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Ta có: \(x^2+y^2-z^2\)
\(=\left(x+y\right)^2-z^2-2xy\)
\(=\left(x+y+z\right)\left(x+y-z\right)-2xy\)
\(=-2xy\)
Ta có: \(x^2+z^2-y^2\)
\(=\left(x+z\right)^2-y^2-2xz\)
\(=\left(x+y+z\right)\left(x+z-y\right)-2xz\)
\(=-2xz\)
Ta có: \(y^2+z^2-x^2\)
\(=\left(y+z\right)^2-x^2-2yz\)
\(=\left(x+y+z\right)\left(y+z-x\right)-2yz\)
\(=-2yz\)
Ta có: \(\dfrac{xy}{x^2+y^2-z^2}+\dfrac{xz}{x^2+z^2-y^2}+\dfrac{yz}{y^2+z^2-x^2}\)
\(=\dfrac{xy}{-2xy}+\dfrac{xz}{-2xz}+\dfrac{yz}{-2yz}\)
\(=\dfrac{1}{-2}+\dfrac{1}{-2}+\dfrac{1}{-2}\)
\(=\dfrac{-3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giả thiết tương đương xy + yz + zx = 0.
Từ đó dễ dàng chứng minh được \(\left(xy\right)^3+\left(yz\right)^3+\left(zx\right)^3=3xy.yz.zx=3x^2y^2z^2\Leftrightarrow\dfrac{\left(xy\right)^3+\left(yz\right)^3+\left(zx\right)^3}{3x^2y^2z^2}=\dfrac{xy}{z^2}+\dfrac{yz}{x^2}+\dfrac{zx}{y^2}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+y^2+z^2=xy+yz+zx\\ \Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\\ \Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\Leftrightarrow x=y=z\\ \text{Mà }x+y+z=-3\Leftrightarrow x=y=z=-1\\ \Leftrightarrow B=1-1+1=1\)
Thực hiện phép tính:(1)/((y-z)(x^2+xz-y^2-yz))+(1)/((z-x)(y^2+zy-z^2-xz))+(1)/((x-y)(x^2+yz-z^2-xy|)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: \(x+y+z=a\Rightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=a^2\)
\(\Rightarrow b+2\left(xy+yz+xz\right)=a^2\Rightarrow xy+yz+xz=\frac{a^2-b}{2}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\Rightarrow\frac{xy+yz+xz}{xyz}=\frac{1}{c}\Rightarrow c\left(xy+yz+xz\right)=xyz\)
Ta có:\(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz\)
\(=a\left(b-\frac{a^2-b}{2}\right)+\frac{3c\left(a^2-b\right)}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
#)Góp ý :
Mời bạn tham khảo :
http://diendantoanhoc.net/topic/160455-%C4%91%E1%BB%81-to%C3%A1n-v%C3%B2ng-2-tuy%E1%BB%83n-sinh-10-chuy%C3%AAn-b%C3%ACnh-thu%E1%BA%ADn-2016-2017/
Mình sẽ gửi link này về chat riêng cho bạn !
Tham khảo qua đây nè :
http://diendantoanhoc.net/topic/160455-%C4%91%E1%BB%81-to%C3%A1n-v%C3%B2ng-2-tuy%E1%BB%83n-sinh-10-chuy%C3%Ân-b%C3%ACnh-thu%E1%BA%ADn-2016-2017
tk cho mk nhé
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