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a, \(\left(2x-1\right)\left(3-2x\right)=6x-4x^2-3+2x=-4x^2+8x-3\)
b, \(\left(x+2\right)+\left(1+x\right)\left(1-x\right)=x+2+1-x^2=-x^2+x+3\)
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: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
a: =9x^2-12x+4-4x^2+14x
=5x^2+2x+4
b: \(=\dfrac{2+x+1+x-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{2x+2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)
a) 3x.(x² - 2)
= 3x.x² + 3x.(-2)
= 3x³ - 6x
b) (6x³ + 2x² - 4x) : 2x
= 6x³ : 2x + 2x² : 2x - 4x : 2x
= 3x² + x - 2
c) 2x(x² - 1)
= 2x.x² - 2x.1
= 2x³ - 2x
a) \(2x\left(x+5\right)-2x^2=2x^2+10x-2x^2=10x\)
b) \(\left(x+3\right)^2+\left(x-1\right)\left(3+2x\right)=x^2+6x+9+3x+2x^2-3-2x\)
\(=3x^2+7x+6\)
a: \(2x\left(x+5\right)-2x^2=2x^2+10x-2x^2=10x\)
b: \(\left(x+3\right)^2+\left(2x+3\right)\left(x-1\right)\)
\(=x^2+6x+9+2x^2-2x+3x-3\)
\(=3x^2+7x+6\)
Ta có: \(\dfrac{2x+3}{1-x^2}+\dfrac{2x+1}{x^2-2x+1}\)
\(=\dfrac{-2x-3}{\left(x-1\right)\left(x+1\right)}+\dfrac{2x+1}{\left(x-1\right)^2}\)
\(=\dfrac{\left(-2x-3\right)\left(x-1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)}+\dfrac{\left(2x+1\right)\left(x+1\right)}{\left(x+1\right)\cdot\left(x-1\right)^2}\)
\(=\dfrac{-2x^2+2x-3x+3+2x^2+2x+x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\dfrac{2x+4}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
thực ra mình cũng cố rồi nhưng mà IQ có hạn nên nghĩ mãi ko ra, thế nên mới phải cầu cứu mấy bạn giỏi hơn đấy =)
