Question

[Trung Vũ]

Cho abc =2 . Rút gọn biểu thức :

M=\(\frac{a}{ab+a+2}\)+\(\frac{b}{bc+b+1}\)+\(\frac{2c}{ac+2c+2}\).Ta được kết quả M = ?

Answer 1

[Doraemon]

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

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

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

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

\(=\frac{1}{b+bc+1}+\frac{b}{b+bc+1}+\frac{bc}{b+bc+1}\)

\(=\frac{b+bc+1}{b+bc+1}=1\)

Answer 2

[Phan Cả Phát]

Theo bài ra , ta có :

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

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

\(\Leftrightarrow\frac{a}{a\left(b+1+bc\right)}+\frac{b}{bc+b+1}+\frac{2bc}{abc+2bc+2b}\)

\(\Leftrightarrow\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{2bc}{2+2bc+2b}\)( Vì abc = 2 )

\(\Leftrightarrow\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{2bc}{2\left(1+bc+b\right)}\)

\(\Leftrightarrow\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{bc}{1+bc+b}\)

\(\Leftrightarrow\frac{1+b+bc}{b+1+bc}=1\)

Vậy M=1

Chúc bạn học tốt =))

Phan Cả Phát xin hết !!!