Bài 1:

a; (\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\)) x \(\dfrac{3}{4}\) = \(\dfrac{1}{4}\)

     \(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) : \(\dfrac{3}{4}\)

      \(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) x \(\dfrac{4}{3}\)

    \(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) =  \(\dfrac{1}{3}\)

      \(\dfrac{1}{4}x\) = \(\dfrac{1}{3}\) + \(\dfrac{1}{8}\)

       \(\dfrac{1}{4}\) \(x\)=  \(\dfrac{8}{24}\) + \(\dfrac{11}{24}\)

          \(\dfrac{1}{4}x=\dfrac{11}{24}\)

           \(x=\dfrac{11}{24}:\dfrac{1}{4}\)

           \(x=\dfrac{11}{24}\times4\)

           \(x=\dfrac{11}{6}\) 

b; \(\dfrac{12}{5}:x\) = \(\dfrac{14}{3}\) x \(\dfrac{4}{7}\)

     \(\dfrac{12}{5}\) : \(x\) = \(\dfrac{8}{3}\)

            \(x\) = \(\dfrac{12}{5}\) : \(\dfrac{8}{3}\)

            \(x\) = \(\dfrac{12}{5}\) x \(\dfrac{3}{8}\)

             \(x\) = \(\dfrac{9}{10}\)

c; \(x+\dfrac{2}{3}\) = 8:4 - 1

   \(x+\dfrac{2}{3}\) = 2 - 1

   \(x+\dfrac{2}{3}\) = 1

  \(x=1-\dfrac{2}{3}\)

  \(x\) = \(\dfrac{1}{3}\)

a; 

c; \(\dfrac{1}{10}\) + \(\dfrac{4}{20}\) + \(\dfrac{9}{30}\)+\(\dfrac{16}{40}+\dfrac{25}{50}+\dfrac{36}{60}+\dfrac{49}{70}+\dfrac{64}{80}+\dfrac{81}{90}\)

= \(\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}\)

= \(\dfrac{1+2+3+4+5+6+7+8+9}{10}\)

=  \(\dfrac{\left(1+9\right)+\left(2+8\right)+\left(3+7\right)+\left(4+6\right)+5}{10}\)

= \(\dfrac{10+10+10+10+5}{10}\)

= \(\dfrac{\left(10+10+10+10\right)+5}{10}\)

= \(\dfrac{10\times4+5}{10}\)

= \(\dfrac{45}{10}\)

= \(\dfrac{9}{2}\)

sao câu 2+2:2 không có dấu = vậy

có vài câu không phải toán lớp 9 đâu

\(1,\\ x+\dfrac{1}{2}=-\dfrac{5}{3}\\ x=-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{13}{6}\\ Vậyx=-\dfrac{13}{6}\)

\(2,\\ \dfrac{1}{3}-x=\dfrac{3}{5}\\ x=\dfrac{1}{3}-\dfrac{3}{5}\\ x=-\dfrac{4}{15}\\ Vậyx=-\dfrac{4}{15}\)

\(3,\\ 3-4+x=\dfrac{7}{2}\\ -1+x=\dfrac{7}{2}\\ x=\dfrac{7}{2}+1\\ x=\dfrac{9}{2}\\ Vậyx=\dfrac{9}{2}\)

\(4,\\ x-\dfrac{4}{3}=-\dfrac{7}{9}\\ x=-\dfrac{7}{9}+\dfrac{4}{3}\\ x=\dfrac{15}{27}\\ Vậyx=\dfrac{15}{27}\)

\(5,\\ x-\left(-\dfrac{7}{3}\right)=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{3}\\ x=-\dfrac{27}{18}\\ Vậyx=-\dfrac{27}{18}\)

\(6,\\ x-\dfrac{1}{5}=\dfrac{9}{10}\\ x=\dfrac{9}{10}+\dfrac{1}{5}\\ x=\dfrac{11}{10}\\ Vậyx=\dfrac{11}{10}\)

\(7,\\ x+\dfrac{5}{12}=\dfrac{3}{8}\\ x=\dfrac{3}{8}-\dfrac{5}{12}\\ x=-\dfrac{1}{24}\\ Vậyx=-\dfrac{1}{24}\)

