\(a,\frac{62}{7}:x=\frac{29}{9}:\frac{3}{56}\)

\(\frac{62}{7}:x=\frac{1624}{27}\)

\(x=\frac{62}{7}:\frac{1624}{27}=\frac{837}{5684}\)

\(b,\frac{1}{5}:x=\frac{1}{5}-\frac{1}{7}\)

\(\frac{1}{5}:x=\frac{2}{35}\)

\(x=\frac{1}{5}:\frac{2}{35}=\frac{7}{2}\)

\(c,\frac{2}{3}.x-\frac{4}{7}=\frac{1}{7}\)

\(\frac{2}{3}.x=\frac{1}{7}+\frac{4}{7}=\frac{5}{7}\)

\(x=\frac{5}{7}:\frac{2}{3}=\frac{15}{14}\)

\(d,\frac{2}{7}-\frac{8}{9}.x=\frac{2}{3}\)

\(\frac{8}{9}.x=\frac{2}{7}-\frac{2}{3}=-\frac{8}{21}\)

\(x=-\frac{8}{21}:\frac{8}{9}=-\frac{3}{7}\)

\(e,\frac{4}{7}+\frac{5}{9}:x=\frac{1}{5}\)

\(\frac{5}{9}:x=\frac{1}{5}-\frac{4}{7}=-\frac{13}{35}\)

\(x=\frac{5}{9}:-\frac{13}{35}=\frac{175}{117}\)

\(i,\frac{2}{5}-\frac{2}{5}.x=\frac{2}{5}\)

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

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

\(x=1-1=0\)

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

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

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

học tốt nha

a, 7/12 - 3/4 . 5/6 

= 7/12 - 5/8

= 14/24- 15/24

= -1/24

b,( 2/1/3 + 1/3/4 ) . 12/13

= ( 6/3 + 7/3 ) . 12/13

= 13/3 . 12/13

=4

c, 12 : ( 3/4 -5/6 ) . 2

= 12 : ( -1/12 ) .2

= 12 . -12 . 2

= -228

d, 7/22 : 3/11 + 7/22 : 4/11 

= 7/22 . 11/3 + 7/22 . 11/4

= 7/22 . ( 11/3 + 11/4 )

....

tiếp theo bạn tự làm nhé!

1+2+3+4+5+6+7+8+9

=(1+9)+(2+8)+(3+7)+(4+6)+5

=10+10+10+10+5

=10×4+5

=40+5

=45

Học tốt

= (1+9) + (2+8) + ( 3+7 ) + (4+6) + 5

= 10+10+10+10+5

= 10*4+5

=40+5

=45

=( 1+ 9) +( 2+8) +(3+ 7) +(4+ 6) + 5

=10 + 10 + 10 + 10 + 5

=10 x 4 + 5

=45

x=12

các bn lm đến đâu cx dc miễn là lm hộ mk cái ạ, ai đang lm vào nhắn tin vs mk để mk bít nha

a; \(-\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{71}{15}< x< -\dfrac{13}{7}+\dfrac{19}{14}-\dfrac{7}{2}\)

              -\(\dfrac{19}{15}\) - \(\dfrac{71}{15}\) < \(x\) < -\(\dfrac{1}{2}\) - \(\dfrac{7}{2}\)

              -6 < \(x\) < -4

             vì \(x\) \(\in\) Z nên \(x\) = -5

- \(\dfrac{19}{6}+\dfrac{-15}{2}+\dfrac{11}{3}\) < \(x\) < - \(\dfrac{5}{4}\) + \(\dfrac{19}{12}\) - \(\dfrac{10}{3}\)

- \(\dfrac{32}{3}+\dfrac{11}{3}< x< \) \(\dfrac{1}{3}\) - \(\dfrac{10}{3}\)

-7 < \(x\) < - 3

Vì \(x\in\) Z nên \(x\in\) {-6;-5;-4}

1 + 2 + 3 + ... + 199 + 200

Ta có : 1 + 2 + 3 + ... + 199 + 200 ( có 200 số )

      = (200 + 1) x 200 : 2 = 20100

3 + 5 + 7 + ... + 97 + 99

Ta có : 3 + 5 + 7 + ... + 97 + 99 (có 49 số )

      = (99 + 3) . 49 : 2 = 2499

A = 1 + 2 + 3 + ... + 199 + 20

A = ( 199 + 1 ) . 199 : 2 +20 = 19920

B = 3 + 5 + 7 +... +  97 + 99

có số số B là : ( 99 - 3 ) :2 + 1 = 49

B = ( 99 + 3 ) . 49 : 2 = 2499

có số số C là : ( 98 - 2 ) : 2 + 1 = 49

C = ( 98 + 2 ) . 49 : 2 = 2450

a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)

    (\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)

     - \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)

     \(x\)   = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))

     \(x=\) - \(\dfrac{1}{2}\)

