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8 tháng 1 2020

Ta có :

+) \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)

\(\Leftrightarrow\)\(\frac{ayz+bxz+cxy}{xyz}=0\) \(\left(1\right)\)

Lại có :

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

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

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

\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{cxy+ayz+bxz}{abc}\right)=0\) \(\left(2\right)\)

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

8 tháng 1 2020

Đặt \(A=\frac{x}{a},B=\frac{y}{b};C=\frac{z}{c}\)

Theo đề bài : \(\left\{{}\begin{matrix}\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=0\\A+B+C=1\end{matrix}\right.\)

Lại có : \(\left(A+B+C\right)^2=A^2+B^2+C^2+2.\left(AB+BC+CA\right)\)

\(=A^2+B^2+C^2+2ABC\left(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\right)\)

\(=A^2+B^2+C^2=1\)

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

28 tháng 4 2018

Ta có \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\Rightarrow ayz+bxz+cxy=0\left(1\right)\)(1) 

lại có \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Rightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{xz}{ac}+\frac{yz}{bc}\right)\) \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1-2\frac{ayz+bxz+cxy}{abc}\left(2\right)\) 

Thay (1) vào (2)  ta được \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(đfcm\right)\)

5 tháng 7 2016

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

Mà \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\Rightarrow ayz+bxz+cxy=0\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.\frac{0}{abc}=1\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.0=1\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(dpcm\right)\)

Chúc bạn học tốt 

1 cái T I C K nha cảm ơn

8 tháng 12 2016

Từ \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)

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

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

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

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

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

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

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

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

24 tháng 12 2018

\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow ayz+bxz+cxy=0\)

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

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

\(=1-2.\frac{cxy+bxz+ayz}{abc}=1-2.0=1\)

13 tháng 8 2017

3) Đặt b+c=x;c+a=y;a+b=z.

=>a=(y+z-x)/2 ; b=(x+z-y)/2 ; c=(x+y-z)/2

BĐT cần CM <=> \(\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}\ge\frac{3}{2}\)

VT=\(\frac{1}{2}\left(\frac{y}{x}+\frac{z}{x}-1+\frac{x}{y}+\frac{z}{y}-1+\frac{x}{z}+\frac{y}{z}-1\right)\)

\(=\frac{1}{2}\left[\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)-3\right]\)

\(\ge\frac{1}{2}\left(2+2+2-3\right)=\frac{3}{2}\)(Cauchy)

Dấu''='' tự giải ra nhá

13 tháng 8 2017

Bài 4 

dễ chứng minh \(\left(a+b\right)^2\ge4ab;\left(b+c\right)^2\ge4bc;\left(a+c\right)^2\ge4ac\)

\(\Rightarrow\left(a+b\right)^2\left(b+c\right)^2\left(a+c\right)^2\ge64a^2b^2c^2\)

rồi khai căn ra \(\Rightarrow\)dpcm. 

đấu " = " xảy ra \(\Leftrightarrow\)\(a=b=c\)

8 tháng 11 2018

chịu khó lắm

Ok

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8 tháng 11 2018

mik đăng dc 5 phút thì 5 phút sau mik lm dk rui 

8 tháng 11 2018

\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\Rightarrow ayz+bxz+cxy=0\)

\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Rightarrow\frac{xbc+yac+zab}{abc}=1\)

\(\Rightarrow xbc+yac+zab=abc\)

\(\Rightarrow\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2+2.xbc.yac+2.yac.zab+2.xbc.zab=\left(abc\right)^2\)

\(\Rightarrow x^2b^2c^2+y^2a^2c^2+z^2a^2b^2+2abc\left(cxy+ayz+bxz\right)=\left(abc\right)^2\)

\(\Rightarrow x^2b^2c^2+y^2a^2c^2+z^2a^2b^2=a^2b^2c^2\)

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

24 tháng 12 2018

\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow ayz+bxz+cxy=0\)

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

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

\(=1-2.\frac{cxy+bxz+ayz}{abc}=1-2.0=1\)

27 tháng 12 2017

bài ở đâu mà hay vậy bạn

Ta có : \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)

\(\Leftrightarrow\frac{ayz}{xyz}+\frac{bxz}{xyz}+\frac{cxy}{xyz}=0\)

\(\Leftrightarrow\frac{ayz+bxz+cxy}{xyz}=0\)

\(\Leftrightarrow ayz+bxz+cxy=0\)

Lại có : \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)

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

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

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

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

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

\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+0=1\)

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

Vậy .............................. 

25 tháng 1 2017

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

\(\frac{yz}{xyz}+\frac{xz}{xyz}+\frac{xy}{xyz}=0\)

\(\frac{yz+xz+xy}{xyz}=0\)

yz + xz + xy = 0

\(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2xz+2yz=x^2+y^2+z^2+2\times\left(xy+xz+yz\right)=x^2+y^2+z^2+2\times0=x^2+y^2+z^2\left(\text{đ}pcm\right)\)

25 tháng 1 2017

a) Từ giả thiết suy ra: xy + yz + zx = 0

Do đó:

\(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)=x^2+y^2+z^2\)

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

Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=a-b+b-c+c-a=0\)

Theo câu a ta có: \(x^2+y^2+z^2=\left(x+y+z\right)^2\)

Suy ra điều phải chứng minh