a) \(\frac{2}{5}x-x=\frac{\left(-2018\right)^0}{5^2}\\ x\left(\frac{2}{5}-1\right)=\frac{1}{25}\\ x\left(\frac{2}{5}-\frac{5}{5}\right)=\frac{1}{25}\\ x\cdot\frac{-3}{5}=\frac{1}{25}\\ x=\frac{1}{25}:\frac{-3}{5}\\ x=\frac{1}{25}\cdot\frac{-5}{3}\\ x=\frac{-1}{15}\)Vậy \(x=\frac{-1}{15}\)

b) \(\left|-1\frac{1}{2}x+2x\right|-\frac{7}{4}=0,5\\ \left|x\left(-1\frac{1}{2}+2\right)\right|-\frac{7}{4}=\frac{1}{2}\\ \left|x\cdot\frac{1}{2}\right|=\frac{1}{2}+\frac{7}{4}\\ \left|x\cdot\frac{1}{2}\right|=\frac{2}{4}+\frac{7}{4}\\ \left|x\cdot\frac{1}{2}\right|=\frac{9}{4}\\ \Rightarrow\left[{}\begin{matrix}x\cdot\frac{1}{2}=\frac{9}{4}\\x\cdot\frac{1}{2}=\frac{-9}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{4}:\frac{1}{2}\\x=\frac{-9}{4}:\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{4}\cdot2\\x=\frac{-9}{4}\cdot2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{2}\\x=\frac{-9}{2}\end{matrix}\right.\)Vậy \(x\in\left\{\frac{9}{2};\frac{-9}{2}\right\}\)

c) \(x+\left(x+\frac{2}{7}\right)+\frac{-5}{11}=\frac{4}{11}\\ x+x+\frac{2}{7}=\frac{4}{11}-\frac{-5}{11}\\ 2x+\frac{2}{7}=\frac{4}{11}+\frac{5}{11}\\ 2x+\frac{2}{7}=\frac{9}{11}\\ 2x=\frac{9}{11}-\frac{2}{7}\\ 2x=\frac{63}{77}-\frac{22}{77}\\ 2x=\frac{41}{77}\\ x=\frac{41}{77}:2\\ x=\frac{41}{77\cdot2}\\ x=\frac{41}{154}\)Vậy \(x=\frac{41}{154}\)

d) \(\left|0,25x-20\%\right|+\frac{3}{8}=1\frac{3}{8}\\ \left|\frac{1}{4}x-\frac{1}{5}\right|=1\frac{3}{8}-\frac{3}{8}\\ \left|\frac{1}{4}x-\frac{1}{5}\right|=1\\ \Rightarrow\left[{}\begin{matrix}\frac{1}{4}x-\frac{1}{5}=1\\\frac{1}{4}x-\frac{1}{5}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=1+\frac{1}{5}\\\frac{1}{4}x=\left(-1\right)+\frac{1}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=\frac{5}{5}+\frac{1}{5}\\\frac{1}{4}x=\frac{-5}{5}+\frac{1}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=\frac{6}{5}\\\frac{1}{4}x=\frac{-4}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{6}{5}:\frac{1}{4}\\x=\frac{-4}{5}:\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{6}{5}\cdot4\\x=\frac{-4}{5}\cdot4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{24}{5}\\x=\frac{-16}{5}\end{matrix}\right.\)Vậy \(x\in\left\{\frac{24}{5};\frac{-16}{5}\right\}\)

a . 7/12 . 6/11 + 7/12 . 5/11 - 2 7/12

= 7/12 . ( 6/11 + 5/11 ) - 31/12

= 7/12 . 1 - 31/12

= 7/12 - 31/12

= -2

b . -5/9 . -6/13 + 5/-9 . -5/13 - 5/9

= -5/9 . ( -6/13 + -5/13 ) - 5/9

= -5/9 . ( -1 ) -5/9

= 5/9 - 5/9

= 0

a, 

3x + 3 - [7x+4] = 7 + [4x-1]

=> 3x + 3 - x - 4 = 7 + 4x - 1

=> 2x - 1 = 6 + 4x

=> 2x - 4x = 6 + 1

=> -2x = 7

=> x = -7/2

b,

3^x+1 + 3^x+3 =810

=> 3^x+1[1 + 3²] = 810

=> 3^x+1 = 810 / 10

=> 3^x+1 = 81

=> x = 4

c, \(1\frac{1}{2}:\left[\frac{1}{2}-\frac{1}{3}\right]-x=5\)

\(\Rightarrow\frac{3}{2}:\frac{1}{6}-x=5\Leftrightarrow9-x=5\)

\(\Leftrightarrow x=4\)

d,

\(2,4:\left[25\%+\frac{x}{40}\right]-\frac{12}{15}=3\frac{1}{5}\)

\(\Rightarrow\frac{12}{5}:\left[\frac{1}{4}+\frac{x}{40}\right]-\frac{12}{15}=\frac{16}{5}\)

\(\Leftrightarrow\frac{12}{5}:\left[\frac{10}{40}+\frac{x}{40}\right]=\frac{16}{5}+\frac{12}{15}\Leftrightarrow\frac{12}{5}:\left[\frac{10}{40}+\frac{x}{40}\right]=4\)

\(\Rightarrow\frac{10+x}{40}=\frac{12}{5}:4\Leftrightarrow\frac{10+x}{40}=\frac{3}{5}\)

