Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\) (\(x,y,z\ne0;x\ne y\ne z\)
\(\Leftrightarrow xy+yz+xz=0\)
\(\Leftrightarrow2yz=yz-xy-xz\)
\(\Leftrightarrow x^2+2yz=\left(x-y\right)\left(x-z\right)\)
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 =))
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)
\(\Rightarrow yz=-xy-zx\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-zx}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)
\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
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)
=> \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(z-y\right)}+\dfrac{xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)
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:
\(A=\frac{yz}{(x-y)(x-z)}+\frac{xz}{(y-x)(y-z)}+\frac{xy}{(z-x)(z-y)}\)
\(A=\frac{-yz(y-z)}{(x-y)(y-z)(z-x)}+\frac{-xz(z-x)}{(x-y)(y-z)(z-x)}+\frac{-xy(x-y)}{x-y)(y-z)(z-x)}\)
\(A=\frac{xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x)}{(x-y)(y-z)(z-x)}\)
\(A=\frac{xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x)}{xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x)}=1\)
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 \(A=0.\)
Từ \(\dfrac{1}{x}\) + \(\dfrac{1}{y}\) + \(\dfrac{1}{z}\) = 0
\(=>yz+xz+xy=0\)
\(=>yz=-xz-xy\)
Ta có : \(x^2+2yz=x^2+yz+yz=x^2+yz-yx-xz=\left(y-x\right)\left(z-x\right)=-\left(x-y\right)\left(z-x\right)\)
Tương tư :
\(y^2+2xz=y^2+xz+xz=y^2+xz-xy-yz=\left(y-x\right)\left(y-z\right)=-\left(x-y\right)\left(y-z\right)\)
\(z^2+2xy=z^2+xy+xy=z^2+xy-yz-xz=\left(z-y\right)\left(z-x\right)=-\left(y-z\right)\left(z-x\right)\)Nên A = \(\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}\)
=\(\dfrac{-yz}{\left(x-y\right)\left(z-x\right)}+\dfrac{-xz}{\left(x-y\right)\left(y-z\right)}+\dfrac{-xy}{\left(y-z\right)\left(z-x\right)}\)
=\(\dfrac{-yz\left(y-z\right)-xz\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
=\(\dfrac{(-y^2z+yz^2-z^2x+x^2z-x^2y+xy^2)+(xyz-xyz)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
=\(\dfrac{\left(xyz-y^2z-z^2x+yz^2\right)+\left(-x^2y+xy^2+x^2z-xyz\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
=\(\dfrac{z\left(xy-y^2-xz+zy\right)-x\left(xy-y^2-xz+zy\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
=\(\dfrac{\left(z-x\right)\left(xy-y^2-xz+zy\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
=\(\dfrac{\left(z-x\right)\left(x-y\right)\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
=1
= =" ... bài này làm dài ..bấm máy mỏi tay lắm...
nhanh gọn lẹ.... A = 0
@phynit em hiểu nguyên tắc đó. cái em càng không hiểu là các bạn bấm chọn. trong khi cái bước rất quan trọng thì đang bỏ lửng
Em suy nghĩ rất nhiều nhiều về cái đề này. không làm nổi-->theo dõi -->
A sẽ giải thích tại sao đặt được nhân tử vậy cho nhé
Ta có:
\(xy\left(x-y\right)+yz\left(y-z\right)+zx\left(z-x\right)\)
\(=xy\left(x-y\right)+y^2z-z^2y+z^2x-zx^2\)
\(=xy\left(x-y\right)+\left(y^2z-zx^2\right)+\left(z^2x-z^2y\right)\)
\(=\left(x-y\right)\left(xy-zx-zy+z^2\right)\)
\(=\left(x-y\right)\left(\left(xy-zx\right)+\left(z^2-zy\right)\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
Cậu ta làm sai thì làm sao ngonhuminh với thầy phynit hiểu được