B

\(\left|x-\dfrac{2}{3}\right|=\dfrac{3}{4}-\dfrac{1}{4}\)

\(x+\dfrac{2}{3}=\dfrac{1}{2}\)

\(x=\dfrac{1}{2}-\dfrac{2}{3}\)

\(x=\dfrac{3}{6}-\dfrac{4}{6}\)

\(x=-\dfrac{1}{6}\)

a, 6x = 613 + 5

6x = 618

x = 618 : 6

x = 103

b,x - 47 = 115

x = 115 + 47

x = 162

c,146 - x = 401 - 315 = 86

x = 146 - 86

x = 60

d,6x + 70 = 575 - 445 = 130

6x = 130 - 70 = 60

x = 60 : 6 

x = 10

e, x - 5 = 15

x = 15 + 5 

x = 20

f,x - 105 = 15 . 21 = 315

x = 315 + 105

x = 420

a, \(\left(250.4\right).73\)

\(=1000.73\)

\(=73000\)

b,\(75.\left(43+64-7\right)\)

\(=75.100\)

\(=7500\)

\(x+2=17\)

\(x=17-2=15\)

câu 4 a

câu 5 b

câu 6 d

\(200+17+\dfrac{7}{100}\)

\(=217+\dfrac{7}{100}\)

\(=\dfrac{21700}{100}+\dfrac{7}{100}\)

\(=\dfrac{21707}{100}\)

a, Do \(\dfrac{1}{n^2}< \dfrac{1}{n^2-1}\) với mọi n ≥ 2 nên 

A < C = \(\dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+\dfrac{1}{4^2-1}+......+\dfrac{1}{n^2-1}\)

Mặc khác :

\(C=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+......+\dfrac{1}{\left(n-1\right).\left(n+1\right)}\)

\(C=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+......+\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)\)

\(C=-\left(1+\dfrac{1}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)< \dfrac{1}{2}.\dfrac{3}{2}=\dfrac{3}{4}< 1\)

Vậy A < 1

b,\(B=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+.....+\dfrac{1}{\left(2n\right)^2}\)

\(B=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{n^2}\right)\)

\(B=\dfrac{1}{2^2}\left(1+A\right)\)

Suy ra \(P< \dfrac{1}{2^2}\left(1+1\right)=\dfrac{1}{2}\)    ; hay \(P< \dfrac{1}{2}\)