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a/x +b/y +c/z =0 ->ayz+bxz+cxz=0
x/a + y/b + z/c=1 ->(x/a +y/b +z/c)^2=1
x^2/a^2 + y^2/b^2 + z^2/c^2 +2(xy/ab +yz/bc +xz/ac)=1
x^2/a^2 + y^2/b^2 + z^2/c^2 =1- 2* ayz+bxz+cxz/abc=1-2*0=1-0=1 =>ĐPCM
k hộ mik nha
#)Giải :
\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\rightarrow ayz+bxz+cxy=0\)
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\rightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1-2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1-2\frac{ayz+bxz+cxy}{abc}=1-2.0=1\left(đpcm\right)\)
#~Will~be~Pens~#
Có ab + bc + ca = 0
=> 2ab + 2bc + 2ca = 0
Lại có a2 + b2 + c2 = 0 (1)
=> a2 + 2ab + b2 + 2bc + c2 + 2ca = 0
=> (a + b + c)2 = 0
=> a + b + c = 0 (2)
Từ (1) và (2) => a = b = c (đpcm)
Ta có: \(\hept{\begin{cases}a^2+b^2+c^2=0\\ab+bc+ca=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2a^2+2b^2+2c^2=0\\2ab+2bc+2ca=0\end{cases}}\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0;\forall a,b,c\\\left(b-c\right)^2\ge0;\forall a,b,c\\\left(c-a\right)^2\ge0;\forall a,b,c\end{cases}}\)\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0;\forall a,b,c\)
Do đó \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\)
\(\Leftrightarrow a=b=c\left(đpcm\right)\)
ban oi a^2+b^2+c^2= a^2+b^2+c^2 là chuyện đương nhiên mà bạn
TA có \(\left(a+b+c\right)^2=0\Rightarrow ab+bc+ca=-\frac{1}{2}\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}\)
=> \(a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)
Mà \(\left(a^2+b^2+c^2\right)^2=1\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)
=> \(a^4+b^4+c^4=\frac{1}{2}\)
^_^
Ta có: a+b+c=0 <=> (a+b+c)2=0 <=> a2+b2+c2+ 2( ab+ac+bc)=0 <=> 2(ab+ac+bc)= -1 ( vì a2+b2+c2=1) <=> ab+ac+bc= -1/2
=> (ab+ac+bc)2= 1/4 <=> a2b2+a2c2+b2c2+2abc(a+b+c)= 1/4 <=> 2(a2b2+a2c2+b2c2)= 1/2 ( vì a+b+c=0) (*)
Lại có: a2+b2+c2=1 <=> (a2+b2+c2)2=1 <=> a4+b4+c4+2(a2b2+a2c2+b2c2)=1 <=> a4+b4+c4= 1/2 ( vì (*))
Vậy,...
viet ra dai lam to cho ban goi y nay: ban nhan ca hai ve cua a/(b+c) + b/(a+c) + c/(a+b) = 1 cho a+b+c xong ban se dc x+x^2/(y+z)+y+y^2/(z+x)+z+z^2/(x+y)=x+y+z chet quen to nham thanh x,y,z ban tu lam dc roi nhe