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![](https://rs.olm.vn/images/avt/0.png?1311)
11) \(\dfrac{5}{7}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{4}{7}\right)+\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{4}{7}\right):\dfrac{7}{5}\)
= \(\dfrac{5}{7}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{4}{7}\right)+\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{4}{7}\right)\cdot\dfrac{5}{7}\)
= \(\dfrac{5}{7}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{4}{7}+\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{4}{7}\right)\)
= \(\dfrac{5}{7}\cdot0\)
=0
12) \(\dfrac{43}{5}\left(\dfrac{17}{3}-\dfrac{16}{9}+2\right)-\dfrac{43}{5}\left(\dfrac{17}{3}-\dfrac{16}{9}\right)\)
= \(\dfrac{43}{5}\left(\dfrac{17}{3}-\dfrac{16}{9}+2-\dfrac{17}{3}+\dfrac{16}{9}\right)\)
= \(\dfrac{43}{5}\cdot2=\dfrac{43}{10}\)
11, 5/7( 1/2-1/3+1/4)+ (1/3-1/2-1/4):7/5
= 5/7.(1/2 - 1/3 + 1/4 )+( 1/3 - 1/2 - 1/4). 5/7
= 5/7.(1/2 - 2/3 + 1/4 + 1/3 - 1/2 - 1/4)
= 5/7 . -1/3
= -5/21
12, 43/5.(17/3 - 16/9 + 2)- 43/5. (17/3 - 16/9)
= 43/5.( 17/3 - 16/9 + 2 - 17/3 + 16/9)
= 43/5 . 2
= 86/5
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{-5}{9}+\dfrac{8}{15}+\dfrac{-2}{11}+\dfrac{4}{-9}+\dfrac{7}{15}\)
=\(\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)+\dfrac{-2}{11}\)
=\(\left(-1\right)+1+\dfrac{-2}{11}\)
=\(\dfrac{-2}{11}\)
b) \(\left(\dfrac{-5}{12}+\dfrac{6}{11}\right)+\left(\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\right)\)
=\(\dfrac{-5}{12}+\dfrac{6}{11}+\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\)
=\(\left(\dfrac{-5}{12}+\dfrac{5}{12}\right)+\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\dfrac{7}{17}\)
=\(0+0+\dfrac{7}{17}\)
=\(\dfrac{7}{17}\)
c) A= \(49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)
A=\(49\dfrac{8}{23}-5\dfrac{7}{32}-14\dfrac{8}{23}\)
A=\(\left(49\dfrac{8}{23}-14\dfrac{8}{23}\right)-5\dfrac{7}{32}\)
A=\(35-5\dfrac{7}{32}\)
A=\(35-\dfrac{167}{32}=\dfrac{953}{32}\)
d) C=\(\dfrac{-3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+2\dfrac{3}{7}\)
C=\(\dfrac{-3}{7}.\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{7}\)
C=\(\dfrac{-3}{7}.1+\dfrac{17}{7}\)
C=\(\dfrac{-3}{7}+\dfrac{17}{7}=2\)
a, `(-5)/9+8/15+(-2)/11+4/(-9)+7/15`
`=-5/9+8/15-2/11-4/9+7/15`
`=(-5/9-4/9)+(8/15+7/15)-2/11`
`=-9/9+15/15-2/11`
`=-1+1-2/11`
`=-2/11`
b, `((-5)/12+6/11)+(7/17+5/11+5/12)`
`=-5/12+6/11+7/17+5/11+5/12`
`=(-5/12+5/12)+(6/11+5/11)+7/17`
`=0+11/11+7/17`
`=1+7/17`
`=17/17+7/17`
`=24/17`
c, `A=49 8/23 - (5 7/32 + 14 8/23)`
`A=49 8/23 - 5 7/32 - 14 8/23`
`A=(49 8/23 - 14 8/23)-5 7/32`
`A=35 - 167/32`
`A=953/32`
d, `C=(-3)/7.5/9+4/9.(-3)/7+2 3/7`
`C=-3/7 . 5/9-4/9 . 3/7+17/7`
`C=-3/7.(5/9+4/9)+17/7`
`C=-3/7 . 1+17/7`
`C=2`
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
a) \(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}...\dfrac{99^2}{99.100}\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{99}{100}\)
\(=\dfrac{1}{100}\)
Vậy giá trị của biểu thức trên là \(\dfrac{1}{100}\).
