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.

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

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

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

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) \(x^4+x^3-8x-8\)

\(=x^3\left(x+1\right)-8\left(x+1\right)\)

\(=\left(x^3+8\right)\left(x+1\right)\)

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

a) \(=x^3\left(x+1\right)-8\left(x+1\right)=\left(x+1\right)\left(x^3-8\right)=\left(x+1\right)\left(x-2\right)\left(x^2+2x+4\right)\)

b) \(=y\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(y-3\right)\)

c) \(=3\left(x-y\right)-a\left(x-y\right)=\left(x-y\right)\left(3-a\right)\)

  a+b+c+d=0 
=>a+b=-(c+d) 
=> (a+b)^3=-(c+d)^3 
=> a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d) 
=> a^3+b^3+c^3+d^3=-3ab(a+b)-3cd(c+d) 
=> a^3+b^3+c^3+d^3=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d)) 
==> a^3 +b^^3+c^3+d^3==3(c+d)(ab-cd) (dpcm)

a: \(\frac{23}{55}=1-\frac{32}{55};\frac{1978}{2010}=1-\frac{32}{2010}\)

Ta có: 55<2010

=>\(\frac{32}{55}>\frac{32}{2010}\)

=>\(-\frac{32}{55}<-\frac{32}{2010}\)

=>\(-\frac{32}{55}+1<-\frac{32}{2010}+1\)

=>\(\frac{23}{55}<\frac{1978}{2010}\)

b: \(\frac{2003\cdot2004-1}{2003\cdot2004}=1-\frac{1}{2003\cdot2004}\)

\(\frac{2004\cdot2005-1}{2004\cdot2005}=1-\frac{1}{2004\cdot2005}\)

Ta có: \(2003\cdot2004<2004\cdot2005\)

=>\(\frac{1}{2003\cdot2004}>\frac{1}{2004\cdot2005}\)

=>\(\frac{-1}{2003\cdot2004}<\frac{-1}{2004\cdot2005}\)

=>\(\frac{-1}{2003\cdot2004}+1<\frac{-1}{2004\cdot2005}+1\)

=>\(\frac{2003\cdot2004-1}{2003\cdot2004}<\frac{2004\cdot2005-1}{2004\cdot2005}\)

c: \(\frac{a}{b}-\frac{a+m}{b+m}=\frac{a\left(b+m\right)-b\left(a+m\right)}{b\left(b+m\right)}=\frac{ab+am-ab-bm}{b\left(b+m\right)}\)

\(=\frac{m\left(a-b\right)}{b\left(b+m\right)}>0\)

=>\(\frac{a}{b}>\frac{a+m}{b+m}\)

d: a<b

=>a-b<0

\(\frac{a}{b}-\frac{a+m}{b+m}=\frac{a\left(b+m\right)-b\left(a+m\right)}{b\left(b+m\right)}=\frac{ab+am-ab-bm}{b\left(b+m\right)}\)

\(=\frac{m\left(a-b\right)}{b\left(b+m\right)}<0\)

=>\(\frac{a}{b}<\frac{a+m}{b+m}\)

a) \(=\left(x-y\right)\left(3+5x\right)\)

b) \(=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-y-3\right)\left(x+y-3\right)\)

c) \(=5\left(x^2-xy-2x+2y\right)=5\left[x\left(x-y\right)-2\left(x-y\right)\right]=5\left(x-y\right)\left(x-2\right)\)

Bạn làm đc bài này chưa chỉ mình với

a: \(=6\sqrt{a}+\dfrac{1}{3}\sqrt{a}-3\sqrt{a}+\sqrt{7}=\dfrac{10}{3}\sqrt{a}+\sqrt{7}\)

b: \(=5a\cdot5b\sqrt{ab}+\sqrt{3}\cdot2\sqrt{3}\cdot ab\sqrt{ab}+9ab\cdot3\sqrt{ab}-5b\cdot9a\sqrt{ab}\)

\(=25ab\sqrt{ab}+12ab\sqrt{ab}+27ab\sqrt{ab}-45ab\sqrt{ab}\)

\(=19ab\sqrt{ab}\)

c: \(=\dfrac{\sqrt{ab}}{b}+\sqrt{ab}-\dfrac{a}{b}\cdot\dfrac{\sqrt{b}}{\sqrt{a}}\)

\(=\sqrt{ab}\left(\dfrac{1}{b}+1\right)-\dfrac{\sqrt{a}}{\sqrt{b}}\)

\(=\sqrt{ab}\)

d: \(=11\sqrt{5a}-5\sqrt{5a}+2\sqrt{5a}-12\sqrt{5a}+9\sqrt{a}\)

\(=-4\sqrt{5a}+9\sqrt{a}\)

a) \(\sqrt{4\left(a-3\right)^2}=2\left(a-3\right)=2a-6\)

b) \(\sqrt{a^2\left(a+1\right)^2}=a\left(a+1\right)=a^2+a\)

c) \(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{\sqrt{8}\left|a\right|}=\dfrac{1}{-\sqrt{8}a}=\dfrac{-\sqrt{8}}{8a}\)

a: \(\sqrt{4\left(a-3\right)^2}=2\cdot\left(a-3\right)=2a-6\)

b: \(\sqrt{a^2\left(a+1\right)^2}=a\left(a+1\right)=a^2+a\)

c: \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\sqrt{\dfrac{2}{16a^2}}=-\dfrac{\sqrt{2}}{4a}\)