y=2x-3 / x+2
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a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)
\(=6x^2-2x-6x^2-2x+18x+6\)
=14x+6
b: Ta có: \(\left(2x-3\right)^2-\left(2x+1\right)\left(2x-1\right)+3\left(2x-3\right)\)
\(=4x^2-12x+9-4x^2+1+6x-9\)
\(=-6x+1\)
c: Ta có: \(\left(x+y-1\right)^2-2\left(x+y-1\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x+y-1-x-y\right)^2\)
=1
a) \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)=6x^2-2x-6x^2-2x+18x+6=14x+6\)
b) \(\left(2x-3\right)^2-\left(1+2x\right)\left(2x-1\right)+3\left(2x-3\right)=4x^2-12x+9-4x^2+1+6x-9=-6x+1\)
c) \(\left(x+y-1\right)^2-2\left(x+y-1\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-1-x-y\right)^2=\left(-1\right)^2=1\)
\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)
\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)
\(=6x^2y\)
\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)
\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)
\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)
1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy
2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3
=6x^2y
3: =(x+y-x+y)^2=(2y)^2=4y^2
4: =(2x+3-2x-5)^2=(-2)^2=4
5: =18^8-18^8+1=1
a, \(\frac{x^2}{x+1}+\frac{2x}{x^2-1}+\frac{1}{x+1}+1\)
\(=\frac{x^2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{2x}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^3-x^2-2x+x-1-x^2-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^3-2x^2-x-2}{\left(x-1\right)\left(x+1\right)}\)
a: =2x^3-3x-5x^3-x^2-x^2
=-3x^3-2x^2-3x
b: =2(x^2+x-6)+x^2-4x+4+x^2+6x+9
=2x^2+2x-12+2x^2+2x+13
=4x^2+4x+1
d: =4x^2-9-x^2-10x-25-x^2-x+2
=2x^2-11x-32
\(\dfrac{4x^2\left(y+z\right)^5}{2x\left(y+z\right)^3}=2x\left(y+z\right)^2\)
a, \(x^3+2x^2+x-xy=x\left(x^2+2x+1-y\right)\)
\(=x\left[\left(x+1\right)^2-y\right]\)
b, \(x^3-y^3+2x^2-2y^2=\left(x-y\right)\left(x^2+xy+y^2\right)+2\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+2\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left[\left(x^2+xy+y^2\right)+2\left(x+y\right)\right]\)
\(=\left(x-y\right)\left(x^2+xy+y^2+2x+2y\right)\)
\(2x\left(x^2-7x-3\right)=2x^3-14x-6x\)
\(4xy^2\left(-2x^3+y^2-7xy\right)=-8x^4y^2+4xy^5-28x^2y^3\)
Hàm số \(y=2x-\dfrac{3}{x}+2\) có tiệm cận đứng là đường thẳng x=0, không có tiệm cận ngang.
Hàm số \(y=2x-\dfrac{3}{x+2}\) có tiệm cận đứng là đường thẳng x= -2, không có tiệm cận ngang.
Hàm số \(y=\dfrac{2x-3}{x+2}\) có tiệm cận ngang là đường thẳng y=2, tiệm cận đứng là đường thẳng x= -2.