Phân tích đa thức thành nhân tử
1) 27+27x+9x2+x3 2) 8 - 27x3
3) x64 + x32 + 1
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`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: \(x^3-6x^2-x+30\)
\(=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-8x+15\right)\)
\(=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
a.
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
b.
\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
c.
\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)
\(=\left(x+3\right)^3\)
d.
\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
e.
\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f.
\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
Chia nhỏ ra cậu ơi :v
Cậu đặt câu hỏi free nên đặt nhỏ ra thì mới có người làm nha để như này dày cộp không ai dám làm đou =(((
\(a,=\left(3x+2y\right)^3\\ b,=\left(4-x\right)^3\\ c,=\left(\dfrac{1}{2}x-3y\right)^3\)
a: \(x^2-4y^2=x^2-\left(2y\right)^2=\left(x-2y\right)\left(x+2y\right)\)
b: \(9x^2-4=\left(3x\right)^2-2^2=\left(3x-2\right)\left(3x+2\right)\)
c: \(16-y^2=4^2-y^2=\left(4-y\right)\left(4+y\right)\)
d: \(\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
e: \(x^3-8=x^3-2^3=\left(x-2\right)\left(x^2+2x+4\right)\)
f: \(27x^3-y^3=\left(3x\right)^3-y^3=\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
k: \(=\left(2-3x\right)\left(4+6x+9x^2\right)\)
i: \(=3\left(x^2-2xy+y^2\right)=3\left(x-y\right)^2\)
\(g,27+27x+9x^2+x^3=\left(3+x\right)^3\\ i,2x^2+2y^2-x^2z+z-y^2z-2=\left(2x^2-x^2z\right)+\left(2y^2-y^2z\right)-\left(2-z\right)=x^2\left(2-z\right)+y^2\left(2-z\right)-\left(2-z\right)=\left(x^2+y^2-1\right)\left(2-z\right)\)
\(k,8-27x^2=2^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
\(l,3x^2-6xy+3y^2=3\left(x^2-2xy+y^2\right)=3\left(x-y\right)^2\)
Câu 2:
a: \(\Leftrightarrow3x^2+2x-1=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
b: \(\Leftrightarrow x^3-4x-x^3-8=4\)
hay x=-3
1) \(27+27x+9x^2+x^3\)
\(=3^3+3.3^2.x+3.3.x^2+x^3\)
\(=\left(3+x\right)^3\)
2) \(8-27x^3\)
\(=2^3-\left(3x\right)^3\)
\(=\left(2-3x\right)\left[2^2+2.3x+\left(3x\right)^2\right]\)
\(=\left(2-3x\right)\left(4+6x+9x^2\right)\)
\(1,27+27+9x^2+x^3\)
\(=\left(3+x\right)^3\)
\(2,8-27x^3\)
\(=\left(2-3x\right).\left(4+6x+9x^2\right)\)
\(3,x^{64}+x^{32}+1\)
\(=\left(x^{64}+2x^{32}+1\right)-x^{32}\)
\(=\left(x^{32}+1\right)^2-x^{32}\)
\(=\left(x^{32}+1-x^{16}\right).\left(x^{32}+1+x^{16}\right)\)
Công nhận câu 3 hơi căng~