}{1+y+yz}=1\)
Chiều về làm tiếp
Bài 1b:Lời giải này chủ yếu nhờ dự đoán trước Min là 2011/2012 đạt được khi x=2012
Ta có \(P=\frac{2012x^2-2.2012x+2012^2}{2012x^2}=\frac{\left(x-2012\right)^2+2011x^2}{2012x^2}\ge\frac{2011x^2}{2012x^2}=\frac{2011}{2012}\)
Bài 2: Dùng phân tích thành bình phương
\(10x^2+y^2+4z^2+6x-4y-4xz+5=\left(9x^2+6x+1\right)+\left(y^2-4y+4\right)+\left(x^2-4xz+4z^2\right)\)
\(=\left(3x+1\right)^2+\left(y-2\right)^2+\left(x-2z\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}3x+1=0\\y-2=0\\x-2z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{3}\\y=2\\z=-\frac{1}{6}\end{cases}}}\)
Bài 3:
a/\(pt\Leftrightarrow\left(x+6\right)\left(x-5\right)\left(x^2-x+1\right)=0\Leftrightarrow x=-6,x=5\)
b/ta phân tích vế trái thành:\(\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\Rightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
c) \(\frac{x+9}{x^2-9}+\frac{1}{x+3}=\frac{x+9}{\left(x-3\right)\left(x+3\right)}+\frac{x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x+9+x-3}{\left(x-3\right)\left(x+3\right)}=\frac{2x+6}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{2}{x-3}\)
d) \(\frac{8m+8}{11n^2}.\frac{22n^2}{m^2+2m+1}\)
\(=\frac{8\left(m+1\right)}{11n^2}.\frac{22n^2}{\left(m+1\right)^2}\)
\(=\frac{8.2}{m+1}=\frac{16}{m+1}\)
e) \(\frac{5x+3}{4xy^3}:\frac{10x+6}{x^2y}\)
\(=\frac{5x+3}{4xy^3}.\frac{x^2y}{2.\left(5x+3\right)}\)
\(=\frac{x}{8y^2}\)
cùng nhau ta qui đồng,
MSC là 9-x2 = (3-x)(3+x) nên
-3(3+x)/MSC ; 2y/MSC ; y(3-y)/MSC
a ) MTC : \(2x\left(x+3\right)\left(x-3\right)\)
\(\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{3-2x}{x^2-9}=\frac{3-2x}{\left(x-3\right)\left(x+3\right)}=\frac{2x\left(3-2x\right)}{2x\left(x+3\right)\left(x-3\right)}\)
b ) MTC : \(2\left(-x\right)\left(x-1\right)^2\)
\(\frac{2x-1}{x-x^2}=\frac{2x-1}{-x\left(x-1\right)}=\frac{2\left(2x-1\right)\left(x-1\right)}{2\left(-x\right)\left(x-1\right)^2}\)
\(\frac{x+1}{2-4x+2x^2}=\frac{x+1}{2\left(x^2-2x+1\right)}=\frac{-x\left(x+1\right)}{2\left(-x\right)\left(x-1\right)^2}\)