Câu hỏi của Jun Nguyễn

Question

Cho a+b=p, a-b=q. Tính a.b, a^3+b^3

Nguyễn Việt Lâm · Accepted Answer

$a+b=p\Rightarrow a^2+2ab+b^2=p^2$

$\Rightarrow a^2-2ab+b^2+4ab=p^2$

$\Rightarrow\left(a-b\right)^2+4ab=p^2$

$\Rightarrow q^2+4ab=p^2\Rightarrow ab=\frac{p^2-q^2}{4}$

$a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]$

$=p\left(q^2+\frac{p^2-q^2}{4}\right)=\frac{p\left(3q^2+p^2\right)}{4}$Câu trả lời: $a+b=p\Rightarrow a^2+2ab+b^2=p^2$

$\Rightarrow a^2-2ab+b^2+4ab=p^2$

$\Rightarrow\left(a-b\right)^2+4ab=p^2$

$\Rightarrow q^2+4ab=p^2\Rightarrow ab=\frac{p^2-q^2}{4}$

$a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]$

$=p\left(q^2+\frac{p^2-q^2}{4}\right)=\frac{p\left(3q^2+p^2\right)}{4}$

Bài 3: Những hằng đẳng thức đáng nhớ