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a) \(3\left(2x-3\right)+5\left(x+2\right)=6x-9+5x+10=11x+1\)
b) \(3x\left(2x-8\right)+\left(6x+2\right)\left(5-x\right)=6x^2-24x+30x-6x^2+10-2x=4x+10\)
c) \(\left(x-3\right)\left(x+3\right)-\left(x-5\right)^2=x^2-9-x^2+10x-25=10x-34\)
d) \(\left(x-y\right)^3-\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-3x^2y+3xy^2-y^3-x^3+y^3=3xy^2-3x^2y\)
\(1,\\ a,=3x^3-2x^2+5x\\ b,=2x^3y^2+\dfrac{2}{9}x^4y^2-\dfrac{1}{3}x^2y^3\\ c,=x^2-2x+6x-12=x^2+4x-12\\ 2,\\ a,\Rightarrow6x-9+4-2x=-3\\ \Rightarrow4x=2\Rightarrow x=\dfrac{1}{2}\\ b,\Rightarrow5x-2x^2+2x^2-2x=13\\ \Rightarrow3x=13\Rightarrow x=\dfrac{13}{3}\\ c,\Rightarrow5x^2-5x-5x^2+7x-10x+14=6\\ \Rightarrow-8x=-8\Rightarrow x=1\\ d,\Rightarrow6x^2+9x-6x^2+4x-15x+10=8\\ \Rightarrow-2x=-2\Rightarrow x=1\)
\(3,\\ A=2x^2+x-x^3-2x^2+x^3-x+3=3\\ B=6x^2-10x+33x-55-6x^2-14x-9x-21=-76\)
\(a.2x\left(x-1\right)-3\left(x^2+4x\right)+x\left(x+2\right)\)
\(=2x^2-2x-3x^2-12x+x^2+2x\)
\(=-12x\)
\(b.\left(2x-3\right)\left(3x+5\right)-\left(x-1\right)\left(6x+2\right)+3-5x\)
\(=6x+10x-9x^2-15-6x^2-2x-6x-2+3-5x\)
\(=-15x^2+3x-14\)
\(c.\left(x-y\right)\left(x^2+xy+y^2\right)-\left(x+y\right)\left(x^2-y^2\right)\)
\(=x^3-y^3-x^3+y^3+x^2y-y^3\)
\(=y^3+x^2y\)
c: \(=\dfrac{x^3+2x^2+x^2+2x-10x-20}{x+2}\)
\(=x^2+x-10\)
a: \(=\dfrac{2x-2x+y}{2\left(2x-y\right)}=\dfrac{y}{2\left(2x-y\right)}\)
b: \(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{6x+2-x^2-x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+5x+2}{2\left(x-1\right)\left(x+1\right)}\)
c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)
\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)
d: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)
\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)
\(a,\left(2x-5\right)\left(5-x\right)=5\left(2x-5\right)-x\left(2x-5\right)=10x-25-2x^2+5x=15x-2x^2-25\\ b,\dfrac{1}{3x-2}-\dfrac{1}{3x+2}=\dfrac{3x+2-3x+2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{4}{\left(3x-2\right)\left(3x+2\right)}\)
\(c,\dfrac{3}{x-3}-\dfrac{6x}{x^2-9}+\dfrac{x}{x+3}=\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x+9-6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2-6x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{x+3}\)
Lời giải:
a.
$(2x-3)^2+(2x+3)(5-2x)=(4x^2-12x+9)-(-4x^2+4x+15)$
$=4x^2-12x+9+4x^2-4x-15$
$=24-8x$
b.
$3(2x-3)+5(x+2)=6x-9+5x+10=11x+1$
c.
$3x(2x-8)+(6x-2)(5-x)=(6x^2-24x)+(-6x^2+32x-10)$
$=6x^2-24x-6x^2-32x+10$
$=8x-10$
d.
$(x-3)(x+3)-(x-5)^2=(x^2-9)-(x^2-10x+25)$
$=x^2-9-x^2+10x-25=10x-34$
e.
$(x-y)^3-(x-y)(x^2+xy+y^2)=(x^3-3x^2y+3xy^2-y^3)-(x^3-y^3)$
$=-3x^2y+3xy^2=3xy(y-x)$
a: ta có: \(\left(2x-3\right)^2+\left(2x+3\right)\left(5-2x\right)\)
\(=4x^2-12x+9+2x-4x^2+15-6x\)
\(=-16x+24\)
b: Ta có: \(3\left(2x-3\right)+5\left(x+2\right)\)
\(=6x-9+5x+10\)
\(=11x+1\)
c: ta có: \(3x\left(2x-8\right)+\left(6x-2\right)\left(5-x\right)\)
\(=6x^2-24x+30x-6x^2-10+2x\)
\(=8x-10\)