a) VT = (a+b)(\(a^2-ab+b^2\)) + \(\left(a-b\right)\left(a^2+ab+b^2\right)=a^3+b^3\)\(+a^3-b^3\) = \(2a^3=VP\) (đpcm)
b, VP =\(\left(a+b\right)\left[\left(a-b\right)^2+ab\right]=\left(a+b\right)\left[a^2-2ab+b^2+ab\right]=\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3=VT\left(đpcm\right)\)
c, Ta có : \(VT=\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)(1)
\(VP=\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2acbd+b^2d^2+a^2d^2-2adbc+b^2c^2=a^2c^2+b^2d^2+a^2d^2+b^2c^2\) (2)
Từ (1) và (2), ta có \(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\left(đpcm\right)\)
a)
\( (a + b)(a^2 - ab + b^2) + (a - b)(a^2 + ab + b^2) = 2a^3 = a^3 + b^3 + a^3 - b^3 = 2a^3\)
b)
\(a^3 + b^3 = (a + b)(a^2 - ab + b^2) = (a + b)(a^2 - (2ab - ab) + b^2) = (a + b)(a^2 - 2ab + b^2 + ab) = (a + b)[(a - b)^2 + ab] \)
a) \(\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a^2+ab+b^2\right)=2a^3\)
\(=a^3+b^3+a^3-b^3\)
\(=2a^3\)
b) \(a^3+b^3=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)
\(\left(a+b\right)\left[\left(a^2-2ab+b^2\right)+ab\right]\)
\(\left(a+b\right)\left(a^2-2ab+b^2+ab\right)\)
\(\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(a^3+b^3\)
a) \(\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(=a^3+b^3+a^3-b^3=2a^3\)
Vậy \(\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a^2+ab+b^2\right)=2a^3\)
a) Ta có VT = (a+b)(a2-ab+b2)+(a-b)(a2+ab+b2) = a3+b3+a3-b3=2a3=VP(đpcm)
b)Mk chỉ cm đc a3+b3=(a+b)[(a-b)2+ab] thôi
Ta có VP = (a+b)(a2-2ab+b2+ab) = (a+b)(a2-ab+b2) = a3+b3 = VT(đpcm)
c)Ta có VP = a2c2+2acbd+b2d2+a2d2-2adbc+b2c2 = a2c2+b2d2+a2d2+b2c2
=a2(c2+d2)+b2(c2+d2)
=(a2+b2)(c2+d2)=VT(đpcm)
