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\(\frac{7}{4}.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
\(=\frac{7}{4}.\left(\frac{33}{12}+\frac{3333\div101}{2020\div101}+\frac{333333\div10101}{303030\div10101}+\frac{33333333\div1010101}{42424242\div1010101}\right)\)
\(=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}\right)\)
= 7/4 . 44/7
= 11
\(\frac{7}{4}.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
\(=\frac{7}{4}.\left(\frac{33}{3.4}+\frac{33.101}{20.101}+\frac{33.10101}{30.10101}+\frac{33.1010101}{42.1010101}\right)\)
\(=\frac{7}{4}.\left(\frac{33}{3.4}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(=\frac{7}{4}.\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)
\(=\frac{7}{4}.33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(=\frac{7}{4}.33.\frac{4}{21}\)
\(=11\)
Tham khảo nhé~
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)=3247x.(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)=3247x.(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)=3247x.(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)=3247x.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)=3247x.33.(31−71)=32
74x.33⋅421=32\frac{7}{4}x.33\cdot\frac{4}{21}=3247x.33⋅214=32
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
Ta có : \(x-128=\left(4\frac{20}{21}-5\right)\left(\frac{4141}{4242}-1\right):\left(\frac{636363}{646464}-1\right)\\ =>x-128=\left(-\frac{1}{21}\right):\left(-\frac{1}{42}\right):\left(-\frac{1}{64}\right)\\ =>x-128=-128\\ =>x=0\)
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=\(\left(\frac{-1}{21}\right):\left(\frac{-1}{42}\right):\left(\frac{-1}{64}\right)\)
\(\Rightarrow x-128=-128\)
\(\Rightarrow x=\left(-128\right)+128\)
\(\Rightarrow x=0\)
bài 1:
\(\frac{7}{4}\left(\frac{33}{42}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(=\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=\frac{231}{4}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{231}{4}\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(=\frac{231}{4}\cdot\frac{4}{21}=11\)
\(\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{1931}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)
= \(\left[\frac{193}{17}.\frac{2}{193}-\frac{193}{17}.\frac{3}{386}+\frac{33}{34}\right]:\left[\frac{1931}{25}.\frac{7}{1931}+\frac{1931}{25}.\frac{11}{3862}+\frac{9}{2}\right]\)
= \(\left[\frac{2}{17}-\frac{3}{17}+\frac{33}{34}\right]:\left[\frac{7}{25}+\frac{11}{50}+\frac{9}{2}\right]\)
= \(\left[\frac{4}{34}-\frac{6}{34}+\frac{33}{34}\right]:\left[\frac{14}{50}+\frac{11}{50}+\frac{225}{50}\right]\)
= \(\frac{31}{34}:2\)
= \(\frac{31}{68}\)
\(B=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(=\dfrac{7}{4}\left(\dfrac{11}{4}+\dfrac{33}{20}+\dfrac{11}{10}+\dfrac{11}{14}\right)\)
\(=\dfrac{7}{4}\cdot\dfrac{11\cdot35+33\cdot7+11\cdot14+11\cdot10}{140}\)
\(=\dfrac{880}{20\cdot4}=11\)
\(C=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right)\cdot\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\dfrac{-41}{21}}\)
\(=\dfrac{\dfrac{25}{108}\cdot\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{-100}{41}}\)
\(=\dfrac{\dfrac{5751+187\cdot27}{108}}{\dfrac{-100}{41}}=100\cdot\dfrac{-41}{100}=-41\)
Lương Thị Ngân Hà
\(A=\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\)\(\left[\left(\frac{7}{1931}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{8}{9}\right]\)
\(\Rightarrow A=\left[\left(\frac{2}{193}.\frac{193}{17}\right)-\left(\frac{3}{386}.\frac{193}{17}\right)+\frac{33}{34}\right]\)\(:\left[\left(\frac{7}{1931}.\frac{1931}{25}\right)+\left(\frac{11}{3862}.\frac{1931}{25}\right)+\frac{8}{9}\right]\)
\(\Rightarrow A=\left[\frac{2}{17}-\frac{3}{34}+\frac{33}{34}\right]:\left[\frac{7}{25}+\frac{11}{50}+\frac{8}{9}\right]\)
\(\Rightarrow A=\left[\frac{4}{34}-\frac{3}{34}+\frac{33}{34}\right]:\left[\frac{126}{450}+\frac{99}{450}+\frac{400}{450}\right]\)
\(\Rightarrow A=1:\frac{625}{450}\)
\(\Rightarrow A=1.\frac{25}{18}=\frac{25}{18}\)
Vậy \(A=\frac{25}{18}\)
[(71931 +113862 ).193125 +89 ]
⇒A=[(2193 .19317 )−(3386 .19317 )+3334 ]:[(71931 .193125 )+(113862 .193125 )+89 ]
⇒A=[217 −334 +3334 ]:[725 +1150 +89 ]
⇒A=[434 −334 +3334 ]:[126450 +99450 +400450 ]
⇒A=1:625450
⇒A=1.2518 =2518
Vậy A=2518