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28 tháng 11 2016

\(\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}+3=0\)

\(\Leftrightarrow\frac{y+z}{x}+1+\frac{x+z}{y}+1+\frac{x+y}{z}+1=0\)

\(\Leftrightarrow\frac{x+y+z}{x}+\frac{x+y+z}{y}+\frac{x+y+z}{z}=0\)

\(\Leftrightarrow\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=0\), luôn đúng

=> đpcm

22 tháng 5 2017

1) \(9x^2+y^2-2z^2-18x+4z-6y+20=0\)

\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)

\(\Leftrightarrow9\left(x^2-2x+1\right)+\left(y-3\right)^2+2\left(z^2+2z+1\right)=0\)

\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)

mà: \(9\left(x-1\right)^2\ge0;\left(y-3\right)^2\ge0;2\left(z+1\right)^2\ge0\)

nên \(_{\hept{\begin{cases}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}}\)

2) Ta có: \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Leftrightarrow\left(\frac{ayz+bxz+cxy}{xyz}\right)=0\Leftrightarrow ayz+bxz+cxy=0\)

Lại có: \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Rightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\Rightarrow\left(\frac{x^2}{a^2}\right)+\frac{y^2}{b^2}+\frac{z^2}{c^2}+\frac{2xy}{ab}+\frac{2yz}{bc}+\frac{2xz}{ac}=1\)

mà : \(\frac{2xy}{ab}+\frac{2yz}{bc}+\frac{2xz}{ac}=\frac{2xyabc^2+2yzbca^2+2xzacb^2}{a^2b^2c^2}=\frac{2abc\left(cxy+ayz+bxz\right)}{a^2b^2c^2}=\frac{2abc\cdot0}{a^2b^2c^2}=0\)

Vậy \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\)

22 tháng 5 2017

1 ) \(9x^2+y^2+2z^2-18x+4z-6y+20=0\)

\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)

\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)

Vì \(\hept{\begin{cases}9\left(x-1\right)^2\ge0\\\left(y-3\right)^2\ge0\\2\left(z+1\right)^2\ge0\end{cases}}\)

\(\Rightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2\ge0\)

Để \(9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\) thì \(\hept{\begin{cases}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}}\)

2 ) Ta có : \(\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)

\(\Leftrightarrow\frac{x^2}{a^2}+\frac{2xy}{ab}+\frac{y^2}{b^2}+\frac{2xz}{ac}+\frac{z^2}{c^2}+\frac{2yz}{bc}=1\)

\(\Leftrightarrow\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)+\left(\frac{2xy}{ab}+\frac{2xz}{ac}+\frac{2yz}{bc}\right)=1\)

\(\Leftrightarrow\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)+\frac{2xyz}{abc}\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)=1\)

\(\Leftrightarrow\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)+\frac{2xyz}{abc}.0=1\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\) (đpcm(

15 tháng 5 2020

Bài 3 thì \(\le1\)

Bài 4 thì \(\ge\frac{3}{4}\) nhé

13 tháng 12 2017

Tiếp tục:\(-A=\frac{x^3+y^3+z^3}{2xyz}\)

thay(1) vào A ta có

\(-A=\frac{y^3+z^3-\left(y+z\right)^3}{2xyz}=\frac{y^3+z^3-y^3-z^3-3yz\left(y+z\right)}{2xyz}\)

\(-A=\frac{3xyz}{2xyz}=\frac{3}{2}\Rightarrow A=\frac{-3}{2}\)

P/s tham khảo bài mình nhé nhớ

13 tháng 12 2017

ta có:\(x+y+z=0\) \(\Rightarrow x=-\left(y+z\right)\)

\(\Rightarrow x^3=-\left(y+z\right)^3\left(1\right)\)\(;x^2=\left(y+z\right)^2\)

\(\Rightarrow y^2+z^2-x^2=-2yz\)

CMTT:\(z^2+x^2-y^2=-2xz;x^2+y^2-z^2=-2xy\)

thay vào A ta có:

