Tìm x:
a)\(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
b)\(\left|x+\dfrac{4}{15}\right|-\left|-3.75\right|=-\left|2.15\right|\)
BT3: Tìm x, biết
13) \(\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{-7}{4}+\dfrac{1}{4}:\dfrac{1}{8}\)
14) \(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
1) \(\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{-7}{4}+\dfrac{1}{4}:\dfrac{1}{8}\)
\(\Leftrightarrow\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{-7}{4}+2\)
\(\Leftrightarrow\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{1}{4}\)
\(\Leftrightarrow-1\dfrac{1}{5}+x=\dfrac{1}{4}.\left(-3\dfrac{3}{5}\right)\)
\(\Leftrightarrow-1\dfrac{1}{5}+x=\dfrac{1}{4}.\left(-\dfrac{18}{5}\right)\)
\(\Leftrightarrow-1\dfrac{1}{5}+x=-\dfrac{9}{10}\)
\(\Leftrightarrow x=\left(-\dfrac{9}{10}\right)-\left(-1\dfrac{1}{5}\right)\)
\(\Leftrightarrow x=\dfrac{3}{10}\)
Tìm x, biết:
a) \(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
b) \(2x.\left(x-\dfrac{1}{7}\right)=0\)
c) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
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...........
1: Tính hợp lí:
a, \(\dfrac{11}{125}\)- \(\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}\)+ \(\dfrac{17}{14}\)
b,\(-1\dfrac{5}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
2: Tìm x:
a, \(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
b,\(|x+\dfrac{4}{15}|-|-3,75|=-|-2,15|\)
c, \(^{7^{x+2}}+2.7^{x-1}=345\)
\(1,\)
\(a,\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}+\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)\)
\(=\dfrac{11}{125}+\left(\dfrac{-1}{2}\right)+\dfrac{1}{2}\)
\(=\dfrac{11}{125}\)
\(b,-1\dfrac{5}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
\(=\dfrac{-12}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
\(=-15.\left[\dfrac{12}{7}+\dfrac{2}{7}+\left(-5\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\right]\)
\(=-15.\left[2+\left(-5\right).\dfrac{1}{105}\right]\)
\(=-15.\left(2-\dfrac{1}{21}\right)\)
\(=-15.\dfrac{41}{21}=\dfrac{-615}{21}\)
\(2,\)
\(a,\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
\(\Leftrightarrow\dfrac{11}{13}-\dfrac{5}{42}+x=\dfrac{-15}{28}+\dfrac{11}{13}\)
\(\Leftrightarrow x=\dfrac{-15}{28}+\dfrac{11}{13}-\dfrac{11}{13}+\dfrac{5}{42}\)
\(\Leftrightarrow x=\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\left(\dfrac{5}{42}+\dfrac{-15}{28}\right)\)
\(\Leftrightarrow x=\dfrac{5}{12}\)
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 \(x=1\)
\(\dfrac{3}{\left(x-4\right).\left(x-7\right)}+\dfrac{6}{\left(x-7\right).\left(x-13\right)}+\dfrac{15}{\left(x-13\right).\left(x-28\right)}-\dfrac{1}{x-28}=\dfrac{-5}{2}\)
\(\Leftrightarrow\dfrac{1}{x-4}-\dfrac{1}{x-7}+\dfrac{1}{x-7}-\dfrac{1}{x-13}+\dfrac{1}{x-13}-\dfrac{1}{x-28}-\dfrac{1}{x-28}=\dfrac{-5}{2}\)
\(\Leftrightarrow\dfrac{1}{x-4}-\dfrac{2}{x-28}=-\dfrac{5}{2}\)
\(\Leftrightarrow\dfrac{x-28-2x+8}{\left(x-4\right)\left(x-28\right)}=\dfrac{-5}{2}\)
\(\Leftrightarrow-5\left(x^2-32x+112\right)=2\left(-x-20\right)\)
\(\Leftrightarrow-5x^2+160x-560=-2x-40\)
\(\Leftrightarrow-5x^2+162x-520=0\)
\(\text{Δ}=162^2-4\cdot\left(-5\right)\cdot\left(-520\right)=15844\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{162-2\sqrt{3961}}{10}\\x_2=\dfrac{162+2\sqrt{3961}}{10}\end{matrix}\right.\)
tính
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
b)\(\dfrac{3}{14}:\dfrac{1}{28}-\dfrac{13}{21}:\dfrac{1}{28}+\dfrac{29}{42}:\dfrac{1}{28}-8\)
c)\(-1\dfrac{5}{7}.15+\dfrac{2}{7}\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)
