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Đặt \(\frac{a}{2002}=\frac{b}{2003}=\frac{c}{2004}=k\)
\(\Rightarrow\hept{\begin{cases}a=2002k\\b=2003k\\c=2004k\end{cases}}\)
\(VT=4\left(a-b\right)\left(b-c\right)=4\left(2002k-2003k\right)\left(2003k-2004k\right)=4\left(-1k\right)\left(-1k\right)=4k^2\)
\(VP=\left(c-a\right)^2=\left(2004k-2002k\right)^2=\left(2k\right)^2=4k^2\)
\(\Rightarrow VT=VP\)
\(\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\left(đpcm\right)\)
4) Ta có :\(\frac{a+1}{2}=\frac{b-1}{3}=\frac{c+2}{4}=\frac{a+b+c+2}{2a+5}=\frac{a+b+c+1-1+2}{2+3+4}=\frac{a+b+c+2}{9}\)(1)
=> 2a + 5 = 9
=> 2a = 4
=> a = 2
Thay a vào (1) ta có :
\(\frac{b-1}{3}=\frac{c+2}{4}=\frac{3}{2}\)
=> \(\hept{\begin{cases}\frac{b-1}{3}=\frac{3}{2}\\\frac{c+2}{4}=\frac{3}{2}\end{cases}}\Rightarrow\hept{\begin{cases}2\left(b-1\right)=9\\2\left(c+2\right)=12\end{cases}}\Rightarrow\hept{\begin{cases}2b-2=9\\2c+4=12\end{cases}}\Rightarrow\hept{\begin{cases}2b=11\\2c=8\end{cases}\Rightarrow\hept{\begin{cases}b=5,5\\c=4\end{cases}}}\)
Vậy a = 2 ; b = 5,5 ; c = 4
5) Đặt \(\frac{a}{2002}=\frac{b}{2003}=\frac{c}{2004}=k\)
=> \(\hept{\begin{cases}a=2002k\\b=2003k\\c=2004k\end{cases}}\)
4(a - b)(b - c) = (c - a)2
=> 4(2002k - 2003k)(2003k - 2004k) = (2002k - 2004k)2
=> 4(-k)(-k) = (-2k)2
=> (-2)2(-k)2 = (-2k)2
=> 22k2 = (2k)2
=> (2k)2 = (2k)2
=> 4(a - b)(b - c) = (c - a)2 (đpcm)
a) \(\text{A : -a+b-c+a+b+c=2b}\)
b)Thay b=-1 vào A=>2 x ( -1)=-2
a, 2b
b,-2
k minh dung nhe ban minh se k cho ban nao k minh
b1 \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{a+b+c}=1\)
\(\frac{a}{b}=1\Rightarrow a=b;\frac{b}{c}=1\Rightarrow b=c;\frac{c}{a}=1\Rightarrow c=a\)
\(\Rightarrow a=b=c\)
b2 \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}=k\)
=> \(k=\frac{1}{2}\)
a) ta có : a+b=4 => (a+b)2=16 =>a2+b2=16-2ab=16-4=12
=> \(a^6+b^6=\left(a^2\right)^3+\left(b^2\right)^3=\left(a^2+b^2\right)\left(a^4-a^2b^2+b^4\right)\)
=12((a2+b2)2-3a2b2=12(122-3.16)=1152
b) \(2\left(a^2+b^2\right)=\left(a+b\right)^2\)
\(\Leftrightarrow2\left(a^2+b^2\right)=a^2+b^2+2ab\\ \Leftrightarrow a^2+b^2-2ab=0\\ \Leftrightarrow\left(a-b\right)^2=0\\ \Leftrightarrow a-b=0\Rightarrow a=b\)