Câu hỏi của Thiện Bình

Question

Cho x2 + y2 =1 . Tính:

a, 2(x6 + y6) - 3 ( x4 + y4 )

b, 2x4 - y4 + x2y2 + 3y2

Hắc Hường · Accepted Answer

Giải:

a) $2\left(x^6+y^6\right)-3\left(x^4+y^4\right)$

$\Leftrightarrow2\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)-3x^4-3y^4$

$\Leftrightarrow2\left(x^4-x^2y^2+y^4\right)-3x^4-3y^4$

$\Leftrightarrow2x^4-2x^2y^2+2y^4-3x^4-3y^4$

$\Leftrightarrow-2x^2y^2-x^4-y^4$

$\Leftrightarrow-\left(x^4+2x^2y^2+y^4\right)$

$\Leftrightarrow-\left(x^2+y^2\right)^2$

$\Leftrightarrow-1$

Vậy ...

b) $2x^4-y^4+x^2y^2+3y^2$

$=x^4-y^4+x^4+x^2y^2+3y^2$

$=\left(x^2-y^2\right)\left(x^2+y^2\right)+x^2\left(x^2+y^2\right)+3y^2$

$=x^2-y^2+x^2+3y^2$

$=2x^2+2y^2$

$=2\left(x^2+y^2\right)$

$=2$

Vậy ...Câu trả lời: Giải:

a) $2\left(x^6+y^6\right)-3\left(x^4+y^4\right)$

$\Leftrightarrow2\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)-3x^4-3y^4$

$\Leftrightarrow2\left(x^4-x^2y^2+y^4\right)-3x^4-3y^4$

$\Leftrightarrow2x^4-2x^2y^2+2y^4-3x^4-3y^4$

$\Leftrightarrow-2x^2y^2-x^4-y^4$

$\Leftrightarrow-\left(x^4+2x^2y^2+y^4\right)$

$\Leftrightarrow-\left(x^2+y^2\right)^2$

$\Leftrightarrow-1$

Vậy ...

b) $2x^4-y^4+x^2y^2+3y^2$

$=x^4-y^4+x^4+x^2y^2+3y^2$

$=\left(x^2-y^2\right)\left(x^2+y^2\right)+x^2\left(x^2+y^2\right)+3y^2$

$=x^2-y^2+x^2+3y^2$

$=2x^2+2y^2$

$=2\left(x^2+y^2\right)$

$=2$

Vậy ...

Violympic toán 9