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25 tháng 5 2019

Đề :

\(199,5:\frac{\frac{3}{5}\cdot x+4}{20}+8,5=\frac{242}{5}\)

\(\Rightarrow199,5:\frac{\frac{3}{5}\cdot x+4}{20}=\frac{242}{5}-8,5\)

\(\Rightarrow199,5:\frac{\frac{3}{5}\cdot x+4}{20}=48,4-8,5\)

\(\Rightarrow199,5:\frac{\frac{3}{5}\cdot x+4}{20}=39,9\)

\(\Rightarrow\frac{\frac{3}{5}\cdot x+4}{20}=199,5:39,9\)

\(\Rightarrow\frac{\frac{3}{5}\cdot x+4}{20}=5\)

\(\Rightarrow\frac{3}{5}\cdot x+4=5\cdot20\)

\(\Rightarrow\frac{3}{5}\cdot x+4=100\)

\(\Rightarrow\frac{3}{5}x=96\Leftrightarrow x=160\)

25 tháng 5 2019

biến đổi là ra x = 160 thôi

\(\left(X+\frac{1}{1.3}\right)+\left(X+\frac{1}{3.5}\right)+...+\left(X+\frac{1}{23.25}\right)=11.X+\)\(\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

\(\Leftrightarrow12X+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)+11X\)\(+\frac{\left(1+\frac{1}{3}+...+\frac{1}{81}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)}{2}\)

\(\Leftrightarrow X+\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\right)=\frac{242}{243}:2\)

\(\Leftrightarrow X+\frac{12}{25}=\frac{121}{243}\)

\(\Leftrightarrow X=\frac{109}{6075}\)

Vậy X=109/6075

Chắc Sai kết quả chứ công thức đúng nha!!!...

Fighting!!!...

28 tháng 5 2019

Đặt: 

 \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{25-23}{23.25}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}=1-\frac{1}{25}=\frac{24}{25}\)

=> \(A=\frac{12}{25}\)

Đặt \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)

   \(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)

=> \(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)=1-\frac{1}{3^5}=\frac{242}{243}\)

=> \(2B=\frac{242}{243}\Rightarrow B=\frac{121}{243}\)

Giải phương trình:

\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)\)

                        \(12x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{242}\right)\)

                                                                            \(12x+\frac{12}{25}=11x+\frac{121}{243}\)

                                                                             \(12x-11x=\frac{121}{243}-\frac{12}{25}\)

                                                                                                  \(x=\frac{109}{6075}\)

18 tháng 9 2017

a,

x.2/7.3/4=5/21

x.3/14=5/21

x=5/21:3/14

x=10/9

b, 

x.1/2=1/3

x=1/3:1/2

x=2/3

c,

x:4/5=25/8:5/4

x:4/5=5/2

x=5/2.4/5=2

18 tháng 9 2017

A, x= 5/25 : 3/4 : 2/7 = 14/15

B, x=1/3 : 1/2 = 2/3

C, x=(25/8 : 5/4)x4/5 = 5/2 x 4/5 = 2

23 tháng 7 2018

\(x\cdot\frac{1}{3}+x\cdot\frac{2}{5}-4=3\)

\(x\cdot\left(\frac{1}{3}+\frac{2}{5}\right)-4=3\)

\(x\cdot\frac{11}{15}=7\)

\(x=\frac{105}{11}\)

23 tháng 7 2018

\(x.\frac{1}{3}+x.\frac{2}{5}-4=3\)

\(\Rightarrow x.\left(\frac{1}{3}+\frac{2}{5}\right)=7\)

\(\Rightarrow x.\left(\frac{5}{15}+\frac{6}{15}\right)=7\)

\(\Rightarrow x.\frac{11}{15}=7\Rightarrow x=7:\frac{11}{15}=\frac{105}{11}\)

Vậy x = 105/11

3 tháng 8 2017

\(\frac{4}{3}+x.\frac{2}{3}-\frac{1}{4}=\frac{53}{12}\)

\(x.\frac{2}{3}-\frac{1}{4}=\frac{53}{12}-\frac{4}{3}\)

\(x.\frac{2}{3}-\frac{1}{4}=\frac{37}{12}\)

\(x.\frac{2}{3}=\frac{37}{12}+\frac{1}{4}\)

\(x.\frac{2}{3}=\frac{10}{3}\)

\(x=\frac{10}{3}:\frac{2}{3}\)

\(x=\frac{30}{6}=5\)

Vậy x = 5

mik ko chắc

3 tháng 8 2017

\(\frac{4}{3}+x.\frac{2}{3}-\frac{1}{4}=4\frac{5}{12}\)

\(\frac{4}{3}+x.\frac{2}{3}=4\frac{5}{12}+\frac{1}{4}\)

\(\frac{4}{3}+x.\frac{2}{3}=\frac{53}{12}+\frac{1}{4}\)

\(\frac{4}{3}+x.\frac{2}{3}=\frac{14}{3}\)

\(x.\frac{2}{3}=\frac{14}{3}-\frac{4}{3}\)

\(x.\frac{2}{3}=\frac{10}{3}\)

\(x=\frac{10}{3}:\frac{2}{3}\)

\(x=5\)

16 tháng 8 2017

Ta có:

\(A=\left(x-\frac{1}{2}\right).\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{1}{3}\)

\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{1}{3}\)

\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{3}\)

\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{10}\right)=\frac{1}{3}\)

\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\frac{9}{10}=\frac{1}{3}\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}.\frac{10}{9}\Leftrightarrow x=\frac{47}{54}\)

\(B=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{96.101}=\frac{1}{10.x}\)

\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)=\frac{1}{10}-\frac{1}{x}\)

