\(\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right).\frac{4}{5}\)

\(=\left(\frac{157}{8}:\frac{7}{12}-\frac{53}{4}:\frac{7}{12}\right).\frac{4}{5}\)

\(=\left[\left(\frac{157}{8}-\frac{53}{4}\right):\frac{7}{12}\right].\frac{4}{5}\)

\(=\left[\frac{51}{8}:\frac{7}{12}\right].\frac{4}{5}\)

\(=\frac{153}{14}.\frac{4}{5}\)

\(=\frac{306}{35}\)

\(\left(\frac{-2}{5}+\frac{3}{7}\right)-\left(\frac{4}{9}+\frac{12}{20}-\frac{13}{35}\right)+\frac{7}{35}\)

\(=\frac{1}{35}-\frac{212}{315}+\frac{7}{35}\)

\(=\frac{1}{35}+\frac{-212}{315}+\frac{7}{35}\)

\(=\frac{9}{315}+\frac{-212}{315}+\frac{63}{315}\)

\(=\frac{-140}{315}=\frac{-4}{9}\)

\(\left(4\frac{5}{57}-3\frac{4}{51}+8\frac{13}{29}\right)-\left(3\frac{5}{57}-6\frac{16}{29}\right)\)

\(=\left(\frac{233}{57}-\frac{157}{51}+\frac{245}{29}\right)-\left(\frac{176}{57}-\frac{190}{29}\right)\)

\(=\frac{1941}{323}-\frac{-5726}{1653}\)

\(=\frac{14011}{1479}\)

a )   ( -1/6 + 5/12 ) + 7/12

= -1/6 + ( 5/12 + 7/12 ) 

= -1/6 + 12/12

= -2/12 + 12/12 

=        -10/12

=        -5/6

\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(A=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{69}-\frac{1}{70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(A=7\cdot\frac{3}{35}=\frac{21}{35}\)

\(A=\frac{7}{10\cdot11}+\frac{7}{11\cdot12}+\frac{7}{12\cdot13}+...+\frac{7}{69\cdot70}\)

\(A=7\left(\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\frac{1}{12\cdot13}+...+\frac{1}{69\cdot70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)=7\cdot\frac{3}{35}=\frac{3}{5}\)

\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\)

\(B=\frac{1}{2}\left(\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+\frac{2}{29\cdot31}+...+\frac{2}{73\cdot75}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)

\(C=\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{2008\cdot2010}\)

\(C=\frac{4}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)

\(C=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(C=2\left(\frac{1}{2}-\frac{1}{2010}\right)=2\cdot\frac{502}{1005}=\frac{1004}{1005}\)

\(B=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

\(2B=\frac{2}{25.27}+\frac{2}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

\(2B=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)

\(2B=\frac{1}{25}-\frac{1}{75}=\frac{2}{75}\)

a: =(5/7+2/7)+(4/3+5/3)=3+1=4

b: =(17/12+7/12)+(29/7-8/7)

=2+3=5

c: =(2/5+3/5)+(6/9+1/3)+(7/4+1/4)

=1+2+1

=4

d: =(1/3+2/3)+(13/17+4/17)+(29/11+4/11)

=1+1+3=5

a. bằng77

b. bằng 13

lớp 2 đã học cái này rồi à

a. bằng77

b. bằng 13

1) \(A=\frac{7}{10\times11}+\frac{7}{11\times12}+\frac{7}{12\times13}+...+\frac{7}{69\times70}\)

    \(A=7\times\left(\frac{1}{10\times11}+\frac{1}{11\times12}+\frac{1}{12\times13}+...+\frac{1}{69\times70}\right)\)

    \(A=7\times\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)

    \(A=7\times\left(\frac{1}{10}-\frac{1}{70}\right)\)

   \(A=7\times\frac{3}{35}\)

   \(A=\frac{3}{5}\)

2) \(B=\frac{1}{25\times27}+\frac{1}{27\times29}+\frac{1}{29\times31}+...+\frac{1}{73\times75}\)

    \(B=\frac{1}{2}\times\left(\frac{2}{25\times27}+\frac{2}{27\times29}+\frac{2}{29\times31}+...+\frac{2}{73\times75}\right)\).

