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\(M=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{2c}{abc+2bc+2b}\)
\(=\frac{a}{a\left(b+1+bc\right)}+\frac{b}{bc+b+1}+\frac{2bc}{2+2bc+2b}\)
\(=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{2bc}{2\left(1+bc+b\right)}\)
\(=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{2bc}{2\left(1+bc+b\right)}\)
\(=\frac{1+b+bc}{b+1+bc}=1\)
Vậy \(M=1.\)
thế abc=2 vào M ta có
M=\(\frac{a}{ab+b+abc}\)+ \(\frac{b}{bc+b+1}\)+ \(\frac{abc^2}{ac+abc^2+abc}\)
M=\(\frac{a}{a\left(bc+b+1\right)}\)+\(\frac{b}{bc+b+1}\)+ \(\frac{abc^2}{ac\left(bc+b+1\right)}\)
M=\(\frac{bc+b+1}{bc+b+1}\)=1
1 nha bạn cho mình nha
Ta có ; \(\frac{a}{ab+a+2}\)+\(\frac{b}{bc+b+1}\)+\(\frac{c}{ac+2c+2}\)
=\(\frac{a}{ab+a+2}\)+\(\frac{ab}{abc+ab+a}\)+\(\frac{c}{ac+2c+abc}\)
=\(\frac{a}{ab+a+2}\)+\(\frac{ab}{a+ab+2}\)+\(\frac{c}{c\left(a+2+ab\right)}\)
=\(\frac{a}{ab+a+2}\)+\(\frac{ab}{a+ab+2}\)+\(\frac{1}{a+ab+2}\)
=\(\frac{a+ab+1}{ab+a+2}\)
Đề bài này hình như có gì sai bạn ạ
đáng ra phải là \(\frac{2c}{ac+2c+2}\) chứ
À xin lỗi nha mình nhập sai. đúng là : \(\frac{2c}{ac+2c+2}\)
\(\frac{a}{ab+a+2}+\frac{b}{bc+b+1}+\frac{2c}{ac+2c+2}\)
\(=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{2c}{ac+2c+abc}\)
\(=\frac{a}{a\left(b+1+bc\right)}+\frac{b}{b+1+bc}+\frac{2c}{c\left(a+ab+2\right)}\)
\(=\frac{1}{b+bc+1}+\frac{b}{b+bc+1}+\frac{2}{a+2+ab}\)
\(=\frac{1}{b+bc+1}+\frac{b}{b+bc+1}+\frac{bc}{b+bc+1}\)
\(=\frac{b+bc+1}{b+bc+1}=1\)
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 !!!
Vì \(abc=2\)nên ta có:
\(M=\frac{a}{ab+a+2}+\frac{b}{bc+b+1}+\frac{2c}{ac+2c+2}\)
\(=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{abc.c}{ac+abc.c+abc}\)
\(=\frac{a}{a\left(b+1+bc\right)}+\frac{b}{bc+b+1}+\frac{abc^2}{ac\left(1+bc+b\right)}\)
\(=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+c+1}\)
\(=\frac{1+b+bc}{bc+c+1}=1\)
a, \(M=\frac{a}{ab+a+2}+\frac{b}{bc+b+1}+\frac{2c}{ca+2c+2}\)
\(=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{bc}{1+bc+b}\)
\(=\frac{1+b+bc}{1+b+bc}=1\)
\(\Rightarrow M=1\)
cho mình xửa lại một chút nha:tính : A=\(\frac{a}{ab+a+2}+\frac{b}{bc+b+1}+\frac{2c}{ca+2c+2}\)
\(M=\frac{b}{bc+b+1}+\frac{a}{ab+a+2}+\frac{2c}{ac+2c+2}\)
\(=\frac{b}{bc+b+1}+\frac{a}{ab+a+abc}+\frac{abc^2}{ac+abc^2+abc}\)
\(=\frac{b}{bc+b+1}+\frac{a}{a\left(bc+b+1\right)}+\frac{abc^2}{ac\left(bc+b+1\right)}\)
\(=\frac{b}{bc+b+1}+\frac{1}{bc+b+1}+\frac{bc}{bc+b+1}\)
\(=\frac{bc+b+1}{bc+b+1}=1\)
Vậy M = 1
= 1 nha