cho a+b+c =0. Tính: \([\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}][\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)
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14 tháng 6 2017
Gọi A= \(\frac{a-b}{c}\)+ \(\frac{b-c}{a}\)+ \(\frac{c-a}{b}\), ta có:
A*\(\frac{c}{a-b}\)= 1+\(\frac{c}{a-b}\)(\(\frac{b-c}{a}\)+\(\frac{c-a}{b}\))
= 1+ \(\frac{c}{a-b}\)* \(\frac{b^2-bc+ac-a^2}{ab}\)= 1 +\(\frac{c}{a-b}\)*\(\frac{\left(a-b\right)\left(c-a-b\right)}{ab}\)= 1+\(\frac{2c^2}{ab}\)= 1-+\(\frac{2c^3}{abc}\)
Tương tụ A* \(\frac{a}{b-c}\)= 1+\(\frac{2a^3}{abc}\)
A*\(\frac{b}{c-a}\)= 1+ \(\frac{2b^3}{abc}\)
Vậy S = 3 +\(\frac{2\left(a^3+b^3+c^3\right)}{abc}\)= 9
ở phần a3 + b3 + c3 thì tổng đấy sẽ bằng 3abc , đoạn đấy mk làm tắt nhé, bạn tự thay vào hehe
6 tháng 7 2021
1, \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)
Do đó \(\left\{{}\begin{matrix}3a=b+c+d\left(1\right)\\3b=a+c+d\left(2\right)\\3c=a+b+d\left(3\right)\\3d=a+b+c\left(4\right)\end{matrix}\right.\)
Từ (1) và (2) \(\Rightarrow3\left(a+b\right)=a+b+2c+2d\Leftrightarrow2\left(a+b\right)=2\left(c+d\right)\Leftrightarrow a+b=c+d\Leftrightarrow\dfrac{a+b}{c+d}=1\)
Tương tự cũng có: \(\dfrac{b+c}{a+d}=1;\dfrac{c+d}{a+b}=1;\dfrac{d+a}{b+c}=1\)
\(\Rightarrow A=4\)
2, Có \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)\(\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{14}{56}=\dfrac{1}{4}\)
Do đó \(\dfrac{x^2}{4}=\dfrac{1}{4};\dfrac{y^2}{16}=\dfrac{1}{4};\dfrac{z^2}{36}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=4\\z^2=9\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm2\\z=\pm3\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(1;2;3\right),\left(-1;-2;-3\right)\)
6 tháng 7 2021
Bài 2 :
a, Ta có : \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)
\(\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=4\\z^2=9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm2\\z=\pm3\end{matrix}\right.\)
Vậy ...
b, Ta có : \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{5+7}=\dfrac{2x+3y-1}{6x}\)
\(\Rightarrow6x=12\)
\(\Rightarrow x=2\)
\(\Rightarrow y=3\)
Vậy ...
6 tháng 11 2019
Đặt \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)
Ta có \(M.\frac{c}{a-b}=1+\frac{c}{a-b}\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\)
\(=1+\frac{c}{a-b}.\frac{b^2-bc+ca-a^2}{ab}\)
\(=1+\frac{c}{a-b}.\frac{\left(b-a\right)\left(a+b-c\right)}{ab}=1+\frac{2c^2}{ab}\)
Tương tự : \(M.\frac{a}{b-c}=1+\frac{2a^2}{bc};M.\frac{b}{c-a}=1+\frac{2b^2}{ca}\)
Do vậy \(A=3+2.\frac{a^3+b^3+c^3}{abc}=9\left(do.a+b+c=0.thi.a^3+b^3+c^3=3abc\right)\)
13 tháng 7 2016
a) Ta có : \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\Leftrightarrow\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{c+a}{b}+1\)
\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)
- TH1: Nếu a + b + c = 0 \(\Rightarrow P=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=\frac{-\left(abc\right)}{abc}=-1\)
- TH2 : Nếu \(a+b+c\ne0\) \(\Rightarrow a=b=c\)
\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
b) Đề bài sai ^^
Đặt \(\left(\frac{a-b}{c};\frac{b-c}{a};\frac{c-a}{b}\right)\rightarrow\left(x;y;z\right)\)
Khi đó:
\(S=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}\)
Ta có:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-cb+ac-a^2}{ab}\cdot\frac{c}{a-b}\)
\(=\frac{\left(b-a\right)\left(b+a\right)-c\left(a-b\right)}{ab}\cdot\frac{c}{a-b}=\frac{\left(b-a\right)\left(b+a-c\right)}{ab}\cdot\frac{c}{a-b}=\frac{c\left(b+a-c\right)}{ab}\)
\(=\frac{2c^2}{ab}=\frac{2c^3}{abc}\)
Một cách tương tự khi đó:\(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}=\frac{2\left(a^3+b^3+c^3\right)}{abc}=\frac{2\cdot3abc}{abc}=6\)
Khi đó:\(S=3+6=9\) Bạn để ý rằng \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\)
sao \(\frac{c\left(b+a-c\right)}{ab}\) lại bằng \(\frac{2c^2}{ab}\)