a: Ta có: \(\dfrac{1}{4}:x=3\dfrac{4}{5}:40\dfrac{8}{15}\)

\(\Leftrightarrow x=\dfrac{1}{4}\cdot\dfrac{\dfrac{608}{15}}{3+\dfrac{4}{5}}\)

\(\Leftrightarrow x=\dfrac{152}{15}:\dfrac{19}{5}=\dfrac{8}{3}\)

b: Ta có: \(\left(x+1\right):\dfrac{5}{6}=\dfrac{20}{3}\)

\(\Leftrightarrow x+1=\dfrac{50}{9}\)

hay \(x=\dfrac{41}{9}\)

c: Ta có: \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

\(\Leftrightarrow x^2-1=63\)

\(\Leftrightarrow x^2=64\)

hay \(x\in\left\{8;-8\right\}\)

c. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\) 

    \(7.9=\left(x-1\right).\left(x+1\right)\) 

    \(63=x^2-1\) 

             \(x^2=63+1\) 

             \(x^2=64\) 

             \(x^2=8^2\)

             \(x=8\)          

a) 7,2 x 94,3 + 7,2 x 4,7 + 7,2

=7,2 x(94,3+4,7+1)

=7,2x100

=720

b)\(\dfrac{7}{9}\times\dfrac{2}{3}+\dfrac{7}{9}\times\dfrac{1}{3}=\dfrac{7}{9}\times\left(\dfrac{2}{3}+\dfrac{1}{3}\right)=\dfrac{7}{9}\times1=\dfrac{7}{9}\)

a)\(x=\left(\dfrac{3}{56}\cdot\dfrac{28}{9}\right):\dfrac{-3}{7}=\dfrac{1}{6}:\dfrac{-3}{7}=-\dfrac{7}{18}\)

b)\(x=\left(\dfrac{7}{15}\cdot\dfrac{5}{3}\right)+\dfrac{3}{16}=\dfrac{7}{9}+\dfrac{3}{16}=\dfrac{139}{144}\)

c)\(x=\left(\dfrac{5}{6}-\dfrac{2}{5}\right).5=\dfrac{13}{6}\)

d)\(=>x\left(\dfrac{3}{4}-\dfrac{2}{5}\right)=\dfrac{1}{6}\cdot\left(\dfrac{3}{7}+\dfrac{5}{7}\right)\)

\(x\cdot\dfrac{7}{20}=\dfrac{4}{21}=>x=\dfrac{4}{21}\cdot\dfrac{20}{7}=\dfrac{80}{147}\)

Bài 4: 

a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)

\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)

\(1,25-x=\dfrac{11}{12}\)

\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)

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

b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)

\(x-\dfrac{7}{6}=\dfrac{13}{12}\)

\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)

\(x=\dfrac{27}{12}=\dfrac{9}{4}\)

c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)

\(4-\left(2x+1\right)=\dfrac{8}{3}\)

\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)

\(2x+1=\dfrac{20}{3}\)

\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)

\(2x=\dfrac{17}{3}\)

\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)

Bài 15:

a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)

\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)

\(=>x=\left(\dfrac{-2}{3}\right)^8\)

b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)

\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)

\(=>x=\left(\dfrac{4}{9}\right)^9\)

c) \(\left(x+4\right)^3=-125\)

\(\left(x+4\right)^3=\left(-5\right)^3\)

\(=>x+4=-5\)

\(x=-5-4\)

\(=>x=-9\)

d) \(\left(10-5x\right)^3=64\)

\(\left(10-5x\right)^3=4^3\)

\(=>10-5x=4\)

\(5x=10-4\)

\(5x=6\)

\(=>x=\dfrac{6}{5}\)

e) \(\left(4x+5\right)^2=81\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(4x+5\right)^2=\left(-9\right)^2\\\left(4x+5\right)^2=9^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+5=-9\\4x+5=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)

Bài 16:

a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)

\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)

\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)

b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)

\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)

\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)

c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)

\(=\dfrac{7}{4}.\dfrac{-12}{5}\)

\(=\dfrac{-21}{5}\)

\(#Wendy.Dang\)

Uh, chừa sau k dám học muộn nx

Đăng từng bài `1` thôi cậu ơii

Mn ơi giúp mk với , please !!!

