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\(a,\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a^2+ab+b^2\right)\)\(=\left(a^3+b^3\right)+\left(a^3-b^3\right)=2a^3\Rightarrowđpcm\)
\(b,\left(a+b\right)\left[\left(a-b\right)^2+ab\right]=\left(a+b\right)\left(a^2-2ab+b^2+ab\right)=\left(a+b\right)\left(a^2-ab+b^2\right)\)\(=\left(a^3+b^3\right)\Rightarrowđpcm\)
\(c,\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+a^2d^2+b^2c^2+b^2d^2=\left(a^2c^2+2abcd+b^2d^2\right)+\left(a^2d^2-2abcd+b^2c^2\right)\)\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\Rightarrowđpcm\)
a) (a+b)(a2-ab+b2)+(a-b)(a2+ab+b2)
= a3+b3+a3-b3 = 2a3
b) a3+b3
= (a+b)(a2-ab+b2)
= (a+b)(a2- 2ab+b2)+ab
= (a+b)(a2-b2)+ab
Trước hết , ta khai triển vế trái , sau đó , nhóm các hạng tử .
\(\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+b^2d^2+2abcd+a^2d^2+b^2c^2-2abcd\)
\(=\left(a^2c^2+a^2d^2\right)+\left(b^2c^2+b^2d^2\right)\)
\(=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)\)
\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
Vậy \(\left(ac+bd\right)^2+\left(ad-bc\right)^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\left(ĐPCM\right)\)
A) Ta có :
Vế phải = ( a + b ) ( a2 - 2ab + b2 +ab )
= ( a + b ) ( a2 - ab + b2 )
= a3 + b3 = Vế trái ( điều phải chứng minh )
Chúc bạn học tốt ^^
Câu a) thôi nhé
Ta có (a+b) [(a-b)2+ab] = (a+b)(a2-ab-b2) = a3-a2b + ab2 + ba2 - ab2 +b3
Thu gọn lại ta được a3 + b3
(ac+bd)^2=\(^{a^2c^2+2abcd+b^2d^2}\)
\(\left(ad-bc\right)^2=a^2d^2-2abcd+b^2c^2\)
\(\Rightarrow\left(ac+bd\right)^2-\left(ad-bc\right)^2=a^2c^2+a^2d^2+b^2c^2+b^2d^2\) =vp(dpcm)
\(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\Leftrightarrow a^2c^2+a^2d^2+b^2c^2+b^2d^2=a^2c^2+b^2d^2+a^2d^2+b^2c^2\Leftrightarrow0=0\)Có điều này đúng nên ta có đpcm đúng
\(\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)
\(=\left(ac\right)^2+2acbd+\left(bd\right)^2+\left(ad\right)^2-2adbc+bc^2\)
\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
a: \(\left(a^2-b^2\right)^2+\left(2ab\right)^2\)
\(=a^4-2a^2b^2+b^4+4a^2b^2\)
\(=a^4+2a^2b^2+b^4=\left(a^2+b^2\right)^2\)
b: \(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)
\(=c^2\left(a^2+b^2\right)+d^2\left(a^2+b^2\right)\)
\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
c: \(\left(ax+b\right)^2+\left(a-bx\right)^2+c^2x^2\)
\(=a^2x^2+b^2+a^2+b^2x^2+c^2x^2\)
\(=a^2\left(x^2+1\right)+b^2\left(x^2+1\right)+c^2x^2\)
\(=\left(x^2+1\right)\left(a^2+b^2\right)+c^2x^2\)
Ta có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=4a^2+4b^2+4c^2-4ab-4bc-4ac\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=4a^2+4b^2+4c^2-4ab-4ac-4bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac-4a^2-4b^2-4c^2+4ab+4bc+4ac=0\)
\(\Leftrightarrow-2a^2-2b^2-2c^2+2ab+2ac+2bc=0\)
\(\Leftrightarrow-\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow a=b=c\)(đpcm)
b)
VP=(a+b)[(a-b)2+ab]
=(a+b)(a2-2ab+b2+ab)
=(a+b)(a2-ab+b2)
=a3+b3=VT
Vậy x3+y3=(a+b)[(a-b)2+ab]
c)
VP=(ac+bd)2+(ad-bc)2
=a2c2+2abcd+b2d2+a2d2-2abcd+b2c2
=a2c2+b2d2+a2d2+b2c2
=(a2c2+a2d2)+(b2d2+b2c2)
=a2.(c2+d2)+b2.(c2+d2)
=(a2+b2)(c2+d2)
Vậy (a2+b2)(c2+d2)=(ac+bd)2+(ad-bc)2
tks mem trieu dang