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23 tháng 9 2021

\(=28\left(\dfrac{3}{14}-\dfrac{13}{21}-\dfrac{29}{42}\right)-8\\ =28\cdot\dfrac{-23}{21}-8=-\dfrac{92}{3}-8=-\dfrac{116}{3}\)

\(C=\dfrac{3}{14}:\dfrac{1}{28}-\dfrac{13}{21}:\dfrac{1}{28}-\dfrac{29}{42}:\dfrac{1}{28}-8\)

\(=28\left(\dfrac{3}{14}-\dfrac{13}{21}-\dfrac{29}{42}\right)-8\)

\(=28\left(\dfrac{9}{42}-\dfrac{26}{42}-\dfrac{29}{42}\right)-8\)

\(=28\cdot\dfrac{-46}{42}-8\)

\(=\dfrac{-116}{3}\)

c: \(\dfrac{3}{14}:\dfrac{1}{28}-\dfrac{13}{21}:\dfrac{1}{28}\)

\(=28\left(\dfrac{3}{14}-\dfrac{13}{21}\right)\)

\(=28\cdot\left(\dfrac{9}{42}-\dfrac{26}{42}\right)\)

\(=28\cdot\dfrac{-17}{42}=\dfrac{-34}{3}\)

\(\Leftrightarrow D=1-\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-\dfrac{1}{28}\)

\(\Rightarrow\dfrac{1}{2}D=\dfrac{1}{2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-\dfrac{1}{5.6}-\dfrac{1}{6.7}-\dfrac{1}{7.8}\)

\(\Rightarrow D\dfrac{1}{2}=\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{7}+\dfrac{1}{8}\)

\(\Rightarrow D=\dfrac{1}{8}.2=\dfrac{1}{4}\)

Vậy D=1/4

 

11 tháng 6 2021

\(b,1\dfrac{5}{9}x=\dfrac{28}{29}< =>\dfrac{14}{9}x=\dfrac{28}{29}=>x=\dfrac{\dfrac{28}{29}}{\dfrac{14}{9}}=\dfrac{18}{29}\)

\(c,x:\left(-\dfrac{2}{5}\right)=\dfrac{-15}{16}< =>x=\dfrac{-15}{16}.\dfrac{-2}{5}=\dfrac{3}{8}\)

11 tháng 6 2021

b, \(\dfrac{14}{9}.x=\dfrac{28}{29}\)

\(x=\dfrac{28}{29}:\dfrac{14}{9}\)

\(\Rightarrow x=\dfrac{18}{29}\)

c, \(x=\dfrac{-15}{16}.\left(\dfrac{-2}{5}\right)\)

\(\Rightarrow x=\dfrac{3}{8}\)

7 tháng 10 2017

a. \(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)

\(\Rightarrow\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{-113}{364}\right)=\dfrac{113}{364}\)

\(\Rightarrow\left(\dfrac{5}{42}-x\right)=\dfrac{11}{13}-\dfrac{113}{364}\)

\(\Rightarrow\left(\dfrac{5}{42}-x\right)=\dfrac{15}{28}\)

\(\Rightarrow x=\dfrac{5}{42}-\dfrac{15}{28}=\dfrac{-5}{12}\)

Vậy..............

b. \(2x.\left(x-\dfrac{1}{7}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{7}\end{matrix}\right.\)

Vậy............

c. \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Rightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\Rightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(\Rightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}=\dfrac{-5}{7}\)

Vậy...........

18 tháng 4 2020

\(A=1-\frac{1}{10}-\frac{1}{15}-\frac{1}{3}-\frac{1}{28}-\frac{1}{6}-\frac{1}{21}\)

\(=1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-\frac{1}{28}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{2.3}-\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}\)

\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}\)\(=\frac{1}{8}\)

\(\Rightarrow A=\frac{1}{8}.2=\frac{1}{4}\)

Vậy tổng của biểu thức cần tính là \(\frac{1}{4}\)

10 tháng 10 2018

A= \(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{9}{11}=\dfrac{1}{3}-\dfrac{7}{9}=\dfrac{3}{9}-\dfrac{7}{9}=-\dfrac{4}{9}\)

12 tháng 10 2022

\(B=\left(\dfrac{1}{5}+\dfrac{2}{15}+\dfrac{2}{3}\right)+\left(-\dfrac{2}{7}+\dfrac{1}{42}-\dfrac{13}{28}-\dfrac{1}{4}\right)\)

\(=\dfrac{3+2+10}{15}+\dfrac{-2\cdot12+2-13\cdot3-21}{84}\)

=1-82/84

=2/84=1/42

\(C=\dfrac{1}{50}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\right)\)

\(=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

\(=\dfrac{1}{50}-1+\dfrac{1}{50}=\dfrac{1}{25}-1=-\dfrac{24}{25}\)

\(D=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)

a: =>5/42-x=11/13-15/28+11/13=421/364

=>x=-1193/1092

b: =>\(\dfrac{7}{2}-2x=7+\dfrac{6}{5}-3-\dfrac{2}{5}-1-\dfrac{4}{5}=3\)

=>2x=1/2

=>x=1/4

c: =>|2x-1/3|=-1/3(vô lý)

d: =>2x-1=-3

=>2x=-2

hay x=-1

e: =>2x=16

hay x=8

\(\Leftrightarrow\dfrac{1}{x-4}-\dfrac{1}{x-7}+\dfrac{1}{x-7}-\dfrac{1}{x-13}+\dfrac{1}{x-13}-\dfrac{1}{x-28}-\dfrac{1}{x-28}=\dfrac{-5}{2}\)

\(\Leftrightarrow\dfrac{1}{x-4}-\dfrac{2}{x-28}=-\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{x-28-2x+8}{\left(x-4\right)\left(x-28\right)}=\dfrac{-5}{2}\)

\(\Leftrightarrow-5\left(x^2-32x+112\right)=2\left(-x-20\right)\)

\(\Leftrightarrow-5x^2+160x-560=-2x-40\)

\(\Leftrightarrow-5x^2+162x-520=0\)

\(\text{Δ}=162^2-4\cdot\left(-5\right)\cdot\left(-520\right)=15844\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{162-2\sqrt{3961}}{10}\\x_2=\dfrac{162+2\sqrt{3961}}{10}\end{matrix}\right.\)

24 tháng 12 2022

\(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)
\(=\dfrac{1}{3}+\left(\dfrac{12}{67}-\dfrac{79}{67}\right)+\left(\dfrac{13}{41}+\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\left(-1\right)+1=\dfrac{1}{3}+0=\dfrac{1}{3}\)
\(\left(\dfrac{15}{4}-5x\right).\left(9x^2-4\right)=0\)
\(\left[{}\begin{matrix}\dfrac{15}{4}-5x=0\\9x^2-4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}5x=\dfrac{15}{4}\\9x^2=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{2}{3}\end{matrix}\right.\)