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\(\left(a+b-2c\right)^2+\left(b+c-2a\right)^2+\left(c+a-2b\right)^2=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)
\(\Leftrightarrow\hept{\begin{cases}a+b-2c=a-b\\b+c-2a=b-c\\c+a-2b=c-a\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2b-2c=0\\2c-2a=0\\2a-2b=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}b-c=0\\c-a=0\\a-b=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}b=c\\c=a\\a=b\end{cases}}\)
\(\Leftrightarrow a=b=c\)( đpcm )
\(\Rightarrow\hept{\begin{cases}a+b-2c=a-b\\b+c-2a=b-c\\c+a-2b=a-c\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2b-2c=0\\2c-2a=0\\2a-2b=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}b-c=0\\c-a=0\\a-b=0\end{cases}\Rightarrow\hept{\begin{cases}b=c\\c=a\\a=b\end{cases}\Rightarrow}a=b=c\left(dpcm\right)}\)
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{2a+b}{2a-b}=\dfrac{2bk+b}{2bk-b}=\dfrac{2k+1}{2k-1}\)
\(\dfrac{2c+d}{2c-d}=\dfrac{2dk+d}{2dk-d}=\dfrac{2k+1}{2k-1}\)
=>\(\dfrac{2a+b}{2a-b}=\dfrac{2c+d}{2c-d}\)
b: \(\dfrac{2a+b}{a-2b}=\dfrac{2bk+b}{bk-2b}=\dfrac{2k+1}{k-2}\)
\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{2k+1}{k-2}\)
=>\(\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}\)
a) 6-3a-2b+ab = (6-3a) + (-2b+ab) =3(2-a) -b(2-a) = (2-a)(3-b)
b) (2a -3) (1+a) - (1-a) (3+2a) = 2a+2a^2 -3 -3a +3a +2a^2 -3 -2a = 4a^2 -6 = 2(2a^2-3)