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

Vì \(abc=2\)nên ta có:

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

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

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

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

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

câu trả lời;

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

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

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

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

moi hok lop 6

\(A=\dfrac{x-4+5}{\sqrt{x}-2}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+5}{\sqrt{x}-2}=\sqrt{x}+2+\dfrac{5}{\sqrt{x}-2}\)

\(=\sqrt{x}-2+\dfrac{5}{\sqrt{x}-2}+4\ge2\sqrt{\dfrac{5\left(\sqrt{x}-2\right)}{\sqrt{x}-2}}+4=4+2\sqrt{5}\)

\(A_{min}=4+2\sqrt{5}\) khi \(9+4\sqrt{5}\)

b.

Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{l}{z}\right)\Rightarrow xyz=1\)

\(B=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\)

\(B_{min}=\dfrac{3}{2}\) khi \(x=y=z=1\Rightarrow a=b=c=1\)

Thay abc = 2 vào biểu thức A ta được:

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

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

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

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

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

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

Vậy \(M=1.\)

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

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

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

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

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