b) \(\left(\dfrac{2}{175}-\dfrac{7}{25}+\dfrac{3}{5}\right).\left(\dfrac{4}{11}+\dfrac{3}{121}-\dfrac{47}{121}\right)\)
\(=\left(\dfrac{2}{175}-\dfrac{7}{25}+\dfrac{3}{5}\right).\left(\dfrac{44}{121}+\dfrac{3}{121}-\dfrac{47}{121}\right)\)
\(=\left(\dfrac{2}{175}-\dfrac{7}{25}+\dfrac{3}{5}\right).\dfrac{0}{121}\)
\(=\left(\dfrac{2}{175}-\dfrac{7}{25}+\dfrac{3}{5}\right).0\)
\(=0\)
Vậy giá trị của biểu thức trên là 0.
c) \(-\dfrac{2}{5}\left(\dfrac{15}{17}-\dfrac{9}{15}\right)-\dfrac{2}{5}\left(\dfrac{2}{17}+\dfrac{-2}{5}\right)\)
\(=-\dfrac{2}{5}\left[\left(\dfrac{15}{17}-\dfrac{9}{15}\right)+\left(\dfrac{2}{17}+\dfrac{-2}{5}\right)\right]\)
\(=-\dfrac{2}{5}\left(\dfrac{15}{17}-\dfrac{9}{15}+\dfrac{2}{17}+\dfrac{-2}{5}\right)\)
\(=-\dfrac{2}{5}\left(1-1\right)\)
\(=-\dfrac{2}{5}.0\)
\(=0\)
Vậy giá trị của biểu thức trên là 0.
Chúc bạn học tốt!!!
\(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^3}{3.4}...\dfrac{99^2}{99.100}\)
\(=\dfrac{1.1}{1.2}.\dfrac{2.2}{2.3}.\dfrac{3.3}{3.4}....\dfrac{99.99}{99.100}\)
\(=\dfrac{1.1.2.2.3.3.....99.99}{1.2.2.3.3.4....99.100}\)
\(=\dfrac{1.2.3...99}{1.2.3....99}.\dfrac{1.2.3....99}{2.3.4....100}=1.\dfrac{1}{100}=\dfrac{1}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\dfrac{1,6:\left(1\dfrac{3}{5}.1,25\right)}{0,64-\dfrac{1}{25}}+\dfrac{\left(1,08-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(5\dfrac{5}{9}-2\dfrac{1}{4}\right).2\dfrac{1}{4}}+0,6.0,5:\dfrac{2}{5}\)
\(M=\dfrac{1,6:\left(\dfrac{8}{5}.1,25\right)}{0,64-\dfrac{1}{25}}+\dfrac{\left(1,08-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(\dfrac{50}{9}-\dfrac{9}{4}\right).\dfrac{35}{17}}+0,6.0,5:\dfrac{2}{5}\)
\(M=\dfrac{1,6:2}{0,64-\dfrac{1}{25}}+\dfrac{1:\dfrac{4}{7}}{\dfrac{119}{36}.\dfrac{35}{17}}+0,6.0,5:\dfrac{2}{5}\)
\(M=\dfrac{0,8}{0,6}+\dfrac{1,75}{\dfrac{245}{36}}+0,6.0,5:\dfrac{2}{5}\)
\(M=\dfrac{4}{3}+\dfrac{9}{35}+0,6.0,5:\dfrac{2}{5}\)
\(M=\dfrac{167}{105}+0,6.0,5:\dfrac{2}{5}\)
\(M=\dfrac{167}{105}+\dfrac{3}{10}:\dfrac{2}{5}\)
\(M=\dfrac{167}{105}+\dfrac{3}{4}\)
\(M=\dfrac{983}{420}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+\dfrac{20}{-17}+\dfrac{-2}{9}+\dfrac{21}{56}\)
\(=\left(\dfrac{3}{17}-\dfrac{20}{17}\right)+\left(\dfrac{2}{9}-\dfrac{2}{9}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)
\(=-1+1=0\)
b) Ta có: \(\left(\dfrac{9}{16}+\dfrac{8}{-27}\right)+\left(1+\dfrac{7}{16}+\dfrac{-19}{27}\right)\)
