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Bài 1:
27x3 - 8 : (6x + 9x2 +4)
= (3x - 2) (9x2 + 6x + 4) : (9x2 + 6x + 4)
= 3x - 2
Bài 3:
a, 81x4 + 4 = (9x2)2 + 36x2 + 4 - 36x2
= (9x2 + 2)2 - (6x)2
= (9x2 + 6x + 2)(9x2 - 6x + 2)
b, x2 + 8x + 15 = x2 + 3x + 5x + 15
= x(x + 3) + 5(x + 3)
= (x + 3)(x + 5)
c, x2 - x - 12 = x2 + 3x - 4x - 12
= x(x + 3) - 4(x + 3)
= (x + 3) (x - 4)
Câu 1:
(27x3 - 8) : (6x + 9x2 + 4)
= (3x - 2)(9x2 + 6x + 4) : (6x + 9x2 + 4)
= 3x - 2
Câu 2:
a) (3x - 5)(2x+ 11) - (2x + 3)(3x + 7)
= 6x2 + 33x - 10x - 55 - 6x2 - 14x - 9x - 21
= -76
⇒ đccm
b) (2x + 3)(4x2 - 6x + 9) - 2(4x3 - 1)
= 8x3 + 27 - 8x3 + 2
= 29
⇒ đccm
Câu 3:
a) 81x4 + 4
= (9x2)2 + 22
= (9x2 + 2)2 - (6x)2
= (9x2 - 6x + 2)(9x2 + 6x + 2)
b) x2 + 8x + 15
= x2 + 3x + 5x + 15
= x(x + 3) + 5(x + 3)
= (x + 3)(x + 5)
c) x2 - x - 12
= x2 - 4x + 3x - 12
= x(x - 4) + 3(x - 4)
= (x - 4)(x + 3)
a) x3 + 2x2 + x
= x3 + x2 + x2 + x
= x2 ( x + 1 ) + x ( x + 1 )
= ( x2 + x ) ( x + 1 )
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
1) x2-4x+5+y2+2y=0
<=>x2-4x+4+y2+2y+1=0
<=>(x-2)2+(x+1)2=0
<=>x-2=0 và x+1=0
<=>x=2 và x=-1
2)2p.p2-(p3-1)+(p+3)2p2-3p5
<=>2p3-p3+1+2p3+6p2-3p5
<=>3p3+6p2-3p5+1
3)(0.2a3)2-0.01a4(4a2-100)=0,04a6-0,04a6+1
=1
4)a) x(2x+1)-x2(x+20)+(x3-x+3)=2x2+x-x3-20x2+x3-x+3
=-18x2+3(đề sai)
b) x(3x2-x+5)-(2x3+3x-16)-x(x2-x+2)=3x3-x2+5x-2x3-3x+16-x3+x2-2x
=16
Vậy x(3x2-x+5)-(2x3+3x-16)-x(x2-x+2) không phụ thuộc vào x
5)a) x(y-z)+y(z-x)+z(x-y)=xy-xz+yz-xy+xz-yz=0
b) x(y+z-yz)-y(z+x-xz)+z(y-x)=xy+xz-xyz-yz-xy+xyz+yz-xz=0
6)M+(12x4-15x2y+2xy2+7)=0
<=>M =-(12x4-15x2y+2xy2+7)
<=>M =-12x4+15x2y-2xy2-7
a,\(xy+3x-7y-21\)
\(=x\left(y+3\right)-7\left(y+3\right)\)
\(=\left(y+3\right)\left(x-7\right)\)
\(b,2xy-15-6x+5y\)
\(=\left(2xy-6x\right)+\left(-15+5y\right)\)
\(=2x\left(y-3\right)-5\left(3-y\right)\)
\(=2x\left(y-3\right)+5\left(y-3\right)\)
\(=\left(y-3\right)\left(2x+5\right)\)
Bài 1:
a) Sửa đề \(x\left(x+y\right)-3y\left(x+y\right)\)
\(=\left(x+y\right)\left(x-3y\right)\)
b) \(x^2+2019x-xy-2019y\)
\(=x\left(x+2019\right)-y\left(x+2019\right)\)
\(=\left(x+2019\right)\left(x-y\right)\)
c) \(x^2-9y^2-4x+4\)
\(=\left(x^2-4x+4\right)-9y^2\)
\(=\left(x-2\right)^2-\left(3y\right)^2\)
\(=\left(x-2-3y\right)\left(x-2+3y\right)\)
d) \(3x^2-5x+2\)
\(=3x^2-3x-2x+2\)
\(=3x\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(3x-2\right)\)
Bài 2:
a) \(\left(6x^3y^3-27xy^2\right):\left(3x^2y\right)-2xy^2\)
\(=6x^3y^3:3x^2y-27xy^2:3x^2y-2xy^2\)
\(=2xy^2-\dfrac{9y}{x}-2xy^2\)
\(=-\dfrac{9y}{x}\)
b) \(\dfrac{2}{x-2}+\dfrac{1-2x}{x+2}+\dfrac{3x+2}{4-x^2}\)
\(=\dfrac{2}{x-2}+\dfrac{1-2x}{x+2}-\dfrac{3x+2}{x^2-4}\)
\(=\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(1-2x\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2\left(x+2\right)+\left(1-2x\right)\left(x-2\right)-3x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x+4+x-2-2x^2+4x-3x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2x^2+4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2x}{x+2}\)
Bài 3:
a) \(3x\left(2x-3\right)-x\left(6x+4\right)=7-12x\)
\(\Rightarrow6x^2-9x-6x^2-4x=7-12x\)
\(\Rightarrow-13x=7-12x\)
\(\Rightarrow-13x+12x-7=0\)
\(\Rightarrow-x-7=0\)
\(\Rightarrow-x=7\)
\(\Rightarrow x=-7\)
b) \(3\left(x-5\right)-2x^2+10x=0\)
\(\Rightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)