    \(B=\frac{1}{2}\times\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)

    \(B=\frac{1}{2}\times\left(\frac{1}{25}-\frac{1}{75}\right)\)

    \(B=\frac{1}{2}\times\frac{2}{75}\)

    \(B=\frac{1}{75}\)

3) \(C=\frac{4}{2\times4}+\frac{4}{4\times6}+\frac{4}{6\times8}+...+\frac{4}{2008\times2010}\)

    \(C=\frac{4}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+...+\frac{2}{2008\times2010}\right)\)

    \(C=2\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

    \(C=2\times\left(\frac{1}{2}-\frac{1}{2010}\right)\)

    \(C=2\times\frac{502}{1005}\)

    \(C=\frac{1004}{1005}\)

_Chúc bạn học tốt_

a) \(x=\dfrac{1}{4}+\dfrac{2}{13}\)

\(x=\dfrac{13}{52}+\dfrac{8}{52}\)

⇒ \(x=\dfrac{21}{52}\)

b) \(\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}\)

\(\dfrac{x}{3}=\dfrac{14}{21}+\dfrac{-3}{21}\)

\(\dfrac{x}{3}=\dfrac{11}{21}\)

⇒ \(x=\dfrac{11.3}{21}=\dfrac{33}{21}\)

⇒ \(x=\dfrac{11}{7}\)

c) \(\dfrac{-8}{3}+\dfrac{1}{3}< x< \dfrac{-2}{7}+\dfrac{-5}{7}\)

\(\dfrac{-17}{7}< x< -1\)

⇒ \(-17< x< -7\)

⇒ \(x\in\left\{-16;-15,....;-6\right\}\)

d) \(\dfrac{1}{6}+\dfrac{2}{5}\)

\(=\dfrac{5}{30}+\dfrac{12}{30}\)

\(=\dfrac{17}{30}\)

e) \(\dfrac{3}{5}+\dfrac{-7}{4}\)

\(=\dfrac{12}{20}+\dfrac{-35}{20}\)

\(=\dfrac{-23}{20}\)

f) \(\dfrac{4}{13}+\dfrac{-12}{30}\)

\(=\dfrac{4}{13}+\dfrac{-2}{5}\)

\(=\dfrac{20}{65}+\dfrac{-26}{65}\)

\(=\dfrac{-6}{65}\)

g) \(\dfrac{-3}{29}+\dfrac{16}{58}\)

\(=\dfrac{-6}{58}+\dfrac{16}{58}\)

\(=\dfrac{10}{58}\)

h) \(\dfrac{8}{40}+\dfrac{-36}{45}\)

\(=\dfrac{1}{5}+\dfrac{-4}{5}\)

\(=\dfrac{-3}{5}\)

j) \(\dfrac{-8}{18}+\dfrac{15}{27}\)

\(=\dfrac{-2}{9}+\dfrac{5}{9}\)

\(=\dfrac{3}{9}\)

\(=\dfrac{1}{3}\)

1) \(\frac{25}{12}.x+\frac{11}{15}=\frac{9}{10}\)

=> \(\frac{25}{12}.x=\frac{9}{10}-\frac{11}{15}\)

=> \(\frac{25}{12}.x=\frac{1}{6}\)

=> \(x=\frac{1}{6}:\frac{25}{12}\)

=> \(x=\frac{2}{25}\)

Vậy \(x=\frac{2}{25}\).

3) \(\frac{29}{12}.\left[x\right]-\frac{5}{6}=\frac{3}{8}\)

=> \(\frac{29}{12}.\left[x\right]=\frac{3}{8}+\frac{5}{6}\)

=> \(\frac{29}{12}.x=\frac{29}{24}\)

=> \(x=\frac{29}{24}:\frac{29}{12}\)

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

Vậy \(x=\frac{1}{2}\).

4) \(\left[4x+\frac{3}{4}\right]-\frac{5}{4}=2\)

=> \(\left[4x+\frac{3}{4}\right]=2+\frac{5}{4}\)

=> \(4x+\frac{3}{4}=\frac{13}{4}\)

=> \(4x=\frac{13}{4}-\frac{3}{4}\)

=> \(4x=\frac{5}{2}\)

=> \(x=\frac{5}{2}:4\)

=> \(x=\frac{5}{8}\)

Vậy \(x=\frac{5}{8}\).

5) 2^x + 2^x+3 = 144

⇔ 2^x + 2^x . 2³ = 144

⇔ 2^x . (1 + 2³) = 144

⇔ 2^x . 9 = 144

⇔ 2^x = 144 : 9

⇔ 2^x = 16

⇔ 2^x = 2⁴

=> x = 4

Vậy x = 4.

Chúc bạn học tốt!