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24 tháng 2 2020

mong ai do giai cho minh nhe

\(\left(a\right)\frac{34-x}{30}=\frac{5}{6}\)

\(\frac{34-x}{30}=\frac{25}{30}\)

34 - x = 25

x = 34 - 25 = 9

\(\left(b\right)\frac{x+13}{34}=\frac{12}{17}\)

\(\frac{x+13}{34}=\frac{24}{34}\)

x + 13 = 24

x = 24 - 13 = 11

\(\left(c\right)\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{9}\right)+\left(x+\frac{1}{27}\right)+\left(x+\frac{1}{81}\right)=\frac{56}{81}\)

\(4x+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}=\frac{56}{81}\)

Đặt \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)

Ta có : \(3A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\)

\(3A-A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}-\frac{1}{3}-\frac{1}{9}-\frac{1}{27}-\frac{1}{81}\)

\(2A=1-\frac{1}{81}=\frac{80}{81}\)

\(A=\frac{80}{81}\div2=\frac{40}{81}\)

\(\Rightarrow4x+\frac{40}{81}=\frac{56}{81}\)

\(4x=\frac{56}{81}-\frac{40}{81}\)

\(4x=\frac{16}{81}\)

\(x=\frac{16}{81}\div4=\frac{4}{81}\)

9 tháng 8 2020

a, \(\frac{34-x}{30}=\frac{5}{6}\Leftrightarrow\frac{34-x}{30}=\frac{25}{30}\)

\(\Leftrightarrow34-x=25\Leftrightarrow x=9\)

b, \(\frac{x+13}{34}=\frac{12}{17}\Leftrightarrow\frac{x+13}{34}=\frac{24}{34}\)

\(\Leftrightarrow x+13=24\Leftrightarrow x=11\)

24 tháng 2 2015

ta có:$\frac{x-1}{12}+\frac{x-1}{20}+\frac{x-1}{30}+\frac{x-1}{42}+\frac{x-1}{56}+\frac{x-1}{72}=\frac{16}{9}$

 => x+1(1/12+1/20+1/30+1/42+1/56+1/72)=16/9

=> x+1.2/9=16/9

=> x+1 = (16/9):(2/9)

=> x+1 = 8

=> x = 9

 

5 tháng 4 2017

thông cảm mình ko đánh được dấu ngoặc tròn

[x-1].[1/12+1/20+1/30+1/42+1/56+1/72] =16/9

[x-1].[1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9]=16/9

[x-1].[1/3-1/9]=16/9

[x-1].2/9=16/9

x-1=16/9:2/9

x-1=8 

x=7 

Vậy x=7

30 tháng 5 2017

Ta có : \(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+...+\) \(\frac{1}{x^2+15x+56}=\frac{1}{14}\)

<=>\(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\)+...+ \(\frac{1}{\left(x+7\right)\left(x+8\right)}=\frac{1}{14}\)

<=> \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+7}-\frac{1}{x+8}\)\(\frac{1}{14}\)

<=> \(\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\)

<=> \(\frac{x+8-x-1}{\left(x+1\right)\left(x+8\right)}=\frac{1}{14}\)

<=>\(\frac{7.14}{14\left(x+1\right)\left(x+8\right)}=\frac{\left(x+1\right)\left(x+8\right)}{14\left(x+1\right)\left(x+8\right)}\)

<=> \(x^2+9x+8=98\)<=> \(x^2+9x-90=0\)

<=> (x-6)(x+15) =0 

<=> \(\orbr{\begin{cases}x=6\\x=-15\end{cases}}\)

Vậy phương trình có 2 nghiệm  x  \(\in\left(6,15\right)\)

==============

- Do ko biết viết dấu ngoặc nhọn nên thay = dấu ngoặc tròn

- Đề ko rõ ràng , lần sau nhớ ghi yêu cầu ?  

11 tháng 9 2017

\(\frac{1}{3}-\left(\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{7.8}\right)=x-\frac{5}{18}\)

\(x-\frac{5}{18}=\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\right)\)

\(x-\frac{5}{18}=\frac{1}{3}-\frac{1}{3}+\frac{1}{8}\)

\(x-\frac{5}{18}=0+\frac{1}{8}\)

\(x-\frac{5}{18}=\frac{1}{8}\)

\(x=\frac{1}{8}+\frac{5}{18}\)

\(x=\frac{9}{72}+\frac{20}{72}\)

\(x=\frac{29}{72}\)

11 tháng 9 2017

1/3 - 1/12 - 1/20 - 1/30 - 1/42 - 1/56 = x - 5/18

            1/4 - 1/20 - 1/30 - 1/42 - 1/56 = x - 5/18

                        1/5 - 1/30 - 1/42 - 1/56 = x - 5/18

                                    1/6 - 1/42 - 1/56 = x - 5/18

                                                1/7 - 1/56 = x - 5/18

                                                            1/8 = x - 5/18

                                                                x=1/8+5/18

                                                                x= 29/72

                                                           Vậy : x = 29/72

28 tháng 3 2020

ĐKXĐ : Tự tìm nha : )

Ta có : \(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+...+\frac{1}{x^2+15x+56}=\frac{1}{14}\)

