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a) (a+b)3+(a-b)3=a3+3a2b+3ab2+b3+a3-3a2b+3ab2-b3
=2a3+6ab2
b) (a + b + c)2 + (a − b − c)2 + (b − c − a)2 + (c − a − b)2
=a2+b2+c2+2ab+2bc+2ca+a2+b2+c2-2ab+2bc-2ac+a2+b2+c2-2bc+2ca-2ba+a2+b2+c2-2ca+2ab-2cb
=4a2+4b2+4c2
a) Ta có: \(\left(a+b\right)^3+\left(a-b\right)^3\)
\(=\left(a+b+a-b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2a\cdot\left(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right)\)
\(=2a\cdot\left(a^2+3b^2\right)\)
\(=2a^3+6ab^2\)
\(1,\left(x+2022\right)\left(x-1\right)=x^2+2021x-2022\left(B\right)\\ 2,\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\left(A\right)\)
Bài 1:
\(P=2a^2-2b^2-a^2+2ab-b^2+a^2+2ab+b^2+b^2=2a^2-b^2+4ab\\ Q=\left(2x+3\right)^2+\left(2x-3\right)^2-2\left(2x-3\right)\left(2x+3\right)\\ Q=\left(2x+3-2x+3\right)^2=9^2=81\)
Bài 2:
\(Sửa:A=x^2+2xy+y^2-4x-4y+2=\left(x+y\right)^2-4\left(x+y\right)+4-2\\ A=\left(x+y-2\right)^2-2=\left(3-2\right)^2-2=1-2=-1\)
\(\dfrac{a^2}{a^2-b^2-c^2}=\dfrac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}=\dfrac{a^2}{\left(a-b\right)\left(-c\right)-c^2}=\dfrac{a^2}{c\left(b-a-c\right)}=\dfrac{a^2}{2bc}\\ \Leftrightarrow M=\sum\dfrac{a^2}{a^2-b^2-c^2}=\sum\dfrac{a^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}\\ \Leftrightarrow M=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{2abc}=0\)