\(=\left(\dfrac{9}{16}+\dfrac{7}{16}\right)+\left(\dfrac{-8}{27}-\dfrac{19}{27}\right)+1\)
=1-1+1=1
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =11+3/4-6-5/6+4+1/2+1+2/3
=10+9/12-10/12+6/12+8/12
=10+13/12=133/12
b: \(=2+\dfrac{17}{20}-1-\dfrac{11}{15}+2+\dfrac{3}{20}\)
=3-11/15
=34/15
c: \(=\dfrac{31}{7}:\left(\dfrac{7}{5}\cdot\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}=\dfrac{5}{7}\)
d: \(=\dfrac{29}{8}\cdot\dfrac{36}{29}\cdot\dfrac{15}{23}\cdot\dfrac{23}{5}=\dfrac{9}{2}\cdot3=\dfrac{27}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{0,8\left(\dfrac{4}{5}.1,25\right)}{0,64-\dfrac{1}{25}}+\dfrac{\left(10,8-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(6\dfrac{5}{9}-3\dfrac{1}{4}.2\dfrac{2}{17}\right)}\)
\(=\dfrac{0,8}{0,6}+\dfrac{\dfrac{268}{25}.\dfrac{7}{4}}{\left(\dfrac{59}{9}-\dfrac{13}{4}.\dfrac{36}{17}\right)}\)
\(=\dfrac{4}{3}+\dfrac{\dfrac{469}{25}}{\dfrac{-50}{153}}\)
\(=\dfrac{4}{3}-57,4056\)
\(=-56,07\)
![](https://rs.olm.vn/images/avt/0.png?1311)
d, \(\dfrac{-5}{9}+\dfrac{3}{5}-\dfrac{3}{9}+\dfrac{-2}{5}\)
\(=\left(\dfrac{-5}{9}+\dfrac{-3}{9}\right)+\left(\dfrac{3}{5}+\dfrac{-2}{5}\right)\)
\(=\dfrac{-8}{9}+\dfrac{1}{5}=-\dfrac{31}{45}\)
e) \(\left(\dfrac{1}{9}-\dfrac{9}{17}\right)+\dfrac{3}{6}-\left(\dfrac{12}{17}-\dfrac{1}{2}\right)+\dfrac{5}{9}\)
\(=\dfrac{1}{9}-\dfrac{9}{17}+\dfrac{3}{6}-\dfrac{12}{17}+\dfrac{1}{2}+\dfrac{5}{9}\)
\(=\left(\dfrac{1}{9}+\dfrac{5}{9}\right)+\left(\dfrac{-9}{17}+\dfrac{-12}{17}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2}\right)\)
\(=\dfrac{2}{3}+\left(-1\right)+1\)
\(=\dfrac{2}{3}+0=\dfrac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
G=\(\dfrac{\left(-2\right)}{3}+\dfrac{\left(-5\right)}{7}+\dfrac{2}{3}+\dfrac{\left(-2\right)}{7}\)
\(\Rightarrow G=\dfrac{\left(-2\right)}{3}+\dfrac{2}{3}+\dfrac{\left(-5\right)}{7}+\dfrac{\left(-2\right)}{7}\)
\(\Rightarrow G=\dfrac{\left(-2\right)+2}{3}+\dfrac{\left(-5\right)+\left(-2\right)}{7}\)
\(\Rightarrow G=0+\dfrac{-7}{7}\)
\(\Rightarrow G=-1\)
\(H=\dfrac{\left(-5\right)}{7}\cdot\dfrac{2}{11}+\dfrac{\left(-5\right)}{7}\cdot\dfrac{9}{11}\)
\(\Rightarrow H=\dfrac{\left(-5\right)}{7}\cdot\left(\dfrac{2}{11}+\dfrac{9}{11}\right)\)\(\Rightarrow H=\dfrac{\left(-5\right)}{7}\cdot\left(\dfrac{2+9}{11}\right)\)
\(\Rightarrow H=\dfrac{\left(-5\right)}{7}\cdot1\)
\(\Rightarrow H=\dfrac{-5}{7}\)
\(\dfrac{4}{17}\)![banh banh](https://hoc24.vn/media/cke24/plugins/smiley/images/banh.png)