=> \(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+7\right)\left(x+8\right)}=\frac{1}{14}\)

=> \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+7}-\frac{1}{x+8}=\frac{1}{14}\)

=> \(\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\)

=> \(\frac{x+8}{\left(x+1\right)\left(x+8\right)}-\frac{x+1}{\left(x+8\right)\left(x+1\right)}=\frac{1}{14}\)

=> \(14\left(x+8-x-1\right)=\left(x+1\right)\left(x+8\right)\)

=> \(x^2+x+8x+8=98\)

=> \(x^2+9x-90=0\)

=> \(\left(x+15\right)\left(x-6\right)=0\)

=> \(\left[{}\begin{matrix}x+15=0\\x-6=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-15\\x=6\end{matrix}\right.\) ( TM )

Vậy phương trình trên có nghiệm là \(S=\left\{6,-15\right\}\)

14 tháng 8 2016

\(S=\frac{\left(9\frac{3}{8}:5,2+3,4.2\frac{7}{34}\right):1\frac{9}{16}}{0,31.8\frac{2}{2}-5,61:27\frac{1}{3}}\)\(\Rightarrow S=\frac{\left(\frac{75}{8}.\frac{5}{26}+\frac{17}{5}.\frac{75}{34}\right):\frac{25}{16}}{\frac{31}{100}.9-\frac{561}{100}.\frac{3}{82}}\)\(\Rightarrow S=\frac{\left(\frac{75.5}{8.26}-\frac{17.75}{5.34}\right).\frac{16}{25}}{\frac{31.9}{100}-\frac{561.3}{100.82}}\)

\(\Rightarrow S=\frac{\left(\frac{375}{208}-\frac{15}{2}\right).\frac{16}{25}}{\frac{279}{100}-\frac{1682}{8200}}\)\(\Rightarrow S=\frac{\frac{-1185}{208}.\frac{16}{25}}{\frac{21196}{8200}}\)\(\Rightarrow S=\frac{-237}{65}:\frac{21196}{8200}\)\(\Rightarrow S=\frac{-194340}{137774}\)

\(\Rightarrow x=\frac{2}{3}S\Rightarrow x=\frac{2}{3}.\frac{-194340}{137774}\Rightarrow x=\frac{-388680}{413322}\)

14 tháng 8 2016

\(M=\frac{23\frac{11}{15}-26\frac{13}{20}}{12^2+5^2}:\frac{1-\frac{1}{3}-\frac{1}{42}-\frac{1}{56}}{3^2.13.2}-\frac{19}{37}\)\(\Rightarrow M=\frac{\frac{356}{15}-\frac{533}{20}}{12^2+5^2}:\frac{\frac{5}{8}}{3^2.13.2}-\frac{19}{37}\)

\(\Rightarrow M=\frac{\frac{-35}{12}}{12^2+5^2}.\frac{3^2.13.2}{\frac{5}{8}}-\frac{19}{37}\)\(\Rightarrow M=\frac{-84}{13}-\frac{19}{37}\Rightarrow M=\frac{-3355}{481}\Rightarrow15\%M=\frac{-3355}{481}.15\%\Rightarrow15\%M=\frac{-2013}{1924}\)

AH
Akai Haruma
Giáo viên
20 tháng 3 2020

Lời giải:

PT \(\Leftrightarrow \frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+....+\frac{1}{(x+7)(x+8)}=\frac{1}{14}\)

(ĐK: $x\neq -1;-2;...;-8$)

\(\Leftrightarrow \frac{(x+2)-(x+1)}{(x+1)(x+2)}+\frac{(x+3)-(x+2)}{(x+2)(x+3)}+....+\frac{(x+8)-(x+7)}{(x+7)(x+8)}=\frac{1}{14}\)

\(\Leftrightarrow \frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+....+\frac{1}{x+7}-\frac{1}{x+8}=\frac{1}{14}\)

\(\Leftrightarrow \frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\Leftrightarrow \frac{7}{x^2+9x+8}=\frac{1}{14}\)

\(\Rightarrow x^2+9x+8=98\Leftrightarrow x^2+9x-90=0\Rightarrow x=6\) hoặc $x=-15$ (đều thỏa mãn)

Vậy........

26 tháng 4 2018

\(\frac{20}{9}-x=\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}=\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=\frac{1}{3}-\frac{1}{9}=\frac{2}{9}\)

\(\Rightarrow\frac{20}{9}-x=\frac{2}{9}\Rightarrow x=\frac{20}{9}-\frac{2}{9}=\frac{18}{9}=2\)

Vậy x = 2.

k cho mk nha

\(2\frac{2}{9}-x=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)

\(2\frac{2}{9}-x=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)

\(2\frac{2}{9}-x=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)

\(2\frac{2}{9}-x=\frac{1}{3}-\frac{1}{9}\)

\(2\frac{2}{9}-x=\frac{3}{9}-\frac{1}{9}\)

\(2\frac{2}{9}-x=\frac{2}{9}\)

\(x=2\frac{2}{9}-\frac{2}{9}\)

\(x=2\)

Vậy x = 2