Vậy \(x=-\dfrac{1}{2}\) 

b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)

           \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)

          \(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)

         3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)

         3\(x\)   - 3,7 = - \(\dfrac{19}{2}\)

         3\(x\)         = - \(\dfrac{19}{2}\) + 3,7

          3\(x\)        = - \(\dfrac{29}{5}\)

           \(x\)         = - \(\dfrac{29}{5}\) : 3

           \(x\)        = - \(\dfrac{29}{15}\)

Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\) 

c; \(\dfrac{7}{9}\) : (2 + \(\dfrac{3}{4}\)\(x\)) + \(\dfrac{5}{9}\) = \(\dfrac{23}{27}\)

    \(\dfrac{7}{9}\): (2 + \(\dfrac{3}{4}\)\(x\)) = \(\dfrac{23}{27}\) - \(\dfrac{5}{9}\)

     \(\dfrac{7}{9}\):(2 + \(\dfrac{3}{4}\)\(x\)) = \(\dfrac{8}{27}\)

      2 + \(\dfrac{3}{4}\)\(x\) = \(\dfrac{7}{9}\) : \(\dfrac{8}{27}\)

      2 + \(\dfrac{3}{4}\)\(x\) =  \(\dfrac{21}{8}\)

             \(\dfrac{3}{4}x\) = \(\dfrac{21}{8}\) - 2

             \(\dfrac{3}{4}\)\(x\) =  \(\dfrac{5}{8}\)

               \(x\) = \(\dfrac{5}{8}\) : \(\dfrac{3}{4}\)

              \(x\) =  \(\dfrac{5}{6}\)

Vậy \(x=\dfrac{5}{6}\)

1.\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{4}{23}-\frac{4}{27}\)

\(=\frac{1}{3}-\frac{1}{27}=\frac{9}{27}-\frac{1}{27}=\frac{8}{27}\)

2. Đặt \(A=\frac{3}{14}+\frac{3}{84}+\frac{3}{204}+\frac{3}{374}+\frac{3}{594}+\frac{3}{864}\)

\(\Rightarrow A=\frac{3}{2.7}+\frac{3}{7.12}+...+\frac{3}{27.32}\)

\(\Rightarrow5A=3.\left(\frac{5}{2.7}+\frac{5}{7.12}+...+\frac{5}{27.32}\right)\)

\(\Rightarrow5A=3.\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{27}-\frac{1}{32}\right)\)

\(\Rightarrow5A=3.\left(\frac{1}{2}-\frac{1}{32}\right)\)

\(\Rightarrow5A=3.\frac{15}{32}=\frac{45}{32}\Rightarrow A=\frac{45}{32}:5=\frac{9}{32}\)

3. Đặt \(S=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{340}\)

\(\Rightarrow3S=\frac{3}{10}+\frac{3}{40}+...+\frac{3}{340}\)

\(\Rightarrow3S=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\)

\(\Rightarrow3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)

\(\Rightarrow3S=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\Rightarrow S=\frac{9}{20}:3=\frac{3}{20}\)

Câu 1:

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{27}\)

\(=\frac{8}{27}\)

Câu 1:

\(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)

Áp dụng tính chất \(\frac{b}{a\left[a+b\right]}=\frac{1}{a}-\frac{1}{a+b}\), ta có:

\(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}=\frac{9}{27}-\frac{1}{27}=\frac{8}{27}\)

Câu 2 tương tự nhưng phải phân

\(\frac{3}{14}+\frac{3}{84}+\frac{3}{204}+...+\frac{3}{864}=\frac{3}{2\cdot7}+\frac{3}{7\cdot12}+\frac{3}{12\cdot17}+...+\frac{3}{27\cdot32}\)

Cái này áp dụng công thức \(\frac{a}{b\left[b+c\right]}=\frac{a}{c}\left[\frac{1}{b}-\frac{1}{b+c}\right]\), ta có:

\(\frac{3}{2\cdot7}+\frac{3}{7\cdot12}+\frac{3}{12\cdot17}+...+\frac{3}{27\cdot32}=\frac{3}{5}\left[\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{27}-\frac{1}{32}\right]\)

\(=\frac{3}{5}\left[\frac{1}{2}-\frac{1}{32}\right]=\frac{3}{5}\cdot\frac{15}{32}=\frac{9}{32}\)

Câu 3:

tương tự quy laautj mẫu là 2.5; 5.8 ....

Câu 4: qL mẫu là 1.7; 7.13; ....

Câu 5: \(=\left[1-\frac{1}{2}\right]+\left[1-\frac{1}{6}\right]+...+\left[1-\frac{1}{110}\right]\)

\(=\left[1-\frac{1}{1\cdot2}\right]+\left[1-\frac{1}{2.3}\right]+\left[1-\frac{1}{3\cdot4}\right]+...+\left[1-\frac{1}{10.11}\right]\)

\(=10-\frac{9}{10}=\frac{91}{10}\)