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10 tháng 11 2019

a) 74x.(3312+33332020+333333303030+3333333342424242)=32\frac{7}{4}x.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)=3247​x.(1233​+20203333​+303030333333​+4242424233333333​)=32

74x.(3312+3320+3330+3342)=32\frac{7}{4}x.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=3247​x.(1233​+2033​+3033​+4233​)=32

74x.(333.4+334.5+335.6+336.7)=32\frac{7}{4}x.\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)=3247​x.(3.433​+4.533​+5.633​+6.733​)=32

74x.33.(13−14+14−15+15−16+16−17)=32\frac{7}{4}x.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)=3247​x.33.(31​−41​+41​−51​+51​−61​+61​−71​)=32

74x.33.(13−17)=32\frac{7}{4}x.33.\left(\frac{1}{3}-\frac{1}{7}\right)=3247​x.33.(31​−71​)=32

74x.33⋅421=32\frac{7}{4}x.33\cdot\frac{4}{21}=3247​x.33⋅214​=32

10 tháng 11 2019

b) 13+16+110+115+...+2x.(x−1)=20072009\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{2}{x.\left(x-1\right)}=\frac{2007}{2009}31​+61​+101​+151​+...+x.(x−1)2​=20092007​

26+212+220+230+...+2(x−1).x=20072009\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{\left(x-1\right).x}=\frac{2007}{2009}62​+122​+202​+302​+...+(x−1).x2​=20092007​

22.3+23.4+24.5+25.6+...+2(x−1).x=20072009\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{\left(x-1\right).x}=\frac{2007}{2009}2.32​+3.42​+4.52​+5.62​+...+(x−1).x2​=20092007​

2.(12−13+13−14+14−15+15−16+...+1x−1−1x)=200720092.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{x-1}-\frac{1}{x}\right)=\frac{2007}{2009}2.(21​−31​+31​−41​+41​−51​+51​−61​+...+x−11​−x1​)=20092007​

2.(12−1x)=200720092.\left(\frac{1}{2}-\frac{1}{x}\right)=\frac{2007}{2009}2.(21​−x1​)=20092007​

1−2x=200720091-\frac{2}{x}=\frac{2007}{2009}1−x2​=20092007​

2x=22009\frac{2}{x}=\frac{2}{2009}x2​=20092​

=> x = 2009

8 tháng 11 2016

bài 1:

\(\frac{7}{4}\left(\frac{33}{42}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)

\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

\(=\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

\(=\frac{231}{4}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(=\frac{231}{4}\left(\frac{1}{3}-\frac{1}{7}\right)\)

\(=\frac{231}{4}\cdot\frac{4}{21}=11\)

8 tháng 11 2016

\(\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{1931}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)

= \(\left[\frac{193}{17}.\frac{2}{193}-\frac{193}{17}.\frac{3}{386}+\frac{33}{34}\right]:\left[\frac{1931}{25}.\frac{7}{1931}+\frac{1931}{25}.\frac{11}{3862}+\frac{9}{2}\right]\)

= \(\left[\frac{2}{17}-\frac{3}{17}+\frac{33}{34}\right]:\left[\frac{7}{25}+\frac{11}{50}+\frac{9}{2}\right]\)

= \(\left[\frac{4}{34}-\frac{6}{34}+\frac{33}{34}\right]:\left[\frac{14}{50}+\frac{11}{50}+\frac{225}{50}\right]\)

= \(\frac{31}{34}:2\)

= \(\frac{31}{68}\)

17 tháng 8 2018

\(\frac{7}{4}.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)

\(=\frac{7}{4}.\left(\frac{33}{12}+\frac{3333\div101}{2020\div101}+\frac{333333\div10101}{303030\div10101}+\frac{33333333\div1010101}{42424242\div1010101}\right)\)

\(=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}\right)\)

= 7/4 . 44/7

= 11

17 tháng 8 2018

\(\frac{7}{4}.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)

\(=\frac{7}{4}.\left(\frac{33}{3.4}+\frac{33.101}{20.101}+\frac{33.10101}{30.10101}+\frac{33.1010101}{42.1010101}\right)\)

\(=\frac{7}{4}.\left(\frac{33}{3.4}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

\(=\frac{7}{4}.\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)

\(=\frac{7}{4}.33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

\(=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)\)

\(=\frac{7}{4}.33.\frac{4}{21}\)

\(=11\)

Tham khảo nhé~

18 tháng 6 2019

App giải toán không cần nhập đề chỉ cần chụp ảnh cho cả nhà đây: https://www.facebook.com/watch/?v=485078328966618

\(B=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)

\(=\dfrac{7}{4}\left(\dfrac{11}{4}+\dfrac{33}{20}+\dfrac{11}{10}+\dfrac{11}{14}\right)\)

\(=\dfrac{7}{4}\cdot\dfrac{11\cdot35+33\cdot7+11\cdot14+11\cdot10}{140}\)

\(=\dfrac{880}{20\cdot4}=11\)

\(C=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right)\cdot\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\dfrac{-41}{21}}\)

\(=\dfrac{\dfrac{25}{108}\cdot\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{-100}{41}}\)

\(=\dfrac{\dfrac{5751+187\cdot27}{108}}{\dfrac{-100}{41}}=100\cdot\dfrac{-41}{100}=-41\)

27 tháng 8 2016

Ta có : \(x-128=\left(4\frac{20}{21}-5\right)\left(\frac{4141}{4242}-1\right):\left(\frac{636363}{646464}-1\right)\\ =>x-128=\left(-\frac{1}{21}\right):\left(-\frac{1}{42}\right):\left(-\frac{1}{64}\right)\\ =>x-128=-128\\ =>x=0\)

27 tháng 8 2016

a, Ta có x - 128 =( \(4\frac{20}{21}-5\)):\(\left(\frac{4141}{4242}-1\right):\left(\frac{636363}{646464}-1\right)\)

\(\Rightarrow\)x-128= \(\left(\frac{104}{21}-5\right):\left(\frac{41.101}{42.101}-1\right):\left(\frac{63.10101}{64.10101}-1\right)\)

\(\Rightarrow\)x-128=\(\left(\frac{-1}{21}\right):\left(\frac{-1}{42}\right):\left(\frac{-1}{64}\right)\)

\(\Rightarrow x-128=-128\)

\(\Rightarrow x=\left(-128\right)+128\)

\(\Rightarrow x=0\)