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19 tháng 3 2019

Bài 2:

a) \(\frac{4}{9}+x=\frac{-5}{3}\)

\(\Leftrightarrow x=\frac{-5}{3}-\frac{4}{9}\)

\(\Leftrightarrow x=\frac{-15}{9}-\frac{4}{9}\)\(=\frac{-19}{9}\)

Vậy: \(x=\frac{-19}{9}\)

b) \(2,4:\left(\frac{1}{2}.x-\frac{3}{4}\right)=\frac{3}{10}\)

\(\Leftrightarrow\frac{24}{10}:\left(\frac{1}{2}x-\frac{3}{4}\right)=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{2}x-\frac{3}{4}=\frac{24}{10}:\frac{3}{10}=\frac{24}{10}.\frac{10}{3}\)\(=8\)

\(\Leftrightarrow\frac{1}{2}x=8+\frac{3}{4}=\frac{35}{4}\)

\(\Leftrightarrow x=\frac{35}{4}:\frac{1}{2}=\frac{35}{4}.2=\frac{35}{2}\)

c) \(\frac{x+1}{-8}=\frac{-2}{x+1}\)

\(\Rightarrow\left(x+1\right).\left(x+1\right)=\left(-2\right).\left(-8\right)\)

\(\Leftrightarrow\left(x+1\right)^2=16=4^2=\left(-4\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{3;-5\right\}\)

7 tháng 7 2016

a)Ta có: \(\frac{-2}{5}+\frac{6}{5}.\left(y-\frac{2}{3}\right)=\frac{-4}{15}\)

\(\Rightarrow\frac{6}{5}.\left(y-\frac{2}{3}\right)=\frac{-4}{15}-\frac{-2}{15}\)

\(\Rightarrow\frac{6}{5}.\left(y-\frac{2}{3}=\right)\frac{-2}{5}\)

\(\Rightarrow y-\frac{2}{3}=\frac{-2}{5}:\frac{6}{5}=\frac{-1}{3}\)

\(\Rightarrow y=\frac{-1}{3}+\frac{2}{3}=\frac{1}{3}\)

Vậy x = \(\frac{1}{3}\)

b) Ta có: \(\frac{-2}{5}+\frac{2}{3}x+\frac{1}{6}x=\frac{-4}{15}\)

        \(\Rightarrow\frac{-2}{5}+x.\left(\frac{2}{3}+\frac{1}{6}\right)=\frac{-4}{15}\)

        \(\Rightarrow x.\frac{5}{6}=\frac{-4}{15}-\frac{-2}{15}\)

         \(x.\frac{5}{6}=\frac{-2}{15}\)

\(\Rightarrow x=\frac{-2}{15}:\frac{5}{6}=\frac{-4}{25}\)

Vậy x = \(\frac{-4}{25}\)

c) Ta có: \(\frac{3}{2}x+\frac{-2}{5}-\frac{2}{3}.x=\frac{-4}{15}\)

\(\Rightarrow\frac{3}{2}x-\frac{2}{3}x+\frac{-2}{5}=\frac{-4}{15}\)

\(\Rightarrow x.\left(\frac{3}{2}-\frac{2}{4}\right)=\frac{-4}{15}-\frac{-2}{15}\)

\(\Rightarrow x.\frac{5}{6}=\frac{-2}{15}\)

\(\Rightarrow x=\frac{-2}{15}:\frac{5}{6}=\frac{-4}{25}\)

Vậy x = \(\frac{-4}{25}\)

Ủng hộ tớ nha m.n

25 tháng 3 2019

a) \(\frac{x-3}{3}-1=\frac{x}{-4}\)

\(\Leftrightarrow\frac{x-3}{3}-\frac{3}{3}=\frac{x}{-4}\)

\(\Leftrightarrow\frac{x-6}{3}=\frac{x}{-4}\)

\(\Leftrightarrow-4\left(x-6\right)=3x\)

\(\Leftrightarrow-4x+24=3x\)

\(\Leftrightarrow24=3x+4x\)

\(\Leftrightarrow7x=24\)

\(\Leftrightarrow x=\frac{24}{7}\)

25 tháng 3 2019

b) \(\frac{5}{8}-\left(x-\frac{1}{2}\right)=\frac{-3}{4}\)

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

\(\Leftrightarrow\frac{5}{8}+\frac{4}{8}-x=\frac{-3}{4}\)

\(\Leftrightarrow\frac{9}{8}-x=\frac{-3}{4}\)

\(\Leftrightarrow x=\frac{9}{8}+\frac{3}{4}\)

\(\Leftrightarrow x=\frac{15}{8}\)

27 tháng 4 2018

x thuộc { -0,37; -0,36; ... ; 2; 3 }

27 tháng 4 2018

Bạn có cách giải luôn ko

19 tháng 2 2017

a/ \(\frac{2}{3}+\frac{4}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)

\(\Rightarrow\frac{82}{105}< \frac{x}{105}< \frac{92}{105}\)

\(\Rightarrow82< x< 92\)

\(\Rightarrow x=\left\{83;84;85;86;87;88;89;90;91\right\}\)

b/ \(-\frac{7}{15}+\frac{8}{60}+\frac{24}{90}\le\frac{x}{15}\le\frac{3}{5}+\frac{8}{30}+-\frac{4}{10}\)

\(\Rightarrow-\frac{1}{15}\le\frac{x}{15}\le\frac{7}{15}\)

\(\Rightarrow-1\le x\le7\)

\(\Rightarrow x=\left\{-1;0;1;2;3;4;5;6;7\right\}\)

15 tháng 3 2019

\(x\times\frac{6}{25}=\frac{15}{-13}\)

x=\(\frac{15}{-13}\div\frac{6}{25}\)

x=\(-\frac{125}{26}\)

các câu còn lại làm tương tự nha!!!

15 tháng 3 2019

\(1.x.\frac{6}{25}=\frac{15}{-13}\\ x=\frac{15}{-13}:\frac{6}{25}\\ x=-\frac{125}{26}\)

\(2.x:\frac{4}{10}=\frac{13}{-45}+\frac{8}{15}\\ x:\frac{4}{10}=\frac{11}{45}\\ x=\frac{11}{45}.\frac{4}{10}\\ x=\frac{22}{225}\)

\(3.\frac{3}{8}-\frac{1}{6}.x=\frac{1}{4}\\ \frac{1}{6}.x=\frac{3}{8}-\frac{1}{4}\\ \frac{1}{6}.x=\frac{1}{8}\\ x=\frac{1}{8}:\frac{1}{6}\\ x=\frac{3}{4}\)

\(4.\frac{1}{3}+\frac{1}{2}:x=-4\\ \frac{1}{2}:x=-4-\frac{1}{3}=-\frac{13}{3}\\ x=\frac{1}{2}:\left(-\frac{13}{3}\right)=-\frac{3}{26}\)

\(5.x+\frac{7}{12}=\frac{17}{18}-\frac{1}{9}=\frac{5}{6}\\ x=\frac{5}{6}-\frac{7}{12}\\ x=\frac{1}{4}\)