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a ) \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{4}{35}-\frac{\left(-11\right)}{70}\right|\)
=> \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{4}{35}+\frac{11}{70}\right|\)
=> \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{19}{70}\right|\)
=> \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\frac{19}{70}=\frac{3}{35}\)
=> \(\frac{2}{5}+x+\frac{3}{2}=\frac{3}{7}-\frac{3}{35}=\frac{12}{35}\)
=> \(\frac{2}{5}+x=\frac{12}{35}-\frac{3}{2}=-\frac{81}{70}\)
=> \(x=-\frac{81}{70}-\frac{2}{5}=-\frac{109}{70}\)
b) \(\frac{3}{4}\left(x-8\right)=\frac{5}{7}\left(4-\frac{1}{2}\right)\)
=> \(\frac{3}{4}x-6=\frac{5}{2}\)
=> \(\frac{3}{4}x=\frac{17}{2}\)
=> \(x=\frac{17}{2}:\frac{3}{4}=\frac{34}{3}\)
Câu c,d tự làm nhé
a. \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{4}{35}-\frac{-11}{70}\right|\)
\(\Rightarrow\frac{3}{7}-\left(\frac{19}{10}+x\right)=\frac{5}{14}-\left|\frac{4}{35}+\frac{11}{70}\right|\)
\(\Rightarrow\frac{3}{7}-\frac{19}{10}-x=\frac{5}{14}-\left|\frac{19}{70}\right|=\frac{5}{14}-\frac{19}{70}\)
\(\Rightarrow-\frac{103}{70}-x=\frac{3}{35}\)
\(\Rightarrow x=-\frac{103}{70}-\frac{3}{35}\)
\(\Rightarrow x=-\frac{109}{70}\)
b. \(\frac{3}{4}\left(x-8\right)=\frac{5}{7}\left(4-\frac{1}{2}\right)\)
\(\Rightarrow\frac{3}{4}\left(x-8\right)=\frac{5}{7}.\frac{7}{2}=\frac{5}{2}\)
\(\Rightarrow x-8=\frac{10}{3}\)
\(\Rightarrow x=\frac{34}{3}\)
c. \(\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\)
\(\Rightarrow\frac{3}{2}-1+4x=\frac{2}{3}-7x\)
\(\Rightarrow\frac{1}{2}=\frac{2}{3}-7x-4x=\frac{2}{3}-11x\)
\(\Rightarrow11x=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{66}\)
d. \(4\left(\frac{1}{2}-x\right)-5\left(x-\frac{3}{10}\right)=\frac{7}{4}\)
\(\Rightarrow2-4x-5x+\frac{3}{2}=\frac{7}{4}\)
\(\Rightarrow2-9x=\frac{1}{4}\)
\(\Rightarrow9x=\frac{7}{4}\)
\(\Rightarrow x=\frac{7}{36}\)
\(x-\frac{3}{8}=\frac{1}{6}-\frac{1}{5}\)
=> \(x-\frac{3}{8}=\frac{5}{30}-\frac{6}{30}=-\frac{1}{30}\)
=> \(x=-\frac{1}{30}+\frac{3}{8}\)
=> \(x=\frac{41}{120}\)
\(-\frac{7}{10}\left(x+\frac{1}{3}\right)=\frac{4}{5}\)
=> \(-\frac{7}{10}x-\frac{7}{30}=\frac{4}{5}\)
=> \(-\frac{7}{10}x=\frac{4}{5}+\frac{7}{30}=\frac{31}{30}\)
=> \(x=\frac{31}{30}:\left(-\frac{7}{10}\right)=\frac{31}{30}\cdot\left(-\frac{10}{7}\right)=-\frac{31}{21}\)
\(x-\frac{4}{3}=\frac{5}{6}\Rightarrow x=\frac{5}{6}+\frac{4}{3}=\frac{5}{6}+\frac{8}{6}=\frac{13}{6}\)
Thiếu đề
\(\frac{6}{5}+\left(x-\frac{2}{3}\right)=\frac{4}{7}\)
=> \(\frac{6}{5}+x-\frac{2}{3}=\frac{4}{7}\)
=> \(\frac{6}{5}+x=\frac{4}{7}+\frac{2}{3}=\frac{26}{21}\)
=> \(x=\frac{26}{21}-\frac{6}{5}=\frac{4}{105}\)
5) // 3x + 1 / - 5 / = 2
=> / 3x + 1 / -5 = 2 hoặc -2
TH1: / 3x + 1 / - 5 = 2 => / 3x + 1 / = 2 + 5 = 7 => 3x + 1 = 7 hoặc -7
*1: 3x + 1 = -7 => 3x = -7 - 1 = -8 => x = -8/3
*2: 3x + 1 = 7 => 3x = 7 - 1 = 6 => x = 6 : 3 = 2
TH2: / 3x + 1 / - 5 = -2 => / 3x + 1/ = -2 + 5 = 3 => 3x + 1 = 3 hoặc -3
*1: 3x + 1 = 3 => 3x = 3 - 1 = 2 => x = 2/3
*2: 3x + 1 = -3 => 3x = -3 - 1 = -4 => x = -4/3
Vậy x thuộc { -8/3 ; 2 ; 2/3 ; -4/3 }
7) 3/2 + 4/5 /x - 3/4 / = 7/4
=> 4/5 / x - 3/4 / = 7/4 - 4/5
=> 4/5 / x - 3/4 / = 19/20
=> / x - 3/4 / = 19/20 : 4/5
=> / x - 3/4 / = 19/16
=> / x - 3/4 / = 19/16 hoặc -19/16
a: 2x-3/2+3/4=-4
=>2x-3/4=-4
=>2x=-13/4
hay x=-13/8
b: \(\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)\cdot\left(\dfrac{-3}{2}-\dfrac{10}{3}\right)=\dfrac{2}{5}\)