\(\Rightarrow\frac{10+x}{40}=\frac{24}{40}\Leftrightarrow10+x=24\Rightarrow x=14\)

a) 3x + 3 - ( x + 4 ) = 7 + ( 4x - 1 )

3x + 3 - x - 4 = 7 + 4x - 1

2x - 1 = 6 + 4x

-2x  = 7

\(\Rightarrow\)x = \(\frac{-7}{2}\)

b) 3^x+1 + 3^x+3 = 810

3^x . 3 + 3^x . 3³ = 810

3^x . ( 3 + 3³ ) = 810

3^x . 30 = 810

3^x = 810 : 30

3^x = 27

3^x = 3³

\(\Rightarrow\)x = 3

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

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

\(\frac{3}{2}:\frac{1}{6}-x=5\)

\(9-x=5\)

\(\Rightarrow x=9-5\)

\(\Rightarrow x=4\)

d) 2,4 : ( 25% + \(\frac{x}{40}\)) - \(\frac{12}{15}\)= \(3\frac{1}{5}\)

\(\frac{12}{5}\) : ( \(\frac{1}{4}\)+ \(\frac{x}{40}\)) - \(\frac{12}{15}\)= \(\frac{16}{5}\)

\(\frac{12}{5}:\left(\frac{1}{4}+\frac{x}{40}\right)=\frac{16}{5}+\frac{12}{15}\)

\(\frac{12}{5}:\left(\frac{1}{4}+\frac{x}{40}\right)=4\)

\(\frac{1}{4}+\frac{x}{40}=\frac{12}{5}:4\)

\(\frac{1}{4}+\frac{x}{40}=\frac{3}{5}\)

\(\frac{x}{40}=\frac{3}{5}-\frac{1}{4}\)

\(\frac{x}{40}=\frac{7}{20}\)

\(\Rightarrow\frac{x}{40}=\frac{14}{40}\)

\(\Rightarrow x=14\)

Xét Sn = 1+2+3+4+...+n               (1)

=> Sn= n+(n-1)+...+2+1               (2)

Thấy 1+n = 2+(n-1) = 3+(n-2) = n-1+2=n+1

Lấy (1);(2) và chú ý trên ta có: 

2.Sn = (n+1)+(n+1)+(n+1)+...+(n+1)=n(n+1)  (vì n số hạng giống nhau)

=> Sn= n(n+1)/2 => Sn/n = (n+1)/2

=> P= 1+ S2/2 + S3/3 + S4/4 +...+ Sn/n

P= 1+3/2+4/2+5/2+...+(n+1)/2

P= 2(2+3+4+...+n+n+1) = 2(1+2+...n+n+1) - 2 = 2.S(n+1) - 2

P= 2.(n+1)(n+2)/2 -2 = (n+1)(n+2) -2 = n²+3n

Bài toán chỉ đến S2016/2016  (tức n=2016)

Vậy S= 2016²+3.2016=2016.(2016+3)=2016.2019=4070304

E = 1 + 1/2.(1 + 2) + 1/3.(1 + 2 + 3) + 1/4.(1 + 2 + 3 + 4) + ... + 2016.(1 + 2 + 3 + ... + 2016)

E = 1 + 1/2.(1 + 2).2:2 + 1/3.(1 + 3).3:2 + 1/4.(1 + 4).4:2 + ... + 2016.(1 + 2016).2016:2

E = 2/2 + 3/2 + 4/2 + 5/2 + ... + 2017/2

E = 2+3+4+5+...+2017/2

E = (2 + 2017).2016/2

E = 2019.1008

E = 2 035 152

a/ \(\frac{-9}{10}.\frac{5}{14}+\frac{1}{10}.\left(\frac{-9}{2}\right)+\frac{1}{7}.\left(-\frac{9}{10}\right)\)

= \(-\frac{9}{10}.\left(\frac{5}{14}+\frac{1}{7}\right)+\frac{1}{10}.\left(-\frac{9}{2}\right)\)

= \(-\frac{9}{10}.\frac{1}{2}+\frac{1}{10}.\left(-\frac{9}{2}\right)\)

= \(\frac{-9}{20}+\left(-\frac{9}{20}\right)=\frac{-18}{20}=\frac{-9}{10}\)

b/ \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{11}\right).132\)

\(=\left(\frac{1}{2}.132\right)+\left(\frac{1}{3}.132\right)+\left(\frac{1}{4}.132\right)+\left(\frac{1}{6}.132\right)\)\(+\left(\frac{1}{11}.132\right)\)

\(=66+44+33+22+12=177\)

c/ \(-\frac{2}{3}.\left(\frac{8}{9}.\frac{8}{13}-\frac{8}{27}.\frac{8}{13}+\frac{4}{3}.\frac{22}{39}\right)\)

= \(-\frac{2}{3}.\left[\frac{8}{13}\left(\frac{8}{9}-\frac{8}{27}\right)+\frac{88}{117}\right]\)

= \(-\frac{2}{3}.\left(\frac{8}{13}.\frac{16}{27}+\frac{88}{117}\right)\)

= còn lại làm nốt nha! bận ròy

gidkjbibvvfrxdrfdfsddf

Có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow B< 1-\frac{1}{100}< 1\)

\(\Rightarrow B< 1\Rightarrow B< A\)

Vậy B<A