\(A=\frac{-x^2}{2yz}+\frac{-y^2}{2xz}+\frac{-z^2}{2xy}\)

10 tháng 2 2019

đặt \(\frac{x-y}{z}=a;\frac{y-z}{x}=b;\frac{z-x}{y}=c\)

\(\Rightarrow\)\(\frac{z}{x-y}=\frac{1}{a};\frac{x}{y-z}=\frac{1}{b};\frac{y}{z-x}=\frac{1}{c}\)

Ta có : \(A=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(A=1+\frac{b}{a}+\frac{c}{a}+\frac{a}{b}+1+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}+1=3+\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\)

Ta có :  \(\frac{b+c}{a}=\left(b+c\right)\frac{1}{a}=\left(\frac{y-z}{x}+\frac{z-x}{y}\right)\frac{z}{x-y}=\frac{y^2-yz+xz-x^2}{xy}.\frac{z}{x-y}=\frac{\left(y-x\right)\left(x+y-z\right)}{xy}.\frac{z}{x-y}=\frac{\left(z-x-y\right)z}{xy}=\frac{2z^2}{xy}\)vì x + y + z = 0 \(\Rightarrow\)z = -x - y

Tương tự : \(\frac{a+c}{b}=\frac{2x^2}{yz}\)\(\frac{a+b}{c}=\frac{2y^2}{xz}\)

\(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\frac{2z^2}{xy}+\frac{2x^2}{yz}+\frac{2y^2}{xz}=\frac{2\left(x^3+y^3+z^3\right)}{xyz}=\frac{2.3xyz}{xyz}=6\)( vì x + y + z = 0 \(\Rightarrow\)x3 + y3 + z3 = 3xyz )

Vậy A = 3 + 6 = 9

19 tháng 12 2016

Đặt: \(\frac{x-y}{z}+\frac{y-z}{x}+\frac{z-x}{y}=M\)

Ta có: 

\(M\cdot\frac{z}{x-y}=1+\frac{z}{x-y}\cdot\left(\frac{y-z}{x}+\frac{z-x}{y}\right)=1+\frac{z}{x-y}\cdot\frac{y^2-yz+xz-x^2}{xy}\)

\(=1+\frac{z}{x-y}\cdot\frac{\left(x-y\right)\left(z-x-y\right)}{xy}=1+\frac{2z^2}{xyz}=1+\frac{2z^3}{xyz}\)            (1)

Tương tự ta cũng có:

\(M\cdot\frac{x}{y-z}=1+\frac{2x^3}{xyz}\)              (2)

\(M\cdot\frac{y}{z-x}=1+\frac{2y^3}{xyz}\)            (3)

Từ (1);(2);(3) suy ra

\(M\cdot\left(\frac{z}{x-y}+\frac{x}{y-z}+\frac{y}{z-x}\right)=3+\frac{2\left(x^3+y^3+z^3\right)}{xyz}\)

Mà \(x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)

Nên:

\(M\cdot\left(\frac{z}{x-y}+\frac{x}{y-z}+\frac{y}{z-x}\right)=3+\frac{2\cdot3xyz}{xyz}=9\)

=>đpcm

7 tháng 6 2021

\(\dfrac{x-y}{z^2+1}=\dfrac{x-y}{z^2+xy+yz+zx}=\dfrac{x-y}{z\left(z+y\right)+x\left(z+y\right)}=\dfrac{x-y}{\left(x+z\right)\left(z+y\right)}\)

Tương tự: \(\dfrac{y-z}{x^2+1}=\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}\);\(\dfrac{z-x}{y^2+1}=\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)

Cộng vế với vế \(\Rightarrow VT=\dfrac{x-y}{\left(x+z\right)\left(y+z\right)}+\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}+\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)

\(=\dfrac{\left(x-y\right)\left(x+y\right)+\left(y-z\right)\left(y+z\right)+\left(z-x\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

\(=\dfrac{x^2-y^2+y^2-z^2+z^2-x^2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=0\)(đpcm)