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1) \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)
2) \(x^3-9x^2+6x+16\)
\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)
\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)
\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)\(\left(x+1\right)\left(x-2\right)\left(x-8\right)\)
3) \(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-1\right)\)
4) \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(9x^2-6x+1\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(3x-1\right)^2\)
\(=\left(3x-1\right)\left(9x^2+3x+1-9x+3\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)
gửi phần này trước còn lại làm sau !!! tk mk nka !!!
a) \(x^4+8x+63\)
\(=x^4+4x^3+9x^2-4x^3-16x^2-36x+7x^2+28x+63\)
\(=x^2\left(x^2+4x+9\right)-4x\left(x^2+4x+9\right)+7\left(x^2+4x+9\right)\)
\(=\left(x^2+4x+9\right)\left(x^2-4x+7\right)\)
c) \(\left(x^2+2x+7\right)+\left(x^2-2x+4\right)\left(x^2+2x+3\right)\left(1\right)\)
Ta có : \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)
\(\Rightarrow x^2+2x+4=\dfrac{x^3-8}{x-2}\)
\(\left(1\right)\Rightarrow\left[\left(\dfrac{x^3-8}{x-2}+3\right)\right]+\left(x^2-2x+4\right)\left[\left(\dfrac{x^3-8}{x-2}-1\right)\right]\)
\(=\left[\left(\dfrac{x^3-3x-14}{x-2}\right)\right]+\left(x^2-2x+4\right)\left[\left(\dfrac{x^3-2x-5}{x-2}\right)\right]\)
\(=\dfrac{1}{x-2}\left[x^3-3x-14+\left(x^2-2x+4\right)\left(x^3-2x-5\right)\right]\)
a) \(=x\left(x^2-2xy+y^2\right)=x\left(x-y\right)^2\)
b) \(=\left(x^2+2x\right)+\left(10x+20\right)=x\left(x+2\right)+10\left(x+2\right)=\left(x+2\right)\left(x+10\right)\)
c) đặt \(x^2+x+1=t\)
\(\left(x^2+x+1\right)\left(x^2+x+4\right)+2=t\left(t+3\right)+2=t^2+3t+2=\left(t^2+t\right)+\left(2t+2\right)=t\left(t+1\right)+2\left(t+1\right)=\left(t+1\right)\left(t+2\right)=\left(x^2+x+2\right)\left(x^2+x+3\right)\)
a: \(=x\left(x-3\right)-4y\left(x-3\right)\)
=(x-3)(x-4y)
d: \(=\left(x-2\right)\left(x+2\right)+\left(x+2\right)^2\)
\(=\left(x+2\right)\left(x-2+x+2\right)\)
=2x(x+2)
\(a,=x\left(x-3\right)-4y\left(x-3\right)=\left(x-4y\right)\left(x-3\right)\\ b,=\left(x-1\right)\left(x^2+x+1\right)-4x\left(x-1\right)=\left(x-1\right)\left(x^2-3x+1\right)\\ c,=\left(x-y\right)\left(1-a\right)\\ d,=\left(x-2\right)\left(x-2+x+2\right)=2x\left(x-2\right)\\ e,=x^2\left(x+y\right)-xz\left(x+y\right)=x\left(x-z\right)\left(x+y\right)\\ f,=\left(x-y-2\right)\left(x+y\right)\)
a) x² - 9
= x² - 3²
= (x - 3)(x + 3)
b) 4x² - 1
= (2x)² - 1²
= (2x - 1)(2x + 1)
c) x⁴ - 16
= (x²)² - 4²
= (x² - 4)(x² + 4)
= (x² - 2²)(x² + 4)
= (x - 2)(x + 2)(x + 4)
d) x² - 4x + 4
= x² - 2.x.2 + 2²
= (x - 2)²
e) x³ - 8
= x³ - 2³
= (x - 2)(x² + 2x + 4)
f) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
`1)x^3-7x+6`
`=x^3-x-6x+6`
`=x(x-1)(x+1)-6(x-1)`
`=(x-1)(x^2+x-6)`
`=(x-1)(x^2-2x+3x-6)`
`=(x-1)[x(x-2)+3(x-2)]`
`=(x-1)(x-2)(x+3)`
`2)x^3-9x^2+6x+16`
`=x^3-2x^2-7x^2+14x-8x+16`
`=x^2(x-2)-7x(x-2)-8(x-2)`
`=(x-2)(x^2-7x-8)`
`=(x-2)(x^2-8x+x-8)`
`=(x-2)[x(x-8)+x-8]`
`=(x-2)(x-8)(x+1)`
`3)x^3-6x^2-x+30`
`=x^3+2x^2-8x^2-16x+15x+30`
`=x^2(x+2)-8x(x+2)+15(x+2)`
`=(x+2)(x^2-8x+15)`
`=(x+2)(x^2-3x-5x+15)`
`=(x+2)[x(x-3)-5(x-3)]`
`=(x+2)(x-3)(x-5)`
`4)2x^3-x^2+5x+3`
`=2x^3+x^2-2x^2-x+6x+3`
`=x^2(2x+1)-x(2x+1)+3(2x+1)`
`=(2x+1)(x^2-x+3)`
`5)27x^3-27x^2+18x-4`
`=27x^3-9x^2-18x^2+6x+12x-4`
`=9x^2(3x-1)-6x(3x-1)+4(3x-1)`
`=(3x-1)(9x^2-6x+4)`
