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H
2
6 tháng 7 2018
\(\frac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\frac{\left(3x^3-3x^2\right)-\left(4x^2-4x\right)+\left(x-1\right)}{\left(2x^3-2x^2\right)+\left(x^2-x\right)-\left(3x-3\right)}=\frac{\left(x-1\right).\left(3x^2-4x+1\right)}{\left(x-1\right).\left(2x^2+x-3\right)}\\ \)
\(=\frac{3x^2-4x+1}{2x^2+x-3}=\frac{\left(3x^2-3x\right)-\left(x-1\right)}{\left(2x^2-2x\right)+\left(3x-3\right)}=\frac{\left(x-1\right).\left(3x-1\right)}{\left(x-1\right).\left(2x+1\right)}=\frac{3x-1}{2x+1}\)
6 tháng 7 2018
\(\frac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}\)
\(=\frac{3x^3-3x^2-4x^2+4x+x-1}{2x^3-2x^2+x^2-x-3x+3}\)
\(=\frac{\left(3x^3-3x^2\right)-\left(4x^2-4x\right)+\left(x-1\right)}{\left(2x^3-2x^2\right)+\left(x^2-x\right)-\left(3x-3\right)}\)
\(=\frac{3x^2\left(x-1\right)-4x\left(x-1\right)+\left(x-1\right)}{2x^2\left(x-1\right)+x\left(x-1\right)-3\left(x-1\right)}\)
\(=\frac{\left(3x^2-4x+1\right)\left(x-1\right)}{\left(2x^2+x-3\right)\left(x-1\right)}\)
\(=\frac{3x^2-4x+1}{2x^2+x-3}\)
\(=\frac{3x^2-3x-x+1}{2x^2-2x+3x-3}\)
\(=\frac{\left(3x^2-3x\right)-\left(x-1\right)}{\left(2x^2-2x\right)-\left(3x-3\right)}\)
\(=\frac{3x\left(x-1\right)-\left(x-1\right)}{2x\left(x-1\right)-3\left(x-1\right)}\)
\(=\frac{3x-1}{2x-3}\)
LV
1 tháng 12 2017
c) hang dang thuc ( x -y+z)^2
o duoi phan h hang dang thuc luon
a) phan h nhan tu ra sao cho co tử la (x-1)(3x^2 -4x +1)
mau la (x-1)(2x^2 -x-3)
b ) k nhin dc de
14 tháng 2 2020
Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
DL
1
10 tháng 1 2020
A=\(\frac{x^3-7x+6}{x^3+5x^2-2x-24}\)=\(\frac{x^3-2x^2+2x^2-4x-3x+6}{x^3-2x^2+7x^2-14x+12x-24}\)=\(\frac{x^2\left(x-2\right)+2x\left(x-2\right)-3\left(x-2\right)}{x^2\left(x-2\right)+7x\left(x-2\right)+12\left(x-2\right)}\)=\(\frac{\left(x-2\right)\left(x^2+2x-3\right)}{\left(x-2\right)\left(x^2+7x+12^{^{^{^{^{^{^{^{^{ }}}}}}}}}\right)}\)=\(\frac{\left(x-2\right)\left(x^2-x+3x-3\right)}{\left(x-2\right)\left(x^2+3x+4x+12\right)}\)=\(\frac{\left(x-2\right)\left(x-1\right)\left(x+3\right)}{\left(x-2\right)\left(x+4\right)\left(x+3\right)}\)=\(\frac{x-1}{x+4}\)
TT
17 tháng 8 2020
a) \(4x^2\left(5x^3-2x+3\right)\)
\(=20x^5-8x^3+12x^2\)
b) \(3y^2\left(4y^3+\frac{2}{3}y^2-\frac{1}{3}\right)\)
\(=12y^5+2y^4-y^2\)
c) \(\left(5x^2-4x\right)\left(x-2\right)\)
\(=5x^3-14x^2+8x\)
d) \(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=6x^2+22x-55-6x^2-23x-21\)
\(=-x-76\)
A
17 tháng 8 2020
1, \(4x^2\left(5x^3-2x+3\right)=20x^5-8x^3+12x^2\)
2, \(3y^2\left(4y^3+\frac{2}{3}y^2-\frac{1}{3}\right)=12y^5+2y^4-y^2\)
3, \(\left(5x^2-4x\right)\left(x-2\right)=5x^3-10x^2-4x^2+8x=5x^3-14x^2+8x\)
4, \(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)=6x^2+33x-10x-55-\left(6x^2+14x+9x+21\right)\)
\(=6x^2+23x-55-6x^2-23x-21=-76\)
N
11 tháng 12 2018
a)\(\frac{x^3-x}{3x+3}=\frac{x.\left(x^2-1\right)}{3.\left(x+1\right)}=\frac{x.\left(x-1\right).\left(x+1\right)}{3.\left(x+1\right)}=\frac{x.\left(x+1\right)}{3}=\frac{x^2+x}{3}\)
30 tháng 6 2019
\(3x^4-4x^3+2x\left(x^3-2x^2+7x\right)\)
\(=3x^4-4x^3+2x^4-4x^3+14x^2\)
\(=5x^4-8x^3+14x^2\)
30 tháng 6 2019
3x4 - 4x3 + 2x(x3 - 2x2 + 7x )
= 3x4 - 4x3 + 2x4 _ 4x3 + 14x2
= 5x4 - 8x3 + 14x2