Tính các tích phân sau: ∫ - 4 6 x + 3 - x - 4 d x
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CM
Cao Minh Tâm
2 tháng 1 2019
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AD
23 tháng 7 2023
\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)
\(e,=-4x^2\left(x-1\right)+\left(x-1\right)\\ =\left(1-4x^2\right)\left(x-1\right)\\ =\left(1-2x\right)\left(1+2x\right)\left(x-1\right)\)
\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)
7 tháng 1 2022
Bài 1:
b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)
Bài 2:
a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)
d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)
\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)
e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
HP
4 tháng 7 2016
\(x^2-4x+3=x^2-3x-x+3=x\left(x-3\right)-\left(x-3\right)=\left(x-1\right)\left(x-3\right)\)
\(x^2+5x+4=x^2+4x+x+4=x\left(x+4\right)+\left(x+4\right)=\left(x+1\right)\left(x+4\right)\)
\(x^2-x-6=x^2-3x+2x-6=x\left(x-3\right)+2\left(x-3\right)=\left(x+2\right)\left(x-3\right)\)
\(x^4+4=\left(x^2\right)^2+2.x^2.2+2^2-4x^2=\left(x^2+2\right)^2-4x^2=\left(x^2+2\right)^2-\left(2x^2\right)=\left(x^2+2+2x\right)\left(x^2-2-2x\right)\)
22 tháng 7 2023
`a, 4x^3 - 16x = 4x(x^2-4) = 4x(x-2)(x+2)`
`b, x^4 - y^4 = (x^2-y^2)(x^2+y^2) = (x-y)(x+y)(x^2+y^2)`
`c, xy^2 + x^2y + 1/4y^3`
`= y(xy + x^2 + 1/4y^2)`
`d, x^2 + 2x - y^2 + 1 = (x+1)^2 - y^2`
`= (x+1+y)(x+1-y)`
13 tháng 8 2019
\(a,x^2+9x+20=x^2+4x+5x+20.\)
\(=x\left(x+4\right)+5\left(x+4\right)=\left(x+4\right)\left(x+5\right)\)
\(b,x^4-5x^2+4=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
\(c,x^4+4=x^4+4x^2+4-4x^2\)
\(=\left(x^2-2\right)-\left(2x\right)^2=\left(x^2-2x-2\right)\left(x^2+2x-2\right)\)
\(d,x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(=x\left(x+3\right)\left(x+1\right)\left(x+2\right)+1\)
\(\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
\(=\left(x^2+3x\right)\left(x^2+3x\right)+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x+1\right)^2\)
17 tháng 6 2019
a) (x-2)(x+2)(x^2-10)-72=(x^2-4)(x^2-82)
b) x^8+x^6+x^4+x^2+1=x^2 (x^4+x^3+x^2+1+1/x^2)
c)(x+y)^4+x^4+y^4=(x+y)^4+(x+y)^4=2 (x+y)^4
17 tháng 6 2019
a) (x-2)(x+2)(x^2 - 10) -72
= (x^2 - 4)(x^2 - 10) - 72
= x^4 - 4x^2 -10x^2 + 40 - 72
= x^4 - 14x^2 - 32
= x^4 - 16x^2 + 2x^2 - 32
= x^2(x^2 - 16) + 2(x^2 - 16)
= (x^2 - 16)(x^2 + 2)
= (x-4)(x+4)(x^2 + 2)
c) (x+y)4 + x4 + y4
= 2x4 + 4xy3 + 6x2y2 + 4x3y + 2y3
= 2(y4 + 2xy3 + 3x2y2 + 2x3y + x4)
= 2(y2 + xy + y2)2
DL
19 tháng 8 2019
a) \(4x^4+4x^3-x^2-x=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=\left(4x^3-x\right)\left(x+1\right)=x\left(4x^2-1\right)\left(x+1\right)\)
\(=x\left\{\left(2x\right)^2-1\right\}\left(x+1\right)=x\left(2x-1\right)\left(2x+1\right) \left(x+1\right)\)
c) \(x^4-4x^3+8x^2-16x+16=x^4+8x^2+16-\left(4x^3+16x\right)\)
\(=\left(x^2+4\right)^2-4x\left(x^2+4\right)=\left(x^2-4x+4\right)\left(x^2+4\right)=\left(x-2\right)^2\left(x^2+4\right)\)
DL
19 tháng 8 2019
b) \(x^6-x^4-9x^3+9x^2=x^4\left(x^2-1\right)-\left(9x^3-9x^2\right)\)
\(=x^4\left(x-1\right)\left(x+1\right)-9x^2\left(x-1\right)\)
\(=\left(x^5+x^4-9x^2\right)\left(x-1\right)=\left(x-1\right)x^2\left(x^3+x^2-9\right)\)