Question

Tìm x

1) \(\frac{-7}{4}x\cdot\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)=22\)

2) \(\frac{1}{2016}:2015\cdot x=\frac{-1}{2015}\)

Answer 1

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

Answer 2

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

Answer 3

a) \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

\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

\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

\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

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

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

đến đây thì bn tự lm đk r

Answer 4

b) \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​

\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​

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

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

\frac{2}{x}=\frac{2}{2009}x2​=20092​

=> x = 2009