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Bài 3:
a) \(4x^2+4x+1=\left(2x+1\right)^2\)
b) \(9x^2-12x+4=\left(3x-2\right)^2\)
c) \(ab^2+\dfrac{1}{4}a^2b^4+1=\left(\dfrac{1}{2}ab^2+1\right)^2\)
c: \(=\dfrac{x^3+2x^2+x^2+2x-10x-20}{x+2}\)
\(=x^2+x-10\)
a)\(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4=x^4-y^4\)
b) \(x\left(3x-18\right)-3\left(x-4\right)\left(x-2\right)+8=3x^2-18x-3x^2+18x-24+8=-16\)
a: \(=\dfrac{x+2y}{xy}\cdot\dfrac{2x^2}{\left(x+2y\right)^2}=\dfrac{2x}{y\left(x+2y\right)}\)
b: \(=\dfrac{x\left(4x^2-y^2\right)}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(2x-y\right)^3}\)
\(=\dfrac{x\left(x-y\right)\left(2x+y\right)\left(2x-y\right)}{\left(2x-y\right)^3}\)
\(=\dfrac{x\left(x-y\right)\left(2x+y\right)}{\left(2x-y\right)^2}\)
c: \(=\dfrac{x+3}{x+2}\cdot\dfrac{2x-1}{3\left(x+3\right)}\cdot\dfrac{2\left(x+2\right)}{2\left(2x-1\right)}\)
=1/3
d: \(=\dfrac{x+1}{x+2}:\left(\dfrac{1}{2x}\cdot\dfrac{3x+3}{2x-3}\right)\)
\(=\dfrac{x+1}{x+2}\cdot\dfrac{2x\left(2x-3\right)}{3\left(x+1\right)}=\dfrac{2x\left(2x-3\right)}{3\left(x+2\right)}\)
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: \(3x\left(2x-8\right)+\left(6x-2\right)\left(5-x\right)\)
\(=6x^2-24x+30x-6x^2-10+2x\)
\(=8x-10\)
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: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)
\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)