1) Ta có: \(x^3-7x+6\)
\(=x^3-x-6x+6\)
\(=x\left(x^2-1\right)-6\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x-6\right)\)
\(=\left(x-1\right)\left(x+3\right)\left(x-2\right)\)
2) Ta có: \(x^3-9x^2+6x+16\)
\(=x^3-2x^2-7x^2+14x-8x+16\)
\(=x^2\left(x-2\right)-7x\left(x-2\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-7x-8\right)\)
\(=\left(x-2\right)\left(x-8\right)\left(x+1\right)\)
3) Ta có: \(x^3-6x^2-x+30\)
\(=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-8x+15\right)\)
\(=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
\(\dfrac{1}{2}x^3+4\)
\(=\dfrac{1}{2}\left(x^3+8\right)\)
\(=\dfrac{1}{2}\left(x^3+2^3\right)\)
\(=\dfrac{1}{2}\left(x+2\right)\left(x^2-2\cdot x+2^2\right)\)
\(=\dfrac{1}{2}\left(x+2\right)\left(x^2-2x+4\right)\)
c) \(16-x^2+2xy-y^2=\left(4-x+y\right)\left(4+x-y\right)\)
d) \(\left(x-1\right)^2-4\left(2x-3\right)^2=\left(5-3x\right)\left(5x-7\right)\)
e) \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
e) \(x^2-7=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)
a) x3+4x-5 = x3-x2+x2+4x-5=(x3-x2)+(x2-x)+(5x-5)=x2(x-1)+x(x-1)+5(x-1)=(x2+x+5)(x-1)
b) x3-3x2+4=x3-2x2-x2+4=(x3-2x2)-(x2-4)=x2(x-2)-(x-2)(x+2)=(x2-x+2)(x-2)
c) x3+2x2+3x+2=x3+x2+x2+x+2x+2=(x3+x2)+(x2+x)+(2x+2)=x2(x+1)+x(x+1)+2(x+1)=(x2+x+2)(x+1)
d) bạn xem lại đề đúng ko
e) (x2+3x)2-2(x2+3x)-8=x4+6x3+9x2-2x2-6x-8=x4+6x3+7x2-6x-8=x4-x3+7x3-7x2+14x2-14x+8x-8=(x4-x3)+(7x3-7x2)+(14x2-14x)+(8x-8)=x3(x-1)+7x2(x-1)+14x(x-1)+8(x-1)=(x3+7x2+14x+8)(x-1)=(x3+x2+6x2+6x+8x+8)(x-1)=\(\left[\left(x^3+x^2\right)+\left(6x^2+6x\right)+\left(8x+8\right)\right]\left(x-1\right)\)\(=\left[x^2\left(x+1\right)+6x\left(x+1\right)+8\left(x+1\right)\right]\left(x-1\right)\)\(=\left(x^2+6x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left(x^2+2x+4x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\left(x+1\right)\left(x-1\right)\)\(=\left[x\left(x+2\right)+4\left(x+2\right)\right]\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)
f) (x2+4x+10)2-7(x2+4x+11)+7=(x2+4x+10)2-\(\left[7\left(x^2+4x+11\right)-7\right]\)\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)\)\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)
a) Ta có: \(x^3+4x-5\)
\(=x^3-x+5x-5\)
\(=x\left(x-1\right)\left(x+1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+5\right)\)
b) Ta có: \(x^3-3x^2+4\)
\(=x^3+x^2-4x^2+4\)
\(=x^2\left(x+1\right)-4\left(x-1\right)\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+4\right)\)
\(=\left(x+1\right)\cdot\left(x-2\right)^2\)
c) Ta có: \(x^3+2x^2+3x+2\)
\(=x^3+x^2+x^2+x+2x+2\)
\(=x^2\left(x+1\right)+x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+2\right)\)
d) Ta có: \(x^2+2xy+y^2+2x+2y-3\)
\(=\left(x+y\right)^2+2\left(x+y\right)-3\)
\(=\left(x+y\right)^2+3\left(x+y\right)-\left(x+y\right)-3\)
\(=\left(x+y\right)\left(x+y+3\right)-\left(x+y+3\right)\)
\(=\left(x+y+3\right)\left(x+y-1\right)\)
\(x^3+x^2+4=x^3+2x^2-x^2-2x+2x+4\)
\(=x^2\left(x+2\right)-x\left(x+2\right)+2\left(x+2\right)\)
\(=\left(x^2-x+2\right)\left(x+2\right)\)
\(x^8+64=x^8+16x^4+64-16x^4\)
\(=\left(x^4+8\right)^2-\left(4x^2\right)^2\)
\(=\left(x^4-4x^2+8\right)\left(x^4+4x^2+8\right)\)
\(4a^4+b^4=4a^4+4a^2b^2+b^4-4a^2b^2\)
\(=\left(2a^2+b^2\right)^2-\left(2ab\right)^2\)
\(=\left(2a^2+b^2-2ab\right)\left(2a^2+b^2+2ab\right)\)
\(x^3-2x-4=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x^2+2x+2\right)\left(x-2\right)\)
Chúc bạn học tốt.