Cho x,y,z đôi một khác nhau và \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\). Tính giá trị biểu thức:
A=\(\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}\)
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NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 3 2021
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)
\(\Rightarrow yz=-xy-zx\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-zx}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)
\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
9 tháng 9 2021
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow\dfrac{xy+yz+xz}{xyz}=0\Leftrightarrow xy+yz+xz=0\Leftrightarrow yz=-xy-xz\)
Ta có \(x^2+2yz=x^2+yz-xy-xz=\left(x-y\right)\left(x-z\right)\)
Tương tự \(y^2+2xz=\left(y-x\right)\left(y-z\right);z^2-2xy=\left(z-x\right)\left(z-y\right)\)
\(A=\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(y-z\right)\left(y-x\right)}+\dfrac{xy}{\left(z-x\right)\left(z-y\right)}\\ A=\dfrac{-yz\left(y-z\right)-xz\left(z-x\right)-xy\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{-yz\left(y-z\right)+xz\left(y-z\right)+xz\left(x-y\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{\left(y-z\right)\left(xz-yz\right)+\left(x-y\right)\left(xz-xy\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
19 tháng 12 2020
Bài này ez thôi, làm mãi rồi.
Theo đề bài, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
=>\(\dfrac{xy+yz+xz}{xyz}=0\)
=> xy+yz+zx=0
=> \(\left\{{}\begin{matrix}xy=-yz-zx\\yz=-xy-zx\\zx=-xy-yz\end{matrix}\right.\)
Ta có: x2+2yz=x2+yz-xy-zx=(x-y)(x-z)
y2+2xz=y2+xz-xy-yz=(x-y)(z-y)
z2+2xy=z2+xy-yz-xz=(x-z)(y-z)
=> \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(z-y\right)}+\dfrac{xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)
PT
22 tháng 1 2019
Ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)\(\Rightarrow xy+yz+xz=0\)
\(\Rightarrow\left\{{}\begin{matrix}xy=-yz-xz\\yz=-xy-xz\\xz=-xy-xz\end{matrix}\right.\)
\(\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-xz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự:
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}\\\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-y\right)\left(x-z\right)}\\\dfrac{yz}{x^2+2yz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xy}{\left(x-y\right)\left(x-z\right)}+\dfrac{yz}{\left(x-y\right)\left(x-z\right)}=\dfrac{xz+xy+yz}{\left(x-y\right)\left(x-z\right)}=\dfrac{0}{\left(x-y\right)\left(x-z\right)}=0\)
Vậy \(A=0.\)
TT
2 tháng 5 2021
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\) (\(x,y,z\ne0;x\ne y\ne z\)
\(\Leftrightarrow xy+yz+xz=0\)
\(\Leftrightarrow2yz=yz-xy-xz\)
\(\Leftrightarrow x^2+2yz=\left(x-y\right)\left(x-z\right)\)
CMTT : \(\left\{{}\begin{matrix}y^2+2xz=\left(y-z\right)\left(y-x\right)\\z^2+2xy=\left(z-x\right)\left(z-y\right)\end{matrix}\right.\)
\(A=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(A=\dfrac{y^2z-yz^2-x^2z+xz^2+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(A=\dfrac{z^2\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(A=\dfrac{z^2-xz-yz+xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{x\left(y-z\right)-z\left(y-z\right)}{\left(x-z\right)\left(y-1\right)}=1\)
Thề, gõ máy mệt gấp đôi viết tay =))
LH
13 tháng 3 2017
dài đấy
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\\ < =>xy+yz+xz=0\\ < =>\left\{{}\begin{matrix}xy=-yz-xz\\yz=-xy-xz\\xz=-xy-yz\end{matrix}\right.\)
\(\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-xz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
cmtt
\(=>\left\{{}\begin{matrix}\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}\\\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-y\right)\left(x-z\right)}\end{matrix}\right.\)
A = ...
= \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xy}{\left(x-y\right)\left(x-z\right)}\)
=\(\dfrac{yz+xz+xy}{\left(x-y\right)\left(x-z\right)}\left(1\right)\)
mà xy + yz + xz = 0
=> (1) = 0
=> a = 0
Lời giải:
Từ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow \frac{xy+yz+xz}{xyz}=0\Rightarrow xy+yz+xz=0\)
Suy ra \(yz=-xy-xz\)
\(\Rightarrow x^2+2yz=x^2+yz-xy-xz=x(x-y)-z(x-y)\)
\(\Leftrightarrow x^2+2yz=(x-z)(x-y)\)
\(\Rightarrow \frac{yz}{x^2+2yz}=\frac{yz}{(x-z)(x-y)}\)
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế:
\(A=\frac{yz}{(x-y)(x-z)}+\frac{xz}{(y-x)(y-z)}+\frac{xy}{(z-x)(z-y)}\)
\(A=\frac{-yz(y-z)}{(x-y)(y-z)(z-x)}+\frac{-xz(z-x)}{(x-y)(y-z)(z-x)}+\frac{-xy(x-y)}{x-y)(y-z)(z-x)}\)
\(A=\frac{xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x)}{(x-y)(y-z)(z-x)}\)
\(A=\frac{xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x)}{xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x)}=1\)