\(\frac{1}{3}\) + \(\frac{5}{6}\): \(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)

<=> \(\frac{5}{6}\):\(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)- \(\frac{1}{3}\)

<=> \(\frac{5}{6}\) : \(\left(x-2\frac{1}{5}\right)\) = \(\frac{5}{12}\)

<=> \(\left(x-2\frac{1}{5}\right)\) =    \(\frac{5}{6}\) : \(\frac{5}{12}\)

,<=> \(\left(x-2\frac{1}{5}\right)\)=   2 

<=. x = 2 + \(\frac{11}{5}\)

<=> x = \(\frac{21}{5}\)

\(\frac{-6}{3}\left[x-\frac{1}{4}\right]=2x-1\)

\(-2x-\left[\frac{1}{4}.-2\right]=2x-1\)\

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

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

\(\)\(4x=\frac{3}{2}\)

\(x=\frac{3}{2}:4\)

\(x=\frac{3}{8}\)

\(\frac{-11}{9}\le x+\frac{11}{18}\Leftrightarrow x\ge\frac{-11}{6}\)

\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)

\(\Rightarrow\frac{-13}{9}\le x-\frac{-11}{18}\)

\(\Leftrightarrow\frac{-13}{9}\le x+\frac{11}{18}\)

\(\Rightarrow x\ge\frac{-37}{18}\)

Học tốt nhé !! ^-^

cảm ơn Lâm Việt Phúc nhé

\(\frac{x-1}{4}=\frac{2x+1}{5}\)

\(\Rightarrow5\left(x-1\right)=4\left(2x+1\right)\)

\(\Rightarrow5x-5=8x+4\)

\(\Rightarrow5x-8x=4+5\)

\(\Rightarrow-3x=9\)

\(\Rightarrow x=-3\)

vậy_

\(\frac{x+2}{x-1}=\frac{x-3}{x+1}\)

\(\Rightarrow\left(x+2\right)\left(x+1\right)=\left(x-1\right)\left(x-3\right)\)

\(\Rightarrow x^2+x+2x+2=x^2-3x-x+3\)

\(\Rightarrow x^2+x+2x-x^2+3x+x=3-2\)

\(\Rightarrow7x=1\)

\(\Rightarrow x=\frac{1}{7}\)

vậy_

 A=5-3(2x+1)^2

Ta có : (2x+1)^2\(\ge\)0

\(\Rightarrow\)-3(2x-1)^2\(\le\)0

\(\Rightarrow\)5+(-3(2x-1)^2)\(\le\)5

Dấu = xảy ra khi : (2x-1)^2=0

=> 2x-1=0 =>x=\(\frac{1}{2}\)

Vậy : A=5 tại x=\(\frac{1}{2}\)

Ta có : (x-1)^2 \(\ge\)0

=> 2(x-1)^2\(\ge\)0

=>2(x-1)^2+3 \(\ge\)3

=>\(\frac{1}{2\left(x-1\right)^2+3}\)\(\le\)\(\frac{1}{3}\)

Dấu = xảy ra khi : (x-1)^2 =0

=> x = 1

Vậy : B = \(\frac{1}{3}\)khi x = 1

\(\frac{x^2+8}{x^2+2}\)= \(\frac{x^2+2+6}{x^2+2}=1+\frac{6}{x^2+2}\)

Làm như câu B                   GTNN = 4 khi x =0 

k vs nha