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\(=28\left(\dfrac{3}{14}-\dfrac{13}{21}-\dfrac{29}{42}\right)-8\\ =28\cdot\dfrac{-23}{21}-8=-\dfrac{92}{3}-8=-\dfrac{116}{3}\)
\(C=\dfrac{3}{14}:\dfrac{1}{28}-\dfrac{13}{21}:\dfrac{1}{28}-\dfrac{29}{42}:\dfrac{1}{28}-8\)
\(=28\left(\dfrac{3}{14}-\dfrac{13}{21}-\dfrac{29}{42}\right)-8\)
\(=28\left(\dfrac{9}{42}-\dfrac{26}{42}-\dfrac{29}{42}\right)-8\)
\(=28\cdot\dfrac{-46}{42}-8\)
\(=\dfrac{-116}{3}\)
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...........
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
A= \(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{9}{11}=\dfrac{1}{3}-\dfrac{7}{9}=\dfrac{3}{9}-\dfrac{7}{9}=-\dfrac{4}{9}\)
\(B=\left(\dfrac{1}{5}+\dfrac{2}{15}+\dfrac{2}{3}\right)+\left(-\dfrac{2}{7}+\dfrac{1}{42}-\dfrac{13}{28}-\dfrac{1}{4}\right)\)
\(=\dfrac{3+2+10}{15}+\dfrac{-2\cdot12+2-13\cdot3-21}{84}\)
=1-82/84
=2/84=1/42
\(C=\dfrac{1}{50}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\right)\)
\(=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(=\dfrac{1}{50}-1+\dfrac{1}{50}=\dfrac{1}{25}-1=-\dfrac{24}{25}\)
\(D=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)
\(A=1-\frac{1}{10}-\frac{1}{15}-\frac{1}{3}-\frac{1}{28}-\frac{1}{6}-\frac{1}{21}\)
\(=1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-\frac{1}{28}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{2.3}-\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}\)\(=\frac{1}{8}\)
\(\Rightarrow A=\frac{1}{8}.2=\frac{1}{4}\)
Vậy tổng của biểu thức cần tính là \(\frac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{x-4}-\dfrac{1}{x-7}+\dfrac{1}{x-7}-\dfrac{1}{x-13}+\dfrac{1}{x-13}-\dfrac{1}{x-28}-\dfrac{1}{x-28}=\dfrac{-5}{2}\)
\(\Leftrightarrow\dfrac{1}{x-4}-\dfrac{2}{x-28}=-\dfrac{5}{2}\)
\(\Leftrightarrow\dfrac{x-28-2x+8}{\left(x-4\right)\left(x-28\right)}=\dfrac{-5}{2}\)
\(\Leftrightarrow-5\left(x^2-32x+112\right)=2\left(-x-20\right)\)
\(\Leftrightarrow-5x^2+160x-560=-2x-40\)
\(\Leftrightarrow-5x^2+162x-520=0\)
\(\text{Δ}=162^2-4\cdot\left(-5\right)\cdot\left(-520\right)=15844\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{162-2\sqrt{3961}}{10}\\x_2=\dfrac{162+2\sqrt{3961}}{10}\end{matrix}\right.\)
\(1,\)
\(a,\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}+\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)\)
\(=\dfrac{11}{125}+\left(\dfrac{-1}{2}\right)+\dfrac{1}{2}\)
\(=\dfrac{11}{125}\)
\(b,-1\dfrac{5}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
\(=\dfrac{-12}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
\(=-15.\left[\dfrac{12}{7}+\dfrac{2}{7}+\left(-5\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\right]\)
\(=-15.\left[2+\left(-5\right).\dfrac{1}{105}\right]\)
\(=-15.\left(2-\dfrac{1}{21}\right)\)
\(=-15.\dfrac{41}{21}=\dfrac{-615}{21}\)
\(2,\)
\(a,\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
\(\Leftrightarrow\dfrac{11}{13}-\dfrac{5}{42}+x=\dfrac{-15}{28}+\dfrac{11}{13}\)
\(\Leftrightarrow x=\dfrac{-15}{28}+\dfrac{11}{13}-\dfrac{11}{13}+\dfrac{5}{42}\)
\(\Leftrightarrow x=\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\left(\dfrac{5}{42}+\dfrac{-15}{28}\right)\)
\(\Leftrightarrow x=\dfrac{5}{12}\)
Vậy \(x=\dfrac{5}{12}\)
\(b,\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|-3,75=-2,15\)
\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2,15+3,75=1,6=\dfrac{16}{10}=\dfrac{8}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{8}{5}\\x+\dfrac{4}{15}=\dfrac{-8}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{5}-\dfrac{4}{15}=\dfrac{4}{3}\\x=\dfrac{-8}{5}-\dfrac{4}{15}=\dfrac{-28}{15}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{4}{3};\dfrac{-28}{15}\right\}\)
\(c,7^{x+2}+2.7^{x-1}=345\)
\(\Leftrightarrow7^{x-1}.\left(7^3+2\right)=345\)
\(\Leftrightarrow7^{x-1}.\left(343+2\right)=345\)
\(\Leftrightarrow7^{x-1}.345=345\)
\(\Leftrightarrow7^{x-1}=345:345=1\)
\(\Leftrightarrow x-1=0\)
\(x=0+1=1\)
Vậy \(x=1\)
Ta có : a, 25/7 + 13/21 - 11/7 + 17/21 + 1/3 .
= ( 25/7 - 11/7 ) + ( 13/21 + 17/21 + 1/3 ) .
= 2 + ( 20/21 + 7/21 ) .
= 2 + 9/7 .
= 23/7 .
b, ( 1/3 + 12/67 + 13/41 ) - ( 79/67 - 28/41 ) .
= 1/3 + 12/67 + 13/41 - 79/67 + 28/41 .
= 1/3 + ( 12/67 - 79/67 ) + ( 13/41 + 28/41 ) .
= 1/3 - 1 + 1 .
= 1/3 .
c, ( 11/4 . -5/9 - 4/9 . 11/4 ) . 8/33 .
= [ 11/4 . ( -5/9 - 4/9 ) ] . 8/33 .
= [ 11/4 . ( - 1 ) ] . 8/33 .
= -11/4 . 8/33 .
= -2/3 .
d, 38/45 - ( 8/45 - 17/51 - 3/11 ) .
= 38/45 - 8/45 + 17/51 + 3/11 .
= 2/3 + 17/51 + 3/11 .
= 374/561 + 187/561 + 153/561 .
= 14/11 .
c: \(\dfrac{3}{14}:\dfrac{1}{28}-\dfrac{13}{21}:\dfrac{1}{28}\)
\(=28\left(\dfrac{3}{14}-\dfrac{13}{21}\right)\)
\(=28\cdot\left(\dfrac{9}{42}-\dfrac{26}{42}\right)\)
\(=28\cdot\dfrac{-17}{42}=\dfrac{-34}{3}\)