\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\)

\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\Leftrightarrow B=\frac{1}{5}.\frac{100}{101}=\frac{1}{10}-\frac{1}{x}\)

\(\Leftrightarrow B=\frac{1}{x}=\frac{1}{10}-\frac{20}{101}=-\frac{99}{1010}\Leftrightarrow x=-\frac{1010}{99}\)

16 tháng 8 2017

c) Sai đề nhé bạn vì không có kết quả nên không tìm được x.

d) \(\left(x-5\right).\left(10-9\frac{40}{41}\right):\left(1-\frac{81}{82}\right):\left(1-\frac{204}{205}\right)=2050\)

\(\Rightarrow\left(x-5\right).\frac{1}{41}.82.205=2050\)

\(\Rightarrow\left(x-5\right).2.205=2050\Leftrightarrow x-5=2050:410=5\Leftrightarrow x=10\)

\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}+\frac{1}{3}\right)=\frac{4}{5}.\frac{3}{3}=\frac{4}{5}.1=\frac{4}{5}\)

\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}+\frac{1}{6}\right)=\frac{3}{4}:\frac{2}{3}=\frac{9}{8}\)

\(\frac{2}{3}.\frac{4}{5}-\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}-\frac{1}{3}\right)=\frac{4}{5}.\frac{1}{3}=\frac{4}{15}\)

\(\frac{1}{2}:\frac{3}{4}-\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{3}{4}:\frac{1}{3}=\frac{9}{4}\)

19 tháng 7 2016

\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\left(\frac{2}{3}+\frac{1}{3}\right).\frac{4}{5}=1.\frac{4}{5}=\frac{4}{5}\)

\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{1}{2}.\frac{4}{3}+\frac{1}{6}.\frac{4}{3}=\left(\frac{1}{2}+\frac{1}{6}\right).\frac{4}{3}=\frac{2}{3}.\frac{4}{3}=\frac{8}{9}\)

c,d tương tự 

25 tháng 7 2015

(15/4.4-12/5.5/4)-(7/2:5/2-1/5)-1/5.x=51/5

(15-3)-(7/5-1/5)-1/5.x=51/5

12-6/5-1/5.x=51/5

54/5-1/5.x=51/5

        1/5x=54/5-51/5

         1/5x=3/5

             x=3

7 tháng 11 2016

a)

\(x.\frac{7}{9}=\frac{2}{3}+2\frac{1}{2}\)

\(x.\frac{7}{9}=\frac{19}{6}\)

\(x=\frac{19}{6}:\frac{7}{9}\)

\(x=\frac{57}{14}\)

b) \(\frac{5}{7}+x:\frac{9}{4}=\frac{4}{3}\)

\(x:\frac{9}{4}=\frac{4}{3}-\frac{5}{7}\)

\(x:\frac{9}{4}=\frac{13}{21}\)

\(x=\frac{13}{21}.\frac{9}{4}\)

\(x=\frac{39}{28}\)

7 tháng 11 2016

A,27/14

CÁCH LÀM NHƯ SAU : 

(7/28 + 1/28) + 1/70 + 1/130 + 1/x.(x+3)

8/28 + 1/70 +1/130 +1/x.(x+3)

2/7+1/70+1/130+1/x.(x+3)

(20/70 +1/70)+1/130+1/x.(x+3)

3/10+1/130+1/x.(x+3)

39/130+1/130+1/x.(x+3)

4/13+1/x.(x+3)

Đến đây bn tự làm hộ mình vớ. chúc hok tốt k cho mình nhé

12 tháng 4 2021

\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{x\left(x+3\right)}\)

\(=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{x\left(x+3\right)}\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{x}-\frac{1}{x+3}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{13}+\frac{1}{x}-\frac{1}{x+3}\right)\)

\(=\frac{1}{3}\left(\frac{12}{13}+\frac{1}{x}-\frac{1}{x+3}\right)\)

\(=\frac{1}{3}.\frac{12}{13}+\frac{1}{3}.\frac{1}{x}-\frac{1}{3}.\frac{1}{x+3}\)

\(=\frac{4}{13}+\frac{1}{3x}-\frac{1}{3x+3}\)

\(=\frac{4}{13}+\frac{1}{3x}-\frac{1}{3x+3}\)

\(=\frac{4}{13}+\frac{1}{3x}=\frac{1}{3x+3}\)

\(=\frac{4}{13}+\frac{1}{3x}=\frac{1}{3x+3}\)

\(=\frac{4}{13}+\frac{1}{3x}=\frac{1}{3}.\frac{1}{x+3}\)

\(=\frac{4}{13}=\frac{1}{3}.\frac{1}{x+3}-\frac{1}{3x}\)

\(=\frac{4}{13}=\frac{1}{3}.\frac{1}{x+3}-\frac{1}{3}.\frac{1}{x}\)

\(=\frac{4}{13}=\frac{1}{3}\left(\frac{1}{x+3}-\frac{1}{x}\right)\)

\(=\frac{4}{13}:\frac{1}{3}=\frac{1}{x+1}-\frac{1}{x}\)

\(=\frac{12}{13}=\frac{1}{x+1}-\frac{1}{x}\)

\(=\frac{12}{13}=\frac{x-\left(x+1\right)}{\left(x+1\right)x}\)

\(=\frac{12}{13}=-\frac{1}{x^2+x}\)

\(\Leftrightarrow=12\left(x^2+x\right)=13.\left(-1\right)\)

\(=12\left(x^2+x\right)=-13\)

\(=x^2+x=-\frac{13}{12}\)

\(=x\left(x+1\right)=-\frac{13}{12}\)

.... Chiụ