Câu hỏi của Mie Nguyễn

Question

1/ (x+1)^2 - 16y^2

2/ 2x^2 - 8

3/  3x^2 - 5x - 3y^2 + 5y

4/  3x^2 -7x +6

5/  x^2 + y^2 - 4x^2y^2 + 2xy

6/ x^2 + 1 - y^2 + 2x

Vũ Minh Tuấn · Accepted Answer

1) $\left(x+1\right)^2-16y^2$

$=\left(x+1\right)^2-\left(4y\right)^2$

$=\left(x+1-4y\right).\left(x+1+4y\right)$

2) $2x^2-8$

$=\left[\left(\sqrt{2x}\right)^2\right]^2-\left(\sqrt{8}\right)^2$

$=\left[\left(\sqrt{2x}\right)^2-\sqrt{8}\right].\left[\left(\sqrt{2x}\right)^2+\sqrt{8}\right]$

6) $x^2+1-y^2+2x$

$=x^2+2x+1-y^2$

$=\left(x^2+2x+1\right)-y^2$

$=\left(x+1\right)^2-y^2$

$=\left(x+1-y\right).\left(x+1+y\right)$

Chúc bạn học tốt!Câu trả lời: 1) $\left(x+1\right)^2-16y^2$

$=\left(x+1\right)^2-\left(4y\right)^2$

$=\left(x+1-4y\right).\left(x+1+4y\right)$

2) $2x^2-8$

$=\left[\left(\sqrt{2x}\right)^2\right]^2-\left(\sqrt{8}\right)^2$

$=\left[\left(\sqrt{2x}\right)^2-\sqrt{8}\right].\left[\left(\sqrt{2x}\right)^2+\sqrt{8}\right]$

6) $x^2+1-y^2+2x$

$=x^2+2x+1-y^2$

$=\left(x^2+2x+1\right)-y^2$

$=\left(x+1\right)^2-y^2$

$=\left(x+1-y\right).\left(x+1+y\right)$

Chúc bạn học tốt!

Duyên · Answer

3/ 3x^2 - 5x - 3y^2 + 5y

= ( 3x$^2$ - 3y$^2$ ) - (5x - 5y ) =  3 ( x$^2$ - y$^2$ ) - 5 ( x - y )= 3 ( x - y ) ( x + y ) - 5 ( x - y )

= ( x - y ) ( 3x + 3y - 5 )Câu trả lời: 3/ 3x^2 - 5x - 3y^2 + 5y

= ( 3x$^2$ - 3y$^2$ ) - (5x - 5y ) =  3 ( x$^2$ - y$^2$ ) - 5 ( x - y )= 3 ( x - y ) ( x + y ) - 5 ( x - y )

= ( x - y ) ( 3x + 3y - 5 )

Bài 9: Phân tích đa thức thành nhân tử bằng cách phối hợp nhiều phương pháp