Phân tích đa thức
a, x\(^2\)-4x+5
b, (x\(^2\)+3x+1)(x\(^2\)+3x+2)-6
Giúp e vs e đang cần gấp ạ
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BH
0
13 tháng 8 2018
1) \(3\left(x+4\right)-x^2-4x=3\left(x+4\right)-x\left(x+4\right)=\left(x+4\right)\left(3-x\right)\)
2) \(5x^2-5y^2-10x+10y=5\left(x^2-y^2\right)-10\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-10\left(x-y\right)=\left(x-y\right)\left(5x+5y-10\right)\)
3) \(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)
4) \(ax-bx-a^2+2ab-b^2=x\left(a-b\right)-\left(a^2-2ab+b^2\right)\)
\(=x\left(a-b\right)-\left(a-b\right)^2=\left(a-b\right)\left(x-a+b\right)\)
5) \(x^3-x^2-x+1=x^2\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)\left(x-1\right)\left(x+1\right)=\left(x-1\right)^2\left(x+1\right)\)
6) \(x^2+4x-y^2+4=x^2+4x+4-y^2=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right)\left(x+2+y\right)\)
21 tháng 10 2021
a: \(x^4+3x^3+x^2+3x\)
\(=x\left(x^3+3x^2+x+3\right)\)
\(=x\left(x+3\right)\left(x^2+1\right)\)
c: \(x^2-xy-x+y\)
\(=x\left(x-y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left(x-1\right)\)
18 tháng 7 2019
a) \(3x-3y+x^2-y^2\)
\(=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(3+x+y\right)\)
18 tháng 7 2019
e) \(x^3-3x+2\)
\(=x^3-x-2x+2\)
\(=x\left(x^2-1\right)-2\left(x-1\right)\)
\(=x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left[x\left(x+1\right)-2\right]\)
\(=\left(x-1\right)\left(x^2+x-2\right)\)
\(=\left(x-1\right)\left(x^2-x+2x-2\right)\)
\(=\left(x-1\right)\left[x\left(x-1\right)+2\left(x-1\right)\right]\)
\(=\left(x-1\right)\left(x-1\right)\left(x+2\right)\)
\(=\left(x-1\right)^2\left(x+2\right)\)
16 tháng 10 2016
Rút gọn
\(\left(2x+1\right)\left(4x^2-3x+1\right)+\left(2x-1\right)\left(4x^2+3x+1\right)\)
\(=8x^3-12x^2+2x+4x^2-3x+1+8x^3+12x^2+2x-4x^2-3x-1\)
\(=16x^3-2x\)
Phân tích đa thức thnahf nhân tử
\(4y^2+16y-x^2-8x\)
\(=\left(4y^2-x^2\right)+\left(16y-8x\right)\)
\(=\left(2y-x\right)\left(2y+x\right)+8\left(2y-x\right)\)
\(=\left(2y-x\right)\left(2y+x+8\right)\)
Chứng minh .............
Có: \(x^2+x+1=\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì: \(\left(x+\frac{1}{2}\right)^2\ge0\)
=> \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
Kết luận......
DN
2
4 tháng 9 2021
Ta có: \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)
\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-6\)
\(=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)
\(a,Đề.sai\\ b,=\left(x^2+3x+1\right)^2+\left(x^2+3x+1\right)-6\\ =\left(x^2+3x+1-2\right)\left(x^2+3x+1+3\right)\\ =\left(x^2+3x-1\right)\left(x^2+3x+4\right)\)
(x\(^2\)+3x+1)\(^2\) là sao vậy ạ