2.

\(P=\left(\dfrac{a+6}{3\left(a+3\right)}-\dfrac{1}{a+3}\right).\dfrac{27a}{a+2}=\left(\dfrac{a+3}{3\left(a+3\right)}\right).\dfrac{27a}{a+2}=\dfrac{27a}{3\left(a+2\right)}=\dfrac{9a}{a+2}\)

ĐKXĐ là :

\(a\ne0;-3;-2\)

Vs a = 1 ta có:

=> P=3

1.

\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)

a: ĐKXĐ: a∉{-1/3;-3}

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

=>\(\frac{\left(3a-1\right)\left(a+3\right)+\left(3a+1\right)\left(a-3\right)}{\left(3a+1\right)\left(a+3\right)}=2\)

=>\(2\left(3a+1\right)\left(a+3\right)=\left(3a-1\right)\left(a+3\right)+\left(3a+1\right)\left(a-3\right)\)

=>\(2\left(3a^2+9a+a+3\right)=3a^2+9a-a-3+3a^2-9a+a-3\)

=>\(6a^2+20a+6=6a^2-6\)

=>20a=-12

=>a=-3/5(nhận)

b: ĐKXĐ: a∉{5/2;2/3}

\(\frac{2a-9}{2a-5}+\frac{3a}{3a-2}=2\)

=>\(\frac{2a-5-4}{2a-5}+\frac{3a-2+2}{3a-2}=2\)

=>\(1-\frac{4}{2a-5}+1+\frac{2}{3a-2}=2\)

=>\(\frac{2}{3a-2}=\frac{4}{2a-5}\)

=>\(\frac{4}{6a-4}=\frac{4}{2a-5}\)

=>6a-4=2a-5

=>4a=-1

=>a=-1/4(nhận)

c: ĐKXĐ: a<>-3

\(\frac{10}{3}-\frac{3a-1}{4a+12}-\frac{7a+2}{6a+18}=2\)

=>\(\frac{3a-1}{4a+12}+\frac{7a+2}{6a+18}=\frac{10}{3}-2=\frac43\)

=>\(\frac{3\left(3a-1\right)}{12\left(a+3\right)}+\frac{2\left(7a+2\right)}{12\left(a+3\right)}=\frac43\)

=>\(\frac{9a-3+14a+4}{12\left(a+3\right)}=\frac{4\cdot4\cdot\left(a+3\right)}{12\left(a+3\right)}\)

=>23a+1=16(a+3)=16a+48

=>7a=47

=>a=47/7(nhận)

\(A=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{a^3-1}+\dfrac{1}{a-1}\right]\cdot\dfrac{a\left(a^2+1\right)}{2a}\)

\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{a^2+1}{2}\)

\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{a^2+1}{2}=\dfrac{a^2+1}{2}\)

\(=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{1}{a-1}\right]:\dfrac{2a}{3}\)

\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}\)

\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}=\dfrac{3}{2a}\)

Đây là câu a/

https://hoc24.vn/hoi-dap/question/693692.html?pos=1903228

Còn câu b thì như sau:

Trước hết, nghi ngờ bạn ghi sai đề ở con này \(\dfrac{1}{a^2+7a+9}\) , số 9 phải là số 12 mới hợp lý. Mình tự sửa lại đề, còn nếu đề đúng như bạn chép thì bạn giữ nguyên nó, phần còn lại rút gọn được còn đâu thì quy đồng giải trâu thôi, chẳng cách nào với đề xấu kiểu ấy cả.

\(B=\dfrac{1}{a\left(a+1\right)}+\dfrac{1}{\left(a+1\right)\left(a+2\right)}+\dfrac{1}{\left(a+2\right)\left(a+3\right)}+\dfrac{1}{\left(a+3\right)\left(a+4\right)}+\dfrac{1}{\left(a+4\right)\left(a+5\right)}\)

\(B=\dfrac{1}{a}-\dfrac{1}{a+1}+\dfrac{1}{a+1}-\dfrac{1}{a+2}+\dfrac{1}{a+2}-\dfrac{1}{a+3}+\dfrac{1}{a+3}-\dfrac{1}{a+4}+\dfrac{1}{a+4}-\dfrac{1}{a+5}\)

\(B=\dfrac{1}{a}-\dfrac{1}{a+5}=\dfrac{5}{a\left(a+5\right)}\)

a)    \(VT=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)\(=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\) (vì  a+b+c = 1)

\(=3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\)

\(=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)

C/m  BĐT phụ:   \(\frac{x}{y}+\frac{y}{x}\ge2\)   với  x,y dương

             \(\Leftrightarrow\)\(x^2+y^2\ge2xy\)

            \(\Leftrightarrow\) \(x^2-2xy+y^2\ge0\)

            \(\Leftrightarrow\) \(\left(x-y\right)^2\ge0\)  luôn đúng

Dấu "=" xảy ra   \(\Leftrightarrow\)\(x=y\)

Áp dụng BĐT trên ta có:   \(\frac{a}{b}+\frac{b}{a}\ge2;\) \(\frac{a}{c}+\frac{c}{a}\ge2;\) \(\frac{b}{c}+\frac{c}{b}\ge2\)

\(\Rightarrow\)\(VT=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge3+2+2+2=9\)

Dấu "=" xảy ra  \(\Leftrightarrow\)\(a=b=c\)

Vậy    \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)

Dấu "=" xảy ra  \(\Leftrightarrow\)\(a=b=c\)

a: \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)

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

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

b: \(\frac{4a^2-3a+5}{a^3-1}-\frac{1-2a}{a^2+a+1}+\frac{2}{a-1}\)

\(=\frac{4a^2-3a+5+\left(2a-1\right)\left(a-1\right)+2\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\frac{4a^2-3a+5+2a^2-3a+1+2a^2+2a+2}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{8a^2-4a+8}{\left(a-1\right)\left(a^2+a+1\right)}\)