1. Ta có: \(\dfrac{x}{-7}=\dfrac{y}{4}\Rightarrow\dfrac{2x}{-14}=\dfrac{3y}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta được:

\(\dfrac{2x-3y}{-14-12}=\dfrac{-78}{-26}=3\)

=> \(\left\{{}\begin{matrix}x=-21\\y=12\end{matrix}\right.\)

2. Ta có:

- \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)

- \(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)

=> \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta được:

\(\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)

=> \(\left\{{}\begin{matrix}x=-27\\y=-21\\z=-9\end{matrix}\right.\)

a) \(\Leftrightarrow\dfrac{3}{2}:x=\dfrac{1}{2}\\ \Leftrightarrow x=\dfrac{3}{2}:\dfrac{1}{2}\\ \Leftrightarrow x=3\)

b) \(\Leftrightarrow x=\dfrac{7}{9}-\dfrac{2}{3}\\ \Leftrightarrow x=\dfrac{1}{9}\)

c) \(\Leftrightarrow x=\dfrac{8}{7}:\dfrac{6}{7}\\ \Leftrightarrow x=\dfrac{4}{3}\)

d) \(\Leftrightarrow x=\dfrac{9}{5}-\dfrac{3}{7}\\ \Leftrightarrow x=\dfrac{48}{35}\)

a) x = 3

b) x = \(\dfrac{1}{9}\)

c) x = \(\dfrac{4}{3}\)

d)\(\dfrac{48}{35}\)

Giải:

\(9-3\times\left(x-9\right)=6\) 

      \(3\times\left(x-9\right)=9-6\) 

      \(3\times\left(x-9\right)=3\) 

               \(x-9=3:3\) 

               \(x-9=1\) 

                     \(x=1+9\) 

                     \(x=10\) 

\(4+6\times\left(x+1\right)=70\) 

      \(6\times\left(x+1\right)=70-4\) 

      \(6\times\left(x+1\right)=66\) 

               \(x+1=66:6\) 

               \(x+1=11\) 

                     \(x=11-1\) 

                     \(x=10\) 

\(\dfrac{x}{13}+\dfrac{15}{26}=\dfrac{46}{52}\) 

         \(\dfrac{x}{13}=\dfrac{23}{26}-\dfrac{15}{26}\) 

         \(\dfrac{x}{13}=\dfrac{4}{13}\) 

\(\Rightarrow x=4\) 

\(\dfrac{11}{14}-\dfrac{3}{x}=\dfrac{5}{14}\) 

         \(\dfrac{3}{x}=\dfrac{11}{14}-\dfrac{5}{14}\) 

         \(\dfrac{3}{x}=\dfrac{3}{7}\) 

\(\Rightarrow x=7\) 

\(5\times\left(3+7\times x\right)=40\) 

         \(3+7\times x=40:5\) 

         \(3+7\times x=8\) 

                \(7\times x=8-3\) 

                \(7\times x=5\) 

                      \(x=5:7\) 

                      \(x=\dfrac{5}{7}\) 

\(x\times6+12:3=120\) 

       \(x\times6+4=120\) 

             \(x\times6=120-4\) 

             \(x\times6=116\) 

                   \(x=116:6\) 

                   \(x=\dfrac{58}{3}\) 

\(x\times3,7+x\times6,3=120\) 

    \(x\times\left(3,7+6,3\right)=120\) 

                  \(x\times10=120\) 

                           \(x=120:10\) 

                           \(x=12\) 

\(\left(15\times24-x\right):0,25=100:\dfrac{1}{4}\) 

      \(\left(360-x\right):0,25=400\) 

                   \(360-x=400.0,25\) 

                   \(360-x=100\) 

                             \(x=360-100\) 

                             \(x=260\) 

\(71+65\times4=\dfrac{x+140}{x}+260\) 

\(\left(x+140\right):x+260=71+260\) 

\(x:x+140:x+260=331\) 

    \(1+140:x+260=331\) 

                    \(140:x=331-1-260\) 

                    \(140:x=70\) 

                             \(x=140:70\) 

                             \(x=2\) 

\(\left(x+1\right)+\left(x+4\right)+\left(x+7\right)+...+\left(x+28\right)=155\) 

                      \(10\times x+\left(1+4+7+...+28\right)=155\)

Số số hạng \(\left(1+4+7+...+28\right)\) :

         \(\left(28-1\right):3+1=10\) 

Tổng dãy \(\left(1+4+7+...+28\right)\) :

         \(\left(1+28\right).10:2=145\) 

\(\Rightarrow10\times x+145=155\) 

               \(10\times x=155-145\) 

               \(10\times x=10\) 

                       \(x=10:10\) 

                       \(x=1\) 

Đều theo cách lớp 5 nha em!

 60x [7/12+4/15]

60x153/180

=9180/180

b 1/2x2/3x3/4x4/5x5/6x6/7x7/8x8/9=40320/4032