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5 tháng 10 2021
Bài 2:
a) \(=x^2-36y^2\)
b) \(=x^3-8\)
Bài 3:
a) \(=x^2+2x+1-x^2+2x-1-3x^2+3=-3x^2+4x+3\)
b) \(=6\left(x-1\right)\left(x+1\right)=6x^2-6\)
H
23 tháng 9 2023
\(\dfrac{2x+4}{x^3-1}-\dfrac{2}{x-1}+\dfrac{x+2}{x^2+x+1}\\ =\dfrac{2x+4}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{2}{x-1}+\dfrac{x+2}{x^2+x+1}\\ =\dfrac{2x+4}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{2x+4-2x^2-2x-2+x^2-x+2x-2}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{-x^2+x}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{-x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=-\dfrac{x}{x^2+x+1}\)
H
23 tháng 9 2023
`a, 2/(x+1)` hay `2/(x-1)` cậu nhỉ?
`b,`
\(\dfrac{x-1}{x^2-5x+6}-\dfrac{x-3}{x-2}+\dfrac{x-2}{x-3}\\ =\dfrac{x-1}{\left(x-2\right)\left(x-3\right)}-\dfrac{x-3}{x-2}+\dfrac{x-2}{x-3}\\ =\dfrac{x-1}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)^2}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(x-2\right)^2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x-1-\left(x^2-6x+9\right)+x^2-4x+4}{\left(x-2\right)\left(x-3\right)}\\ =\dfrac{x-1-x^2+6x-9+x^2-4x+4}{\left(x-2\right)\left(x-3\right)}\\ =\dfrac{3x-6}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{3\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\\ =\dfrac{3}{x-3}\)
16 tháng 1 2023
a: \(=\dfrac{x^2-2x+1}{x}:\dfrac{x-1-3x^2+3x-3}{\left(x-1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{x}\cdot\dfrac{\left(x-1\right)\left(x^2-x+1\right)}{-2x^2+4x-4}\)
\(=\dfrac{\left(x-1\right)^3\cdot\left(x^2-x+1\right)}{-2x\left(x^2-2x+2\right)}\)
b: \(=\left[\dfrac{x^2-2x+1}{x^2+x+1}+\dfrac{2x^2-4x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right]:\dfrac{2}{x^2+1}\)
\(=\dfrac{x^3-3x^2+3x+1+2x^2-4x+1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)
\(=\dfrac{x^3+3}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)
24 tháng 12 2021
b: \(=\dfrac{7x-42-x^2+36}{x\left(x-6\right)}=\dfrac{-x^2+7x-6}{x\left(x-6\right)}=\dfrac{-x+1}{x}\)
24 tháng 12 2021
\(\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x-3\right)}=\dfrac{x\left(x+3\right)-3\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+3x-3x-9}{x\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{1}{x}\)
29 tháng 12 2021
c: \(=\dfrac{2\left(x+3\right)}{x\left(3x-1\right)}\cdot\dfrac{-\left(3x-1\right)}{x\left(x+3\right)}=\dfrac{-2}{x^2}\)
24 tháng 8 2021
Bài 3:
a) \(4x^2+4x+1=\left(2x+1\right)^2\)
b) \(9x^2-12x+4=\left(3x-2\right)^2\)
c) \(ab^2+\dfrac{1}{4}a^2b^4+1=\left(\dfrac{1}{2}ab^2+1\right)^2\)
HP
1
16 tháng 1 2023
\(\left(\dfrac{1}{x}+x-2\right):\left(\dfrac{1}{x^2-x}+1-\dfrac{3}{x-1}\right)\)
\(=\dfrac{x^2-2x+1}{x}:\dfrac{1+x^2-x-3x}{x\left(x-1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{x}\cdot\dfrac{x\left(x-1\right)}{x^2-4x+1}=\dfrac{\left(x-1\right)^3}{x^2-4x+1}\)
F
26 tháng 12 2021
a)\(\dfrac{x^2}{x-1}+\dfrac{1-2x}{x-1}\)
=\(\dfrac{x^2+1-2x}{x-1}\)
=\(\dfrac{x^2-2x+1}{x-1}\)
=\(\dfrac{\left(x-1\right)^2}{x-1}\)
= x - 1
F
26 tháng 12 2021
b) \(\dfrac{x}{x-3}\) + \(\dfrac{-9}{x^2-3x}\)
=\(\dfrac{x}{x-3}\)+ \(\dfrac{-9}{x\left(x-3\right)}\)
=\(\dfrac{x.x}{x\left(x-3\right)}\) + \(\dfrac{-9}{x\left(x-3\right)}\)
=\(\dfrac{x^2+3^2}{x\left(x-3\right)}\)
=\(\dfrac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}\)
=\(\dfrac{x+3}{x}\)
#Fiona
a. Câu hỏi của Nguyễn Thị Anh Thư - Toán lớp 8 - Học toán với OnlineMath
a, \(\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)
\(=\left(x-1-x-1\right)\left[\left(x-1\right)^2+\left(x-1\right)\left(x+1\right)+\left(x+1\right)^2\right]+6\left(x^2-1\right)\)
\(=-2\left[x^2-2x+1+x^2-1+x^2+2x+1\right]+6x^2-6\)
\(=-2\left(3x^2+1\right)+6x^2-6=-6x^2-2+6x^2-6=-8\)
b, \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)
\(=\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)\)
\(=\left(x-1\right)\left[\left(x-1\right)^2-\left(x^2+x+1\right)+3x\right]\)
\(=\left(x-1\right)\left(x^2-2x+1-x^2-x-1+3x\right)\)
\(=\left(x-1\right).0=0\)