Phân tích đa thức thành nhân tử: (x-2)^2(2x-5)(2x-3)-5 (3x-2)^2(6x-5)(6x-3)-5 04/09/2021 Bởi Adalynn Phân tích đa thức thành nhân tử: (x-2)^2(2x-5)(2x-3)-5 (3x-2)^2(6x-5)(6x-3)-5
`1)(x-2)^2(2x-5)(2x-3)-5` `= ( x^2−4x+4)(4x^2−16x+15)-5` Đặt `y=x^2−4x+4⇒4y=4.(x^2−4x+4)=4x^2-16x+16` `⇒ 4x^2-16x+16 – 1 = 4y-1` `⇔4x^2-16x+15 = 4y-1` `⇒bt⇔ y.(4y-1)-5` `=4y^2-y-5` `=4y^2+4y-5y-5` `=4y(y+1)-5(y+1)` `=(y+1)(4y-5).` Thay `y=x^2−4x+4` vào lại ta có: `bt⇒(x^2-4x+4+1)[4.(x^2−4x+4)-5]` `=(x^2-4x+4+1)[4x^2−16x+16-5]` `=(x^2-4x+5)(4x^2−16x+11).` `2)(3x-2)^2(6x-5)(6x-3)-5` `=(9x^2−12x+4)(36x^2−48x+15)-5` Đặt `y=9x^2−12x+4⇒4y=4.(9x^2−12x+4)=36x^2−48x+16` `⇒ 36x^2−48x+16 – 1 = 4y-1` `⇔36x^2−48x+15 = 4y-1` `⇒bt⇔ y.(4y-1)-5` `=4y^2-y-5` `=4y^2+4y-5y-5` `=4y(y+1)-5(y+1)` `=(y+1)(4y-5).` Thay `y=9x^2−12x+4` vào lại ta có: `bt⇒(9x^2−12x+4+1)[4.(9x^2−12x+4)-5]` `=(9x^2−12x+5)(36x^2-48x+16-5)` `=(9x^2−12x+5)(36x^2-48x+11).` Bình luận
a/ $\left(x-2\right)^2\left(2x-5\right)\left(2x-3\right)-5$ = $\left(x^2-2. 2x+2^2\right)\left[2x\left(2x-3\right)-5\left(2x-3\right)\right]-5$ = $\left(x^2-4x+4\right)\left(4x^2-6x-10x+15\right)-5$ = $\left(x^2-4x+4\right)\left(4x^2-16x+15\right)-5$ b/ $\left(3x-2\right)^2\left(6x-5\right)\left(6x-3\right)-5$ = $\left[\left(3x\right)^2-2.3x.2+2^2\right]\left[6x\left(6x-3\right)-5\left(6x-3\right)\right]-5$ = $\left(9x^2-12x+4\right)\left(36x^2-18x-30x+15\right)-5$ = $\left(9x^2-12x+4\right)\left(36x^2-48x+15\right)-5$ Bình luận
`1)(x-2)^2(2x-5)(2x-3)-5`
`= ( x^2−4x+4)(4x^2−16x+15)-5`
Đặt `y=x^2−4x+4⇒4y=4.(x^2−4x+4)=4x^2-16x+16`
`⇒ 4x^2-16x+16 – 1 = 4y-1`
`⇔4x^2-16x+15 = 4y-1`
`⇒bt⇔ y.(4y-1)-5`
`=4y^2-y-5`
`=4y^2+4y-5y-5`
`=4y(y+1)-5(y+1)`
`=(y+1)(4y-5).`
Thay `y=x^2−4x+4` vào lại ta có:
`bt⇒(x^2-4x+4+1)[4.(x^2−4x+4)-5]`
`=(x^2-4x+4+1)[4x^2−16x+16-5]`
`=(x^2-4x+5)(4x^2−16x+11).`
`2)(3x-2)^2(6x-5)(6x-3)-5`
`=(9x^2−12x+4)(36x^2−48x+15)-5`
Đặt `y=9x^2−12x+4⇒4y=4.(9x^2−12x+4)=36x^2−48x+16`
`⇒ 36x^2−48x+16 – 1 = 4y-1`
`⇔36x^2−48x+15 = 4y-1`
`⇒bt⇔ y.(4y-1)-5`
`=4y^2-y-5`
`=4y^2+4y-5y-5`
`=4y(y+1)-5(y+1)`
`=(y+1)(4y-5).`
Thay `y=9x^2−12x+4` vào lại ta có:
`bt⇒(9x^2−12x+4+1)[4.(9x^2−12x+4)-5]`
`=(9x^2−12x+5)(36x^2-48x+16-5)`
`=(9x^2−12x+5)(36x^2-48x+11).`
a/ $\left(x-2\right)^2\left(2x-5\right)\left(2x-3\right)-5$
= $\left(x^2-2. 2x+2^2\right)\left[2x\left(2x-3\right)-5\left(2x-3\right)\right]-5$
= $\left(x^2-4x+4\right)\left(4x^2-6x-10x+15\right)-5$
= $\left(x^2-4x+4\right)\left(4x^2-16x+15\right)-5$
b/ $\left(3x-2\right)^2\left(6x-5\right)\left(6x-3\right)-5$
= $\left[\left(3x\right)^2-2.3x.2+2^2\right]\left[6x\left(6x-3\right)-5\left(6x-3\right)\right]-5$
= $\left(9x^2-12x+4\right)\left(36x^2-18x-30x+15\right)-5$
= $\left(9x^2-12x+4\right)\left(36x^2-48x+15\right)-5$