Câu hỏi của Jatsumin

Question

Cho x+y=1. Tính giá trị của:

a) $P=x^3+y^3+3xy$

b) $Q=x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)$

Thiên Hàn · Accepted Answer

a) Ta có:

$x+y=1$

$\Rightarrow\left(x+y\right)^3=1$

$\Rightarrow x^3+y^3+3xy\left(x+y\right)=1$

$\Rightarrow x^3+y^3+3xy=1$

$\Rightarrow P=1$

b) $Q=x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)$

$Q=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2\left(x+y\right)$

$Q=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2\left(x+y\right)$

Thay x + y = 1 vào Q

$Q=1-3xy+3xy\left(1-2xy\right)+6x^2y^2$

$Q=1-3xy+3xy-6x^2y^2+6x^2y^2$

$Q=1$Câu trả lời: a) Ta có:

$x+y=1$

$\Rightarrow\left(x+y\right)^3=1$

$\Rightarrow x^3+y^3+3xy\left(x+y\right)=1$

$\Rightarrow x^3+y^3+3xy=1$

$\Rightarrow P=1$

b) $Q=x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)$

$Q=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2\left(x+y\right)$

$Q=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2\left(x+y\right)$

Thay x + y = 1 vào Q

$Q=1-3xy+3xy\left(1-2xy\right)+6x^2y^2$

$Q=1-3xy+3xy-6x^2y^2+6x^2y^2$

$Q=1$

Bài 3: Những hằng đẳng thức đáng nhớ