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(–a – b)2 = [(– 1).(a + b)]2 = (–1)2(a + b)2 = 1.(a + b)2 = (a + b)2 (đpcm)
\(BĐVT:\left(a+b\right)^2+\left(a-b\right)^2=a^2+2ab+b^2+a^2-2ab+b^2\)
\(=a^2+b^2+a^2+b^2\)
\(=2\left(a^2+b^2\right)\left(BVP\right)\left(đpcm\right)\)
\(a^2+b^2\) = (a+b)\(^2\) - 2ab
ta có
(a+b)\(^2\) - 2ab
= a\(^2\) + 2ab + b\(^2\) - 2ab
= a\(^2\) + b\(^2\) ( đpcm)
a) VT = ( a + b + a − b ) ( a + b − a + b ) 4 = 2 a . 2 b 4 = 4 = VP => đpcm.
b) VP = x 2 + 2 xy + y 2 + x 2 – 2 xy + y 2 = 2 ( x 2 + y 2 ) = VT => đpcm.
a) Ta có:
\(VT=\left(a-b\right)^2\)
\(=a^2-2ab+b^2\)
\(=a^2+2ab+b^2-4ab\)
\(=\left(a+b\right)^2-4ab=VP\left(dpcm\right)\)
b) Ta có:
\(VT=\left(x+y\right)^2+\left(x-y\right)^2\)
\(=x^2+2xy+y^2+x^2-2xy+y^2\)
\(=\left(x^2+y^2\right)+\left(x^2+y^2\right)\)
\(=2\left(x^2+y^2\right)=VP\left(dpcm\right)\)
\(\left(-a-b\right)^2=\left(-a\right)^2-2.\left(-a\right).b+b^2\)
\(=a^2+2ab+b^2\)(1)
\(\left(a+b\right)^2=a^2+2ab+b^2\)(2)
Từ (1) và (2) => \(\left(-a-b\right)^2=\left(a+b\right)^2\)
\(\left(-a-b\right)\)\(2\)\(=\)\(\left(-a\right)\)\(2\)\(-\)\(2\)\(.\)\(\left(-a\right)\)\(.\)\(b\)\(+\)\(b^2\)
\(=\)\(a^2\)\(+\)\(2\)\(.\)\(ab\)\(+\)\(b^2\)\(\left(1\right)\)
\(\left(a+b\right)\)\(=\)\(a\)\(+\)\(2\)\(.\)\(ab\)\(+\)\(b\)\(\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)ta có :
\(\left(-a-b\right)\)\(^2\)\(=\)\(\left(a+b\right)\)\(^2\)
(-a-b)2=[-1(a+b)]2=(-1)2(a+b)2=(a+b)2 (đpcm)
(–a – b)2 = [(– 1).(a + b)]2
= (–1)2(a + b)2
= 1.(a + b)2
= (a + b)2 (đpcm)