\(M=\frac{2018a}{ab+2018a+2018}+\frac{b}{bc+b+2018}+\frac{c}{ac+c+1}\)

\(\Rightarrow M=\frac{2018a}{ab+2018a+2018}+\frac{ab}{a\left(bc+b+2018\right)}+\frac{abc}{ab\left(ac+c+1\right)}\)

\(\Rightarrow M=\frac{2018a}{ab+2018a+2018}+\frac{ab}{ab+2018a+2018}+\frac{1}{ab+2018a+2018}\)

\(\Rightarrow M=\frac{2018a+ab+1}{2018a+ab+1}=1\)

Do : \(abc=2018\)nên : \(a,b,c\ne0\)

Ta có : \(M=\frac{2018a}{ab+2018a+2018}+\frac{b}{bc+b+2018}+\frac{c}{ac+c+1}\)

\(=\frac{2018a}{ab+2018a+2018}+\frac{ab}{abc+ab+2018a}+\frac{abc}{a^2bc+abc+ab}\)

\(=\frac{2018a}{ab+2018a+2018}+\frac{ab}{2018+ab+2018a}+\frac{2018}{2018+ab+2018a}\)

\(=\frac{2018a+ab+2018}{ab+2018a+2018}=1\)

2x - 1/5 = 6/5x - 1/2

=> 2x - 6/5x = 1/2 + 1/5

=> 4/5x = 7/10

=> x = 7/10 : 4/5

=> x = 7/8

\(2x-\frac{1}{5}=\frac{6}{5}x-\frac{1}{2}\)

\(\Rightarrow2x-\frac{6}{5}x=-\frac{1}{2}+\frac{1}{5}\)

\(\Rightarrow x\left(2-\frac{6}{5}\right)=\frac{-5+2}{10}\)

\(\Rightarrow\frac{4}{5}x=-\frac{3}{10}\)

\(\Rightarrow x=-\frac{3}{10}:\frac{4}{5}=-\frac{3}{10}.\frac{5}{4}=-\frac{3}{8}\)

Đặt : \(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=M\)

\(\Rightarrow\left(x+y+z\right).M=\frac{1}{672}.2017\)

\(\Rightarrow1+\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}=\frac{2016}{672}+\frac{1}{672}\)

\(\Rightarrow3+\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=3+\frac{1}{672}\)

\(\Rightarrow\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=\frac{1}{672}\)

Nhân cả 2 vế với \(x+y+z\),ta được:

\(\left(x+y+z\right)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{1}{672}\cdot2017\)

\(\Rightarrow\frac{x+y+z}{x+y}+\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}=\frac{2017}{672}\)

\(\Rightarrow3+\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\frac{2017}{672}\)

\(\Rightarrow C=\frac{1}{672}\)

\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot....\cdot\frac{30}{62}\cdot\frac{31}{64}=2^x\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.....\cdot\frac{30}{31}\cdot\frac{31}{32}\right)=2^x\)

\(\Leftrightarrow\frac{1}{32}=2^{x+1}\)

Làm nốt.

ko làm được câu này hay câu b ib với tớ nha.khẳng định tối giải.

Kiến thức cơ bản :v 

GT : \(\left(x_1a-y_1b\right)^{2n}+\left(x_2a-y_2b\right)^{2n}+\left(x_3a+y_3b\right)^{2n}+...+\left(x_ma-y_mb\right)^{2n}\le0\)

Có : \(\left(x_1a-y_1b\right)^{2n}+\left(x_2a-y_2b\right)^{2n}+\left(x_3a-y_3b\right)^{2n}+...+\left(x_ma-y_mb\right)^{2n}\ge0\)

\(\Rightarrow\)\(x_1a-y_1b=x_2a-y_2b=x_3a-y_3b=...=x_ma-y_mb=0\)

\(\Rightarrow\)\(x_1a=y_1b\)\(;\)\(x_2a=y_2b\)\(;\)\(x_3a=y_3b\)\(;\)\(...\)\(;\)\(x_ma=y_mb\)

\(\Rightarrow\)\(\frac{x_1}{y_2}=\frac{x_2}{y_2}=\frac{x_3}{y_3}=...=\frac{x_m}{y_m}=\frac{b}{a}\) \(\left(1\right)\)

Tính chất dãy tỉ số bằng nhau : 

\(\frac{x_1}{y_1}=\frac{x_2}{y_2}=\frac{x_3}{y_3}=...=\frac{x_m}{y_m}=\frac{x_1+x_2+x_3+...+x_m}{y_1+y_2+y_3+...+y_m}=\frac{b}{a}\) ( đpcm ) 

Phùng Minh Quân:tại sao dòng thứ hai lại đổi dấu \(\le\rightarrow\ge\)?