\(\Leftrightarrow-\dfrac{2}{3}x-\dfrac{3}{5}=\dfrac{2}{5}:\dfrac{-29}{6}=\dfrac{-2}{5}\cdot\dfrac{6}{29}=\dfrac{-12}{145}\)
=>2/3x+3/5=12/145
=>2/3x=-15/29
hay x=-45/58
c: \(\dfrac{x}{2}-\left(\dfrac{3}{5}x-\dfrac{13}{5}\right)=-\left(\dfrac{7}{10}x+\dfrac{7}{5}\right)\)
=>1/2x-3/5x+13/5=-7/10x-7/5
=>-1/10x+7/10x=-7/5-13/5
=>3/5x=-2
hay x=-2:3/5=-10/3
a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
\(\Rightarrow\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=-4x+1\end{cases}}\Rightarrow\orbr{\begin{cases}4x-\frac{3}{2}x-1=\frac{1}{2}\\-4x-\frac{3}{2}x+1=\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{2}x=\frac{3}{2}\\-\frac{11}{2}x=-\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
phần b ở đề bài mình ghi sai, là bằng 0 chứ ko phải bằng 10
Bài 1:
a; \(\dfrac{7}{8}\) + \(x\) = \(\dfrac{4}{7}\)
\(x\) = \(\dfrac{4}{7}\) - \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{32}{56}\) - \(\dfrac{49}{56}\)
\(x=-\) \(\dfrac{49}{56}\)
Vậy \(x=-\dfrac{49}{56}\)
b; 6 - \(x\) = - \(\dfrac{3}{4}\)
\(x\) = 6 + \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{24}{4}+\dfrac{3}{4}\)
\(x=\dfrac{27}{4}\)
Vậy \(x=\dfrac{27}{4}\)
c; \(\dfrac{1}{-5}\) + \(x\) = \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{3}{4}\) + \(\dfrac{1}{5}\)
\(x=\dfrac{15}{20}\) + \(\dfrac{4}{20}\)
\(x=\dfrac{19}{20}\)
Vậy \(x=\dfrac{19}{20}\)
Bài 1:
d; - 6 - \(x\) = - \(\dfrac{3}{5}\)
\(x\) = - 6 + \(\dfrac{3}{5}\)
\(x=-\dfrac{30}{5}\) + \(\dfrac{3}{5}\)
\(x=-\dfrac{27}{5}\)
Vậy \(x=-\dfrac{27}{5}\)
e; - \(\dfrac{2}{6}\) + \(x\) = \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{6}\)
\(x\) = \(\dfrac{15}{21}\) + \(\dfrac{1}{3}\)
\(x=\dfrac{15}{21}\) + \(\dfrac{7}{21}\)
\(x=\dfrac{22}{21}\)
Vậy \(x=\dfrac{22}{21}\)
f; - 8 - \(x\) = - \(\dfrac{5}{3}\)
\(x\) = \(-\dfrac{5}{3}\) + 8
\(x\) = \(\dfrac{-5}{3}\) + \(\dfrac{24}{3}\)
\(x\) = \(\dfrac{-19}{3}\)
Vậy \(x=-\dfrac{19}{3}\)
Vũ Hồng Linh bạn check lại bài đầu dùm =_="
\(\left[-\frac{1}{3}\right]^3\cdot x=\frac{1}{81}\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{3}\right]^3\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{27}\right]\)
\(\Leftrightarrow x=\frac{1}{81}\cdot(-27)=-\frac{1}{3}\)
\(\left[x-\frac{1}{2}\right]^3=\frac{1}{27}\)
\(\Leftrightarrow\left[x-\frac{1}{2}\right]^3=\left[\frac{1}{3}\right]^3\)
=> Làm nốt
Mấy bài kia cũng làm tương tự
(- \(\dfrac{1}{3}\))3.\(x\) = \(\dfrac{1}{81}\)
\(x=\dfrac{1}{81}\) : (- \(\dfrac{1}{3}\))3
\(x\) = - (\(\dfrac{1}{3}\))4 :(\(\dfrac{1}{3}\))3
\(x=-\dfrac{1}{3}\)
Vậy \(x=-\dfrac{1}{3}\)
1) \(\left|x+\frac{4}{5}\right|+\frac{7}{5}=\frac{3}{5}\)
\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{3}{5}-\frac{7}{5}\)
\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{-4}{5}\)
\(x+\frac{4}{5}=\pm\frac{4}{5}\)
\(TH1:x+\frac{4}{5}=\frac{4}{5}\)
\(\Rightarrow x=\frac{4}{5}-\frac{4}{5}=0\)
\(TH2:x+\frac{4}{5}=\frac{-4}{5}\)
\(\Rightarrow x=\frac{-4}{5}-\frac{4}{5}=\frac{-8}{5}\)
Vậy x ∈ {0; \(\frac{-8}{5}\)}
[4/5] 5 x = [-4/5]^7
\(\left(\frac{4}{5}\right)^5.x=\left(\frac{-4}{5}\right)^7\)
\(\Rightarrow x=\left(\frac{-1.4}{5}\right)^7:\left(\frac{4}{5}\right)^5\)
\(\Rightarrow x=\left(-1\right)^7.\left(\frac{4}{5}\right)^7:\left(\frac{4}{5}\right)^5\)
\(\Rightarrow x=\left(-1\right).\left(\frac{4}{5}\right)^2\)
\(\Rightarrow x=\left(-1\right).\frac{16}{25}\)
\(\Rightarrow x=\frac{-16}{25}\)