Lời giải:

a.

 \(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)

\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)

\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)

b.

\(\frac{1}{x+2}=\frac{x(2-x)}{x(x+2)(2-x)}=\frac{x(2-x)}{x(4-x^2)}\)

\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)

c.

\(\frac{4x^2-3x+5}{x^3-1}\)

\(\frac{1-2x}{x^2+x+1}=\frac{(1-2x)(x-1)}{(x-1)(x^2+x+1)}=\frac{-2x^2+3x-1}{x^3-1}\)

\(-2=\frac{-2(x^3-1)}{x^3-1}\)

\(a,=30x^3-6x^2-12x\\ b,=-6x^5-2x^3+x^2\)

a: \(=30x^3-6x^2-12x\)

b: \(=-6x^5-2x^3+x^2\)

xy + 2x + y + 2 = y(x + 1) + 2(x + 1) = (x + 1).(y + 2)

x(x - 1) + x(x + 3) = x(x - 1 + x + 3) = x. ( 2x + 2) = 2x.(x + 1)

\(-4x^2+8x-4=-4\left(x^2-2x+1\right)=-4\left(x-1\right)^2\)

c: \(-4x^2+8x-4\)

\(=-4\left(x^2-2x+1\right)\)

\(=-4\left(x-1\right)^2\)

a) \(xy+2x+y+2\)

\(=\left(xy+y\right)+\left(2x+2\right)\)

\(=y\left(x+1\right)+2\left(x+1\right)\)

\(=\left(y+2\right)\left(x+1\right)\)

b) \(x\left(x-1\right)+x\left(x+3\right)\)

\(=x^2-x+x^2+3x\)

\(=2x^2+2x\)

\(=2x\left(x+1\right)\)

c) \(-4x^2+8x-4\)

\(=-\left[\left(2x\right)^2-2.2x.2+2^2\right]\)

\(=-\left(2x-2\right)^2\)

Ta có \(A\left(x\right)=\dfrac{1}{3}x+1=0\Leftrightarrow x=-1:\dfrac{1}{3}=-3\)

\(B\left(x\right)=-\dfrac{3}{4}x+\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}\left(-\dfrac{3}{4}\right)=4\)

\(C=\left(2x-4\right)\left(x+1\right)=0\Leftrightarrow x=2;x=-1\)

\(D\left(x\right)-4x\left(x-2\right)=0\Leftrightarrow x=0;x=2\)

Mình nghĩ ra câu C rồi bạn nào giúp mình nghĩ nốt câu A,B hộ mình nhé mình cảm ơn!

a:6x-5-9x^2

=-(9x^2-6x+5)

=-(9x^2-6x+1+4)

=-(3x-1)^2-4<=-4

=>A>=2/-4=-1/2

Dấu = xảy ra khi x=1/3

b: \(B=\dfrac{4x^2-6x+4-1}{2x^2-3x+2}=2-\dfrac{1}{2x^2-3x+2}\)

2x^2-3x+2=2(x^2-3/2x+1)

=2(x^2-2*x*3/4+9/16+7/16)

=2(x-3/4)^2+7/8>=7/8

=>-1/2x^2-3x+2<=-1:7/8=-8/7

=>B<=-8/7+2=6/7

Dâu = xảy ra khi x=3/4

\(\dfrac{x+10}{x-2}+\dfrac{x-18}{x+2}+\dfrac{x+2}{x^2-4}=\dfrac{\left(x+10\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-18\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+12x+20+x^2-16x-36+x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2-3x-14}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(2x^2+4x\right)-\left(7x+14\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x\left(x+2\right)-7\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(2x-7\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-7}{x-2}\)

\(2a^2-3b-6a+ab=a\left(2a+b\right)-3\left(2a+b\right)=\left(a-3\right)\left(2a+b\right)\)

\(\left(a-b\right)^2-4=\left(a-b\right)^2-2^2=\left(a-b-2\right)\left(a-b+2\right)\)

\(4x^2-12xy+3x-9y=4x\left(x-3y\right)+3\left(x-3y\right)=\left(x-3y\right)\left(4x+3\right)\)