tham khảo bài của mình tại  http://olm.vn/hoi-dap/question/133172.html

\(=\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}=\left(\frac{1}{3}+\frac{3}{5}+\frac{1}{15}\right)+\left(-\frac{3}{4}-\frac{2}{9}-\frac{1}{36}\right)+\frac{1}{64}\)

= 1 + -1 + 1/64 

= 0 +1/64 

= 1/64

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\Rightarrow x+1=0\)

=>x=-1

vậy x=-1

\(\frac{7}{3}:\left(4.x-1\right)^2-\frac{1}{4}=\frac{1}{3}\)

\(\frac{7}{3}:\left(4.x-1\right)^2=\frac{1}{3}+\frac{1}{4}\)

\(\frac{7}{3}:\left(4.x-1\right)^2=\frac{7}{12}\)

\(\left(4.x-1\right)^2=\frac{7}{3}:\frac{7}{12}\)

\(\left(4.x-1\right)^2=4\)

\(\left(4.x-1\right)^2=2^2\)

\(4.x-1=2\)

\(4.x=2+1\)

\(4.x=3\)

\(x=3:4\)

\(x=0,75\)

\(A=5+\left|\frac{1}{3}-x\right|\)

          Vì \(\left|\frac{1}{3}-x\right|\ge0\Rightarrow\left|\frac{1}{3}-x\right|+5\ge5\)

        Dấu = xảy ra khi \(\frac{1}{3}-x=0\Rightarrow x=\frac{1}{3}\)

Vậy Min A = 5 khi \(x=\frac{1}{3}\)

\(\frac{x}{2000}+\frac{x+1}{2001}+\frac{x+2}{2002}+\frac{x+3}{2003}=4\)

\(\Leftrightarrow\left(\frac{x}{2000}-1\right)+\left(\frac{x+1}{2001}-1\right)+\left(\frac{x+2}{2002}-1\right)+\left(\frac{x+3}{2003}-1\right)=4-4=0\)

\(\Leftrightarrow\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}+\frac{x-2000}{2003}=0\)

\(\Leftrightarrow\left(x-2000\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)

\(\Leftrightarrow x-2000=0\)  ( do \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\ne0\) )

\(\Leftrightarrow x=2000\)

Vậy x = 2000

Đây là cách của lớp 7 nha

@@ Học tốt

\(\frac{x}{2000}\)- 1+\(\frac{x+1}{2001}\)-1+\(\frac{x+2}{2002}\)-1+\(\frac{x+3}{2003}\)-1=0

<=>\(\frac{x-2000}{2000}\)+ \(\frac{x-2000}{2001}\)+ \(\frac{x-2000}{2002}\)+ \(\frac{x-2000}{2003}\)=0

<=>\(\left(x-2000\right)\)\(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)\)=0

Do \(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)\)khác 0

=> \(x-2000=0\)<=> \(x=2000\)

\(\Leftrightarrow\frac{x}{2000}-1+\frac{x+1}{2001}-1+\frac{x+2}{2002}-1+\frac{x+3}{2003}-1=0\)

\(\Leftrightarrow\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}+\frac{x-2000}{2003}=0\)

\(\Leftrightarrow\left(x-2000\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)

\(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}>0\)

\(\Rightarrow x-2000=0\)\(\Rightarrow x=2000\)