Câu hỏi của Vương nguyên

Question

cho a^3 +3ab^2 =2006

b^3+3a^2b=2005

tính P=a^2-b^2

Aug.21 · Accepted Answer

Ta có :$\left(a^3+3ab^2\right)^2=a^6+6a^4b^2+9a^2b^4=2006^2$

$\left(b^3+3a^2b\right)^2=b^6+6a^2b^4+9a^4b^2=2005^2$

$\Rightarrow\left(a^3+3ab^2\right)^2-\left(b^3+3a^2b\right)^2=a^6-3a^4b^2+3a^2b-b^6$

$=2006^2-2005^2$

Hay $\left(a^2-b^2\right)^3=4011$

Vậy $P=a^2-b^2=^3\sqrt{4011}$

Câu trả lời:

Ta có :$\left(a^3+3ab^2\right)^2=a^6+6a^4b^2+9a^2b^4=2006^2$

$\left(b^3+3a^2b\right)^2=b^6+6a^2b^4+9a^4b^2=2005^2$

$\Rightarrow\left(a^3+3ab^2\right)^2-\left(b^3+3a^2b\right)^2=a^6-3a^4b^2+3a^2b-b^6$

$=2006^2-2005^2$

Hay $\left(a^2-b^2\right)^3=4011$

Vậy $P=a^2-b^2=^3\sqrt{4011}$

Nguyễn Tấn Phát · Answer

Theo đề bài ta có:

$a^3+3ab^2=2006$

$b^3+3a^2b=2005$

$\Rightarrow a^3+3ab^2-3a^2b-b^3=2006-2005$

$\Leftrightarrow a^3-3a^2b+3ab^2-b^3=1$

$\Leftrightarrow\left(a-b\right)^3=1$

$\Leftrightarrow a-b=1$

Ta có:

$P=a^2-b^2$

$P=\left(a-b\right)\left(a+b\right)$

$P=1\left(a+b\right)$

VẬY $P=a+b$