\(8,\\ x+\dfrac{5}{4}=\dfrac{7}{6}\\ x=\dfrac{7}{6}-\dfrac{5}{4}\\ x=-\dfrac{9}{24}\\ Vậyx=-\dfrac{9}{24}\)

\(9,\\ x-\dfrac{2}{7}=\dfrac{1}{35}\\ x=\dfrac{1}{35}+\dfrac{2}{7}\\ x=\dfrac{11}{35}\\ Vậyx=\dfrac{11}{35}\\ 10,\\ x-\dfrac{1}{5}=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}+\dfrac{1}{5}\\ x=-\dfrac{1}{2}\\ Vậyx=-\dfrac{1}{2}\)

ahiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii hãy ấn a

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`@` `\text {Ans}`

`\downarrow`

`1)`

\(x+\dfrac{1}{2}=\dfrac{5}{3}\)

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

`\Rightarrow`\(x=\dfrac{7}{6}\)

Vậy, `x =`\(\dfrac{7}{6}\)

`2)`

\(\dfrac{3}{5}-x=\dfrac{1}{3}\)

`\Rightarrow`\(x=\dfrac{3}{5}-\dfrac{1}{3}\)

`\Rightarrow`\(x=\dfrac{4}{15}\)

Vậy, `x =`\(\dfrac{4}{15}\)

`3)`

\(\dfrac{3}{4}+x=\dfrac{7}{2}\)

`\Rightarrow`\(x=\dfrac{7}{2}-\dfrac{3}{4}\)

`\Rightarrow`\(x=\dfrac{11}{4}\)

Vậy, \(x=\dfrac{11}{4}\)

`4)`

\(x-\dfrac{4}{3}=\dfrac{7}{9}\)

`\Rightarrow`\(x=\dfrac{7}{9}+\dfrac{4}{3}\)

`\Rightarrow`\(x=\dfrac{19}{9}\)

Vậy, `x=`\(\dfrac{19}{9}\)

`5)`

\(x-\dfrac{5}{6}=\dfrac{7}{3}\)

`\Rightarrow`\(x=\dfrac{7}{3}+\dfrac{5}{6}\)

`\Rightarrow x =`\(\dfrac{19}{6}\)

Vậy, `x=`\(\dfrac{19}{6}\)

`6)`

\(x-\dfrac{1}{5}=\dfrac{9}{10}\)

`\Rightarrow x=`\(\dfrac{9}{10}+\dfrac{1}{5}\)

`\Rightarrow x=`\(\dfrac{11}{10}\)

Vậy, `x=`\(\dfrac{11}{10}\)

450 nha !

= 450

(1+9)+(2+8)+(3+7)+(4+6)+(5+5)+(6+4)+(7+3)+(8+2)+(10+90)+(20+80)+(30+70)+(40+60)+(50+50)+(40+60)+(30+70)+(20+80)                            =10+10+10+10+10+10+10+10+100+100+100+100+100+100+100+100=(10x8)+(100x8)                                                                                       = 80 + 800                                                                                       =880 

Tick vs ah

a) \(\text{​​}/3x-5/-\frac{1}{7}=\frac{1}{3}\)                           b)\(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)

  \(/3x-5/=\frac{10}{21}\)                                           \([x.\left(\frac{3}{5}-\frac{2}{3}-1\right)]=\frac{-5}{21}.7\)

 \(\Rightarrow3x-5=\frac{10}{21}hay3x-5=\frac{-10}{21}\)         \(\left[x.\frac{-16}{15}\right]=\frac{-5}{3}\)

\(3x=\frac{115}{21}\)                \(3x=\frac{95}{21}\)                         \(x=\frac{25}{16}\)

\(x=\frac{115}{63}\)                  \(x=\frac{95}{63}\)                             Vậy x = \(\frac{25}{16}\)

                      Vậy x \(\in\left\{\frac{115}{63};\frac{95}{63}\right\}\)

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. \(4\left(\frac{1}{2}-x\right)-5\left(x-\frac{3}{10}\right)=\frac{7}{4}\)

\(\Rightarrow2-4x-5x+\frac{3}{2}=\frac{7}{4}\)

\(\Rightarrow2-9x=\frac{1}{4}\)

\(\Rightarrow9x=\frac{7}{4}\)

\(\Rightarrow x=\frac{7}{36}\)