M\(=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)

\(M=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{2bc}{b\left(ac+2c+2\right)}\)

M = \(\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{b+1+bc}+\dfrac{2bc}{abc+2bc+2b}\)

M=\(\dfrac{1}{b+1+bc}+\dfrac{b}{b+1+bc}+\dfrac{2bc}{2+2bc+2b}\)

M = \(\dfrac{1+b}{b+1+bc}+\dfrac{2bc}{2\left(1+bc+b\right)}\)

M = \(\dfrac{1+b}{b+1+bc}+\dfrac{bc}{b+1+bc}=\dfrac{1+b+bc}{b+1+bc}=1\)

Thay abc = 2 vào biểu thức A ta được:

\(A=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{abc\cdot c}{ac+abc+abc}\\ A=\dfrac{1}{b+1+bc}+\dfrac{b}{bc+b+1}+\dfrac{bc}{1+bc+b}\\ A=\dfrac{1+b+bc}{1+b+bc}\\ A=1\)

\(B=\left(ab+bc+ca\right)\left(\dfrac{ab+bc+ca}{abc}\right)-abc\left(\dfrac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}\right)\)

\(=\dfrac{\left(ab+bc+ca\right)^2-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)

\(=\dfrac{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)

\(=2\left(a+b+c\right)\)

\(a,A=\dfrac{-3\left(2n-3\right)-8}{2n-3}=-3-\dfrac{8}{2n-3}\in Z\\ \Leftrightarrow2n-3\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\\ \Leftrightarrow n\in\left\{1;2\right\}\left(n\in Z\right)\)

\(b,\dfrac{ab}{a+2b}=\dfrac{3}{2}\Leftrightarrow\dfrac{a+2b}{ab}=\dfrac{2}{3}\Leftrightarrow\dfrac{1}{b}+\dfrac{2}{a}=\dfrac{2}{3}\\ \dfrac{bc}{b+2c}=\dfrac{4}{3}\Leftrightarrow\dfrac{b+2c}{bc}=\dfrac{3}{4}\Leftrightarrow\dfrac{1}{c}+\dfrac{2}{b}=\dfrac{3}{4}\\ \dfrac{ca}{c+2a}=3\Leftrightarrow\dfrac{c+2a}{ca}=\dfrac{1}{3}\Leftrightarrow\dfrac{1}{a}+\dfrac{2}{c}=\dfrac{1}{3}\)

Cộng vế theo vế \(\Leftrightarrow\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}=\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{1}{3}=\dfrac{7}{4}\)

\(\Leftrightarrow3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{7}{4}\\ \Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{7}{12}\\ \Leftrightarrow\dfrac{ab+bc+ca}{abc}=\dfrac{7}{12}\\ \Leftrightarrow T=\dfrac{12}{7}\)

\(\frac{a}{ab+a+2}\)+ \(\frac{b}{bc+b+1}\)+ \(\frac{2c}{ac+2c+2}\)

= \(\frac{a}{ab+a+2}\)+ \(\frac{ab}{a\left(bc+b+1\right)}\)+ \(\frac{2abc}{ab\left(ac+2c+2\right)}\)

= \(\frac{a}{ab+a+2}\)+ \(\frac{ab}{abc+ab+a}\)+ \(\frac{2abc}{a^2bc+2abc+2ab}\)

= \(\frac{a}{ab+a+2}\)+ \(\frac{ab}{ab+a+2}\)+ \(\frac{2}{ab+a+2}\)   (vì  abc = 2  )

= \(\frac{ab+a+2}{ab+a+2}\)= 1

tại sao lại nhân vs a và ab z bn

bn ý nhân cả tử và mẫu vs cùng 1 số

\(\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)

\(=\dfrac{a}{ab+a+2}+\dfrac{ab}{abc+ab+a}+\dfrac{2c}{ac+2c+abc}\)

\(=\dfrac{a}{ab+a+2}+\dfrac{ab}{2+ab+a}+\dfrac{2}{a+2+ab}\)

\(=\dfrac{ab+a+2}{ab+a+2}=1\)

ta có : \(A=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)

\(=\dfrac{a}{abc+ab+a}+\dfrac{b}{bc+b+1}+\dfrac{abc^2}{abc^2+abc+ac}\)