Có: \(\frac{1}{x^2+3x+2}=\frac{1}{\left(x+1\right)\left(x+2\right)}=\frac{\left(x+2\right)-\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\frac{1}{x+1}-\frac{1}{x+2}\)

\(\frac{1}{x^2+5x+6}=\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{\left(x+3\right)-\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}=\frac{1}{x+2}-\frac{1}{x+3}\)

\(\frac{1}{x^2+9x+20}=\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{\left(x+5\right)-\left(x+4\right)}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+4}-\frac{1}{x+5}\)

1/x^2+7x+12=1/(x+3)(x+4)= 1/x+3-1/x+4

Vật ta có: VT= 1/x+1-1/x+3 

VP=2/x-4- (1/x-3+1/x+5)

Đến đoạn này mình lag (sai thông cảm)

a, \(A=-x^2+2x-3=-\left(x^2-2x+1-1\right)-3=-\left(x-1\right)^2-2\le-2< 0\forall x\)

Vậy ta có đpcm 

b, \(C=-x^2+4x-7=-\left(x^2-4x+4-4\right)-7=-\left(x-2\right)^2-3\le-3< 0\forall x\)

Vậy ta có đpcm 

c, \(D=-2x^2-6x-5=-2\left(x^2+\frac{2.3}{2}x+\frac{9}{4}-\frac{9}{4}\right)-5\)

\(=-2\left(x+\frac{3}{2}\right)^2-\frac{1}{2}\le-\frac{1}{2}< 0\forall x\)

Vậy ta có đpcm 

d, \(E=-3x^2+4x-4=-3\left(x^2-\frac{4}{3}x+\frac{4}{9}-\frac{4}{9}\right)-4\)

\(=-3\left(x-\frac{2}{3}\right)^2-\frac{8}{3}\le-\frac{8}{3}< 0\forall x\)

Vậy ta có đpcm 

e, tự làm nhé 

ĐKXĐ :  a;b;c \(\ne0\)

Khi đó \(\frac{ab}{b}=\frac{bc}{c}=\frac{ca}{a}\)

<=> \(a.\frac{b}{b}=b.\frac{c}{c}=c.\frac{a}{a}\)

<=> \(a=b=c\)

Từ: \(\frac{ab}{b}=\frac{bc}{c}=\frac{ca}{a}\Leftrightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\left(đk: a,b,c>0; a+b+c\ne0\right)\)

Có: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\left(a+b+c\ne0\right)\Leftrightarrow a=b=c\)

\(3|3x+1|+4=13\Leftrightarrow3|3x+1|=13-4=9\Leftrightarrow|3x+1|=3\)

\(\Leftrightarrow\orbr{\begin{cases}3x+1=3\\3x+1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=2\\3x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{4}{3}\end{cases}}}\)

Vậy \(x\in\left\{\frac{2}{3};-\frac{4}{3}\right\}\)

không OK

CON KHONG BIET CO OI

a : 3 dư 1 \(\Rightarrow a-1⋮3\)

b : 3 sư 2 \(\Rightarrow b-2⋮3\)

\(\Rightarrow\left(a-1\right)\left(b-2\right)=ab-\left(2a+b\right)+2⋮3\)

Ta có \(a-1⋮3\Rightarrow2a-2⋮3\)

\(\Rightarrow2a-2+b-2=2a+b-4=2a+b-1-3⋮3\Rightarrow2a+b-1⋮3\)

Từ \(ab-\left(2a+b\right)+2=ab-\left(2a+b-1\right)+1⋮3\)

Mà \(2a+b-1⋮3\Rightarrow ab+1⋮3\) => ab : 3 dư 2