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1) a) \(\left(a-b\right)^2-\left(a+b\right)^2=\left(a-b-a-b\right)\left(a-b+a+b\right)\)
\(=-2b\left(2a\right)=-4ab\)
b) ta có : \(\left(a+2b\right)^2+\left(b-a\right)^2-\left(a-b\right)^2=\left(a+2b\right)^2+\left(b-a\right)-\left(b-a\right)^2\)
\(=\left(a+2b\right)^2\)
2) ta có : \(\left(a-b\right)^2=\left(-\left(b-a\right)\right)^2=\left(b-a\right)^2\left(đpcm\right)\)
3) \(\left(a-b\right)^4=\left(a-b\right)^2\left(a-b\right)^2=\left(a^2-2ab+b^2\right)\left(a^2-2ab+b^2\right)\)
\(=a^4-2a^3b+a^2b^2-2a^3b+4a^2b^2-2ab^3+b^2a^2-2ab^3+b^4\)
\(=a^4-4a^3b+6a^2b^2-4ab^3+b^4\)
1)
\(\left(a+2b\right)^2+\left(b-a\right)^2-\left(a-b\right)^2\)
\(=\left(a^2+2a.2b+\left(2b\right)^2\right)+\left(b^2-2ba+a^2\right)-\left(a^2-2ab+b^2\right)\)
\(=a^2+4ab+4b^2+b^2-2ab+a^2-a^2+2ab-b^2\)
\(=a^2+4ab+4b^2\)
a. \(a^3+a^2c-abc+b^2c+b^3\)
<=> \(\left(a^3+b^3\right)+c\left(a^2-ab+b^2\right)\)
<=> (\(\left(a+b\right)\left(a^2-ab+b^2\right)+c\left(a^2-ab+b^2\right)\)
<=> \(\left(a+b+c\right)\left(a^2-ab+b^2\right)\)
vì a+b+c =0 => đpcm
b. 2(a+1)(b+1)=(a+b)(a+b+2)
<=> \(2\left(ab+a+b+1\right)=\)\(a^2+ab+2a+ab+b^2+2b\)
<=> \(2ab+2a+2b+2=a^2ab+2a+ab+b^2+2b\)
<=> \(a^2+b^2=2\)=> đpcm
1,
Ta có
a + 2b + 3c = 14
=> 2a +4b +6c = 28
Mà a2 + b2 + c2 = 14
Nên a2 + b2 + c2 - 2a - 4b -6c =14 - 28
=> a2 +b2 +c2 -2a -4b - 6c + 14=0
=> (a2 - 2a +1) + (b2 -4b +4 ) + ( c2 - 6c + 9) = 0
=> (a-1)2 + ( b-2 )2 +(c-3)2 =0
=> \(\hept{\begin{cases}a-1=0\\b-2=0\\c-3=0\end{cases}}\Rightarrow\hept{\begin{cases}a=1\\b=2\\c=3\end{cases}}\)
Vậy abc = 6
c) \(\frac{a\left(a^2-ab+b^2\right)}{b\left(a+b\right)\left(a^2-ab+b^2\right)}\)
=\(\frac{a}{b\left(a+b\right)}\)
1. Ta có : x + y + z = 0 \(\Rightarrow\)( x + y + z )2 = 0 \(\Rightarrow\)x2 + y2 + z2 = - 2 ( xy + yz + xz )\(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\frac{-2\left(xy+yz+xz\right)}{2\left(x^2+y^2+z^2\right)-2\left(yz+xz+xy\right)}\)
\(S=\frac{-2\left(xy+yz+xz\right)}{-4\left(xy+yz+xz\right)-2\left(yz+xz+xy\right)}=\frac{-2\left(xy+yz+xz\right)}{-6\left(xy+yz+xz\right)}=\frac{1}{3}\)