Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Hướng dẫn :\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)
Thay vào:\(x^2+2yz=x^2+yz+yz=x^2+yz-xy-zx=x\left(x-y\right)-z\left(x-y\right)=\left(x-y\right)\left(x-z\right)\)
Tương tự thay vào mà quy đồng
Em(mình) thử nhé, ko chắc đâu
3/ Ta có \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\)
\(=\left[ab\left(a+b\right)+abc\right]+\left[bc\left(b+c\right)+abc\right]+\left[ca\left(c+a\right)+ca\right]-abc\)
\(=\left(a+b+c\right)ab+\left(a+b+c\right)bc+\left(a+b+c\right)ca-abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)= -abc
Suy ra \(P=\frac{-abc}{abc}=-1\)
Vậy..
\(\frac{x^2-yz}{yz}+1+\frac{y^2-zx}{zx}+1+\frac{z^2-xy}{xy}+1=3\Leftrightarrow\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3\)
\(\Leftrightarrow\frac{1}{xyz}\left(x^3+y^3+z^3\right)=3\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)
Tới đây bạn thay vào nhé :)
a) \(A=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}+\frac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{2\left(y-z\right)\left(z-x\right)+2\left(x-y\right)\left(z-x\right)+2\left(x-y\right)\left(y-z\right)+\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left[\left(x-y\right)+\left(y-z\right)+\left(z-x\right)\right]^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(x-y+y-z+z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)
Áp dụng: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
b)Ta có: \(\frac{x^2}{y+z}+x=\frac{x^2+x\left(y+z\right)}{y+z}=\frac{x^2+xy+xz}{y+z}=\frac{x\left(x+y+z\right)}{y+z}\)
Tương tự: \(\frac{y^2}{x+z}+y=\frac{y^2+xy+zy}{x+z}=\frac{y\left(x+y+z\right)}{x+z}\)
\(\frac{z^2}{x+y}+z=\frac{z^2+xz+zy}{x+y}=\frac{z\left(x+y+z\right)}{x+y}\)
Suy ra: \(A+\left(x+y+z\right)\)
\(=\frac{x\left(x+y+z\right)}{y+z}+\frac{y\left(x+y+z\right)}{z+x}+\frac{z\left(x+y+z\right)}{x+y}+\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}+1\right)\)
\(=2.\left(x+y+z\right)\)
Nên \(A=2.\left(x+y+z\right)-\left(x+y+z\right)=x+y+z\)
Mình có sai chỗ nào không nhỉ?
\(\frac{x^2-yz}{\left(x+y\right)\left(x+z\right)}+\frac{y^2-zx}{\left(y+z\right)\left(y+x\right)}+\frac{z^2-xy}{\left(z+x\right)\left(z+y\right)}=\frac{\left(x^2-yz\right)\left(y+z\right)+\left(y^2-zx\right)\left(x+z\right)+\left(z^2-xy\right)\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)
=\(\frac{x^2y+x^2z+xy^2+y^2z+xz^2+yz^2-x^2y-x^2z-xy^2-y^2z-xz^2-yz^2}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}=\frac{0}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}=0\)
lik.e nhé!
1)\(A=\frac{b\left(2a\left(a+5b\right)+\left(a+5b\right)\right)}{a-3b}.\frac{a\left(a-3b\right)}{ab\left(a+5b\right)}=\frac{b\left(a+5b\right)\left(2a+1\right).a\left(a-3b\right)}{\left(a-3b\right).ab\left(a+5b\right)}\)
\(A=2a+1\)=>lẻ với mọi a thuộc z=> dpcm
2) từ: x+y+z=1=> xy+z=xy+1-x-y=x(y-1)-(y-1)=(y-1)(x-1)
tường tự: ta có tử của Q=(x-1)^2.(y-1)^2.(z-1)^2=[(x-1)(y-1)(z-1)]^2=[-(z+y).-(x+y).-(x+y)]^2=Mẫu=> Q=1
3) kiểm tra lại xem đề đã chuẩn chưa