=\(\dfrac{10}{11}.\dfrac{-1}{2}\)
=\(\dfrac{-5}{11}\)
b;
B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8
B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8
B = \(\dfrac{2}{7}\) - 8
B = \(\dfrac{2}{7}-\dfrac{56}{7}\)
B = - \(\dfrac{54}{7}\)
c; C = -1\(\dfrac{5}{7}\).15 + \(\dfrac{2}{7}\)(-15) + (-105).(\(\dfrac{2}{3}\) - \(\dfrac{4}{5}\) + \(\dfrac{1}{7}\))
C = - 15.(- 1 - \(\dfrac{5}{7}\) + \(\dfrac{2}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\) + \(1\))
C = -15.[(1 - 1) - (\(\dfrac{5}{7}\) - \(\dfrac{2}{7}\)) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]
C = -15.[0 - \(\dfrac{3}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]
C = -15 . [- \(\dfrac{45}{105}\) + \(\dfrac{490}{105}\) - \(\dfrac{588}{105}\)]
C = -15. [ \(\dfrac{445}{105}\) - \(\dfrac{588}{105}\)]
C = - 15.(- \(\dfrac{143}{105}\))
C = \(\dfrac{143}{7}\)
a) \(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
b) \(\left(\dfrac{7}{2}-2x\right)x3\dfrac{2}{5}+1\dfrac{4}{5}=7\dfrac{6}{5}\)
c) \(\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{3}-2\)
d) (2x-1)3= -27
e) \(\dfrac{16}{2x}=1\)
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
\(\dfrac{11}{8}.\left[\left(-\dfrac{5}{11}:\dfrac{13}{8}-\dfrac{5}{11}:\dfrac{13}{15}\right)+\dfrac{-6}{33}\right]+\dfrac{-3}{4}\)
\(\dfrac{11}{8}\cdot\left[\left(-\dfrac{5}{11}:\dfrac{13}{8}-\dfrac{5}{11}:\dfrac{13}{15}\right)+\dfrac{-6}{33}\right]+\dfrac{-3}{4}\)
\(=\dfrac{11}{8}\cdot\left[\left(-\dfrac{5}{11}\cdot\dfrac{8}{13}-\dfrac{5}{11}\cdot\dfrac{15}{13}\right)-\dfrac{2}{11}\right]-\dfrac{3}{4}\)
\(=\dfrac{11}{8}\cdot\left[-\dfrac{5}{11}\cdot\left(\dfrac{8}{13}+\dfrac{15}{13}\right)-\dfrac{2}{11}\right]-\dfrac{3}{4}\)
\(=\dfrac{11}{8}\cdot\left(-\dfrac{5}{11}\cdot\dfrac{23}{13}-\dfrac{2}{11}\right)-\dfrac{3}{4}\)
\(=\dfrac{11}{8}\cdot\left(-\dfrac{115}{143}-\dfrac{2}{11}\right)-\dfrac{3}{4}\)
\(=\dfrac{11}{8}\cdot\dfrac{-141}{143}-\dfrac{3}{4}\)
\(=-\dfrac{141}{104}-\dfrac{3}{4}\)
\(=-\dfrac{219}{104}\)
tìm x:
a) \(-8:\left(4\dfrac{1}{5}x+\dfrac{3}{10}\right)=4\dfrac{4}{9}\)
b) \(4\dfrac{2}{3}-\left(\dfrac{3}{5}:x\right)=-20\%\)
c) \(\left(\dfrac{13}{10}x-15\right).-\dfrac{5}{14}=3\)
d) \(\left(5,5x-44\right):\left(-\dfrac{3}{2}\right)=30\)
e) \(\dfrac{11}{14}:\left(2x+\dfrac{5}{7}\right)=-\dfrac{11}{18}\)
\(a.-8:\left(4\dfrac{1}{5}x+\dfrac{3}{10}\right)=4\dfrac{4}{9}\)
\(4\dfrac{1}{5}x+\dfrac{3}{10}=\left(-8\right):4\dfrac{4}{9}\)
\(4\dfrac{1}{5}x+\dfrac{3}{10}=\dfrac{-9}{5}\)
\(4\dfrac{1}{5}x=\dfrac{-9}{5}-\dfrac{3}{10}\)
\(4\dfrac{1}{5}x=\dfrac{-21}{10}\)
\(x=\dfrac{-21}{10}:\dfrac{21}{5}\)
\(x=\dfrac{-1}{2}\)
Vay \(x=\dfrac{-1}{2}\).
\(b.4\dfrac{2}{3}-\left(\dfrac{3}{5}:x\right)=-20\%\)
\(\dfrac{14}{3}-\left(\dfrac{3}{5}:x\right)=\dfrac{-1}{5}\)
\(\dfrac{3}{5}:x=\dfrac{14}{3}-\dfrac{-1}{5}\)
\(\dfrac{3}{5}:x=\dfrac{73}{15}\)
\(x=\dfrac{3}{5}:\dfrac{73}{15}\)
\(x=\dfrac{9}{73}\)
Vay \(x=\dfrac{9}{73}\).
Câu c; d; e tương tự nhé.
Tính hợp lý :
a,\(\left(\dfrac{136}{15}-\dfrac{28}{5}+\dfrac{62}{10}\right).\dfrac{21}{24}\)
b,\(\dfrac{3}{11}.\dfrac{4}{11}+\left(\dfrac{3}{13}.\dfrac{4}{11}-\dfrac{1}{13}\right)\)
a: \(=\dfrac{2\cdot136-28\cdot6+62\cdot3}{30}\cdot\dfrac{7}{8}=\dfrac{290}{30}\cdot\dfrac{7}{8}=\dfrac{29}{3}\cdot\dfrac{7}{8}=\dfrac{203}{24}\)
b: \(=\dfrac{3}{11}\cdot\dfrac{4}{11}+\dfrac{3}{13}\cdot\dfrac{4}{11}-\dfrac{1}{13}\)
\(=\dfrac{4}{11}\left(\dfrac{3}{11}+\dfrac{3}{13}\right)-\dfrac{1}{13}\)
\(=\dfrac{4}{11}\cdot\dfrac{72}{143}-\dfrac{1}{13}=\dfrac{167}{1573}\)