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4 tháng 1 2017

Kết quả là 0,1904761905

tk nha, 100% lun nhé

4 tháng 1 2017

mk nghĩ 0,1904761905

đúng thì kick nha

7 tháng 8 2015

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

\(\frac{7}{4x}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=22\)

\(\frac{7}{4x}\left[33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right]=22\)

\(\frac{7}{4x}\left[33.\left(\frac{35}{420}+\frac{21}{420}+\frac{14}{420}+\frac{10}{420}\right)\right]=22\)

\(\frac{7}{4x}\left[33.\frac{4}{21}\right]=22\)

\(\frac{7}{4x}.\frac{44}{7}\)=22

\(\frac{11}{x}=22\)

x=11:22

x=\(\frac{1}{2}\)

b,\(\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right).x=1\)

Đặt A\(=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

Ta có :\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

\(\Rightarrow4A=4.\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right)\)

\(\Rightarrow4A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}=\frac{32}{64}+\frac{16}{64}+\frac{8}{64}+\frac{4}{64}+\frac{2}{64}+\frac{1}{64}\)

\(\Rightarrow4A=\frac{32+16+8+4+2+1}{64}=\frac{63}{64}\)

\(\Rightarrow A=\frac{63}{64}:4=\frac{63}{256}\)

\(\Rightarrow\frac{63}{256}.x=1\)

\(\Leftrightarrow x=1:\frac{63}{256}=\frac{256}{63}\)

8 tháng 9 2017

bài 2 

a] = 3 x \(\frac{4343}{7171}\)\(\frac{17372}{7171}\)\(\frac{172}{71}\)

b] = \(\frac{1}{33}\)\(\frac{44}{7}\)\(\frac{1}{3}\)\(\frac{4}{7}\)=\(\frac{4}{21}\)

bài 1 

a] y là 9

b] <=> 64y + 36y = 700 - 75 - 225

<=> 100y = 400

<=> y = 4

8 tháng 9 2017

trên lớp cô sửa rồi nên mình giải luôn:

1) Tìm y

a) y3 + 3y = 12 x 11

    y3 + 3y = 132

    y x 10 + 3 + 3 x 10 + y = 132

   ( y x 10 + y ) + ( 3 x 10 + 3 ) = 132

     11 x y + 33 = 132 

     11 x y = 132 - 33

     11 x y = 99

            y = 99 : 11

            y = 9

b) 64 x y + 225 = 700 - 75 - 36 x y

    64 x y + 225 = 625 - 36 x y

    64 x y + 36 x y = 625 -225

    64 x y + 36 x y = 400

    ( 64 + 36 ) x y = 400

    100 x y = 400

              y =  400 : 100

              y = 4

2) Tính

a) \(\frac{4343}{7171}+\frac{4343}{7171}+\frac{4343}{7171}+\frac{4343}{7171}\)

\(=\frac{4343}{7171}\times4\)

\(=\frac{43}{71}\times4\)

\(=\frac{172}{71}\)

b) A = \(\frac{1}{33}\times\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)

   Ta có:

   \(\frac{3333}{2020}=\frac{3333:101}{2020:101}=\frac{33}{20}\)

   \(\frac{333333}{303030}=\frac{333333:10101}{303030:10101}=\frac{33}{30}\)

  \(\frac{33333333}{42424242}=\frac{33333333:1010101}{42424242:1010101}=\frac{33}{42}\)

   A = \(\frac{1}{33}\times\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

   A = \(\frac{1}{33}\times33\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

  A = 1 x \(\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

  A = 1 x \(\left(\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}\right)\)

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

  A = 1 x \(\left(\frac{1}{3}-\frac{1}{7}\right)\)

  A = 1 x \(\left(\frac{7}{21}-\frac{3}{21}\right)\)

  A = 1 x \(\frac{4}{21}\)

  A = \(\frac{4}{21}\)

13 tháng 6 2018

Sửa đề:

\(\frac{1}{33}\times\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)

\(=\frac{1}{33}\times\frac{33}{12}+\frac{1}{33}\times\frac{3333}{2020}+\frac{1}{33}\times\frac{333333}{303030}+\frac{1}{33}\times\frac{33333333}{42424242}\)

\(=\frac{1}{12}+\frac{33\times101}{33\times101\times20}+\frac{33\times10101}{33\times10101\times30}+\frac{33\times1010101}{33\times1010101\times42}\)

\(=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(=\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}\)

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

\(=\frac{1}{3}-\frac{1}{7}\)

\(=\frac{4}{21}\)

