\(\dfrac{30}{100}\)= \(\dfrac{200x}{100}\)+5

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25 tháng 6 2017

1) \(x+\dfrac{30}{100}x=-1,31\)

\(\Leftrightarrow x+\dfrac{3}{10}x=-\dfrac{131}{100}\)

\(\Leftrightarrow100x+30x=-131\)

\(\Leftrightarrow130x=-131\)

\(\Leftrightarrow x=-\dfrac{131}{130}\)

Vậy \(x=-\dfrac{131}{130}\)

b) \(\left(4,5-2x\right)\cdot\left(-1\dfrac{4}{7}\right)=\dfrac{11}{4}\)

\(\Leftrightarrow\left(\dfrac{9}{2}-2x\right)\cdot\left(-\dfrac{4}{7}\right)=\dfrac{11}{4}\)

\(\Leftrightarrow-\dfrac{18}{7}+\dfrac{8}{7}x=\dfrac{11}{4}\)

\(\Leftrightarrow-72+32x=77\)

\(\Leftrightarrow32x=77+72\)

\(\Leftrightarrow32x=149\)

\(\Leftrightarrow x=\dfrac{149}{32}\)

Vậy \(x=\dfrac{149}{32}\)

25 tháng 6 2017

sao k làm hết cho bạn ấy v anh

26 tháng 2 2017

uk

26 tháng 3 2017

1.

\(\dfrac{30}{100}.x+\dfrac{1}{4}=\dfrac{1}{5}.x-\dfrac{1}{2}\)

\(\dfrac{3}{10}.x=\dfrac{1}{5}.x-\dfrac{1}{2}-\dfrac{1}{4}\)

\(\dfrac{3}{10}.x=\dfrac{1}{5}.x-\dfrac{1}{4}\)

\(\dfrac{1}{5}.x-\dfrac{3}{10}.x=\dfrac{1}{4}\)

\(\left(\dfrac{1}{5}-\dfrac{3}{10}\right).x=\dfrac{1}{4}\)

\(\dfrac{-1}{10}.x=\dfrac{1}{4}\)

\(x=\dfrac{1}{4}:\dfrac{-1}{10}\)

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

2.

\(\left(\dfrac{1}{7.9}+\dfrac{1}{9.11}+...+\dfrac{1}{31.33}\right).x=\left(0,25-3,5\right).\dfrac{27}{3}\)

\(\dfrac{2}{2}.\left(\dfrac{1}{7.9}+\dfrac{1}{9.11}+...+\dfrac{1}{31.33}\right).x=-3,25.9\)

\(\dfrac{1}{2}.\left(\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{2}{31.33}\right).x=-29,25\)

\(\dfrac{1}{2}.\left(\dfrac{1}{7}-\dfrac{1}{33}\right).x=-29,25\)

\(\dfrac{1}{2}.\dfrac{26}{231}.x=-29,25\)

\(\dfrac{13}{231}.x=-29,25\)

\(x=-29,25:\dfrac{13}{231}\)

\(x=\dfrac{-2079}{4}\)

tick mink nha :)

31 tháng 3 2017

ukm thanks you

1: =>7/3x=3+1/3-8-2/3=-5-1/3=-16/3

=>x=-16/3:7/3=-7/16

2: =>1/3|x-2|=4/5+3/7=28/35+15/35=43/35

=>|x-2|=129/35

=>x-2=129/35 hoặc x-2=-129/35

=>x=199/35 hoặc x=-59/35

a: =>3/2x=5/6

hay x=5/6:3/2=5/6x2/3=10/18=5/9

b: \(\Leftrightarrow219\cdot1-7\left(x+1\right)=100\)

=>7(x+1)=119

=>x+1=17

hay x=16

6 tháng 5 2017

Ta có :

\(D=\dfrac{1}{5}-\dfrac{1}{5^2}+\dfrac{1}{5^3}-\dfrac{1}{5^4}+\dfrac{1}{5^5}-..........-\dfrac{1}{5^{100}}+\dfrac{1}{5^{101}}\)

\(5D=1-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+\dfrac{1}{5^4}-\dfrac{1}{5^5}+..........+\dfrac{1}{5^{100}}\)

\(5D+D=\left(1-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+.........+\dfrac{1}{5^{100}}\right)+\left(\dfrac{1}{5}-\dfrac{1}{5^2}+..............-\dfrac{1}{5^{100}}+\dfrac{1}{5^{101}}\right)\)\(6D=1-\dfrac{1}{5^{101}}\)

\(D=\dfrac{1-\dfrac{1}{5^{101}}}{6}\)