6 tháng 8 2015

x-(20/11*13+20/13*15+20/15*17+...+20/553*55)=3/7

14 tháng 8 2015

(a-b)(a-b)+(b-c)(b-c)+(c-a)(c-a)=(a+b-2c)(a+b-2c)+(b+c-2a)(b+c-2a)+(c+a-2b)(c+a-2b)

Cm:a=b=c

8 tháng 8 2015

a, (x+2)+(x+4)+(x+6)+...+(x+100)=6000

(x+x+x+...+x)+(2+4+6+...+100)=6000

50.x+2550=6000

50.x=6000-2550

50.x=3450

x=3450:50

x=69

b, 1+2+3+4+...+x=15

10+...+x=15

x=15-10

x=5

Nho **** cho minh nha

 

26 tháng 2 2016

Ta có: (x+x+x+...+x) + (2+4+6+...+100) = 6000

Ta thấy vế phải có: (100-2):2+1=50(số hạng)

Tổng của vế phải: [(2+100)*50]:2=2550

\(\Rightarrow\)có 50 số x

\(\Rightarrow\)50*x + 2550 = 6000

\(\Rightarrow\)50*x=6000-2550

\(\Rightarrow\)50*x=3450

\(\Rightarrow\)x=3450:50

\(\Rightarrow\)x=69

   Vậy x=69

Mình đúng nè, nhớ k nha

9 tháng 2 2018

\(B=\)\(\frac{3+33+333+3333+33333}{4+44+444+4444+44444}\)

\(B=\frac{3.1+3.11+3.111+3.1111+3.11111}{4.1+4.11+4.111+4.1111+4.11111}\)

\(B=\frac{3.\left(1+11+111+1111+11111\right)}{4.\left(1+11+111+1111+11111\right)}\)

\(B=\frac{3}{4}\)

\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)

\(A.2=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right).2\)

\(A.2=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)

=>\(A.2-A=\left(\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right)\)

\(A=\frac{2}{3}-\frac{1}{192}\)

\(A=\frac{127}{192}\)

\(\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)

Đặt \(C=\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)

      \(C=\frac{1995.1990.1997.1993.997}{1997.1993.1994.1995.995}\)

      \(C=\frac{1990.997}{1994.995}\)

      \(C=\frac{995.2+997}{997.2+995}=1\)

9 tháng 2 2018

\(B=\frac{3+33+333+3333+ 33333}{4+44+444+4444+44444}\)

\(\Rightarrow B=\frac{3\left(1+11+111+1111+11111\right)}{4\left(1+11+111+1111+11111\right)}=\frac{3}{4}\)

19 tháng 9 2015

A = \(\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{4242}+\frac{3333}{3030}\right)\)

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

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

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

A = \(\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{7}{4}.33.\frac{4}{21}\)

=> A = \(\frac{1}{3}.33\)

=> A = 11

10 tháng 7 2017

\(M=\frac{7}{4}\times\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}+\frac{3333}{5656}\right)\)

\(M=\frac{7}{4}\times\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}+\frac{33}{56}\right)\)

\(M=\frac{7}{4}\times\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}+\frac{33}{56}\right)\)

\(M=\frac{7}{4}\times\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}+\frac{33}{7.8}\right)\)

\(M=\frac{7}{4}\times\left[33\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\right]\)

\(M=\frac{7}{4}\times\left[33\times\left(\frac{1}{3}-\frac{1}{8}\right)\right]\)

\(M=\frac{7}{4}\times\left(33\times\frac{5}{24}\right)=\frac{7}{4}\times\frac{55}{8}=\frac{385}{32}\)

10 tháng 7 2017

Bài 3 : 

b) Ta có 1+ 2 + 3 +4 + ...+ x =15

Nên \(\frac{x\left(x+1\right)}{2}=15\)

\(x\left(x+1\right)=30\)

=> \(x\left(x+1\right)=5.6\)

=> x = 5

19 tháng 6

Bài 2:

h; \(\dfrac{2}{3}\)\(x\)  + 50% + \(x\) = \(\dfrac{1}{10}\)

    \(\dfrac{2}{3}\)\(x\) + \(\dfrac{1}{2}\)  + \(x\) = \(\dfrac{1}{10}\)

    (\(\dfrac{2}{3}\)\(x\) + \(x\)) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)

     \(x\) \(\times\) (\(\dfrac{2}{3}\) + 1) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)

      \(x\) \(\times\) \(\dfrac{5}{3}\) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)

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

      \(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{-2}{5}\)

      \(x\)         = \(\dfrac{-2}{5}\)\(\dfrac{5}{3}\)

      \(x\)         =   - \(\dfrac{6}{25}\) 

Lớp 5 chưa học số âm em nhé.