a: x+2/5=1/2

=>x=1/2-2/5=5/10-4/10=1/10

b; x-2/5=2/7

=>x=2/7+2/5=10/35+14/35=24/35

c: 3/5-x=1/10

=>x=3/5-1/10=6/10-1/10=5/10=1/2

d: x*3/4=9/20

=>x=9/20:3/4=9/20*4/3=36/60=3/5

e: x:1/7=14

=>x=14*1/7=2

f: =>x+1/4=2/5:1/2=4/5

=>x=4/5-1/4=16/20-5/20=11/20

g: =>x*2/3=9/12+2/3=3/4+2/3=9/12+8/12=17/12

=>x=17/12:2/3=17/12*3/2=51/24=17/8

a, 2/5 + 3/4 : x = -1/2

3/4 : x = -1/2 - 2/5

3/4 : x = -9/10

x = 3/4 : -9/10

x = -5/6

b, 5/7 - 2/3 . x = 4/5 

2/3 . x = 4/5 + 5/7

2/3 . x = 53/35

x = 53/35 : 2/3

x = 159/70

c và d mình làm dược nhưng ko ghi được cái suy ra

\(a,x.\dfrac{-5}{8}=\dfrac{15}{32}\)

\(\Leftrightarrow x=\dfrac{15}{32}:\dfrac{-5}{8}\)

\(\Leftrightarrow x=\dfrac{15}{32}.\dfrac{-8}{5}\)

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

\(b,\dfrac{3}{10}:x=-\dfrac{9}{20}\)

\(\Leftrightarrow x=\dfrac{3}{10}:\dfrac{-9}{20}\)

\(\Leftrightarrow x=\dfrac{3}{10}.\dfrac{-20}{9}\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)

\(c,-\dfrac{1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)

\(\Leftrightarrow-\dfrac{1}{4}x=\dfrac{3}{4}-\dfrac{4}{5}\)

\(\Leftrightarrow-\dfrac{1}{4}x=-\dfrac{1}{20}\)

\(\Leftrightarrow x=-\dfrac{1}{20}\times\left(-4\right)\)

\(\Leftrightarrow x=\dfrac{1}{5}\)

\(d,-\dfrac{7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}.\dfrac{-5}{12}\)

\(\Leftrightarrow-\dfrac{7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{7}{8}\)

\(\Leftrightarrow\dfrac{2}{3}:x=\dfrac{5}{8}\)

\(\Leftrightarrow x=\dfrac{2}{3}:\dfrac{5}{8}\)

\(\Leftrightarrow x=\dfrac{16}{15}\)

a, \(x\cdot\dfrac{-5}{8}=\dfrac{15}{32}\)

\(x=\dfrac{15}{32}:\dfrac{-5}{8}\)

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

b, \(\dfrac{3}{10}:x=\dfrac{-9}{20}\)

\(x=\dfrac{3}{10}:\dfrac{-9}{20}\)

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

c, \(\dfrac{-1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)

\(\dfrac{-1}{4}x=\dfrac{3}{4}-\dfrac{4}{5}\)

\(\dfrac{-1}{4}x=-\dfrac{1}{20}\)

\(x=-\dfrac{1}{20}:\dfrac{-1}{4}\)

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

d, \(\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}\cdot\dfrac{-5}{12}\)

\(\dfrac{-7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)

\(\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{-7}{8}\)

\(\dfrac{2}{3}:x=\dfrac{5}{8}\)

\(x=\dfrac{2}{3}:\dfrac{5}{8}\)

\(x=\dfrac{16}{15}\)

#YVA6

\(a,x.\dfrac{-5}{8}=\dfrac{15}{32} \)

\(x=\dfrac{15}{32}:\dfrac{-5}{8}\)

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

\(b,\dfrac{3}{10}:x=\dfrac{-9}{20}\)

\(x=\dfrac{3}{10}:\dfrac{-9}{20}\)

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

\(c,\dfrac{-1}{4}.x+\dfrac{4}{5}=\dfrac{3}{4}\)

\(\dfrac{-1}{4}.x=\dfrac{-1}{20}\)

\(x=\dfrac{-1}{20}:\dfrac{-1}{4}\)

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

\(d,\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}.\dfrac{-5}{12}\)

\(\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{-1}{4}\)

\(\dfrac{2}{3}:x=\dfrac{5}{8}\)

\(x=\dfrac{16}{15}\)

\(#yH\)

\(#NBaoNgoc\)

a, - \(\dfrac{2}{5}\) + \(\dfrac{4}{5}\).\(x\) = \(\dfrac{3}{5}\)

               \(\dfrac{4}{5}\).\(x\) = \(\dfrac{3}{5}\)+ \(\dfrac{2}{5}\)

                 \(\dfrac{4}{5}\).\(x\) = 1

                      \(x\) = \(\dfrac{5}{4}\)

b, - \(\dfrac{3}{7}\) - \(\dfrac{4}{7}\): \(x\) = \(\dfrac{2}{5}\)

              \(\dfrac{4}{7}\): \(x\) = - \(\dfrac{3}{7}\) - \(\dfrac{2}{5}\)

                \(\dfrac{4}{7}\): \(x\) = - \(\dfrac{29}{35}\)

                  \(x\) = \(\dfrac{4}{7}\): (- \(\dfrac{29}{35}\) )

                  \(x\) = - \(\dfrac{20}{29}\)

c, \(\dfrac{4}{7}\).\(x\) + \(\dfrac{2}{3}\) = - \(\dfrac{1}{5}\)

     \(\dfrac{4}{7}\).\(x\)         = -\(\dfrac{1}{5}\) - \(\dfrac{2}{3}\)

      \(\dfrac{4}{7}\).\(x\)       = - \(\dfrac{13}{15}\)

           \(x\)     = - \(\dfrac{13}{15}\): \(\dfrac{4}{7}\)

            \(x\)    = - \(\dfrac{91}{60}\)

d, \(\dfrac{5}{7}\): \(x\) - 1 = \(\dfrac{2}{3}\)

     \(\dfrac{5}{7}\): \(x\)     = \(\dfrac{2}{3}\)+ 1

       \(\dfrac{5}{7}\): \(x\)   = \(\dfrac{5}{3}\)

              \(x\)  = \(\dfrac{5}{7}\): \(\dfrac{5}{3}\)

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

c.\(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)

\(\dfrac{5}{7}:x=\dfrac{1}{3}-\dfrac{3}{7}\)

\(\dfrac{5}{7}:x=-\dfrac{2}{21}\)

\(x=\dfrac{5}{7}:-\dfrac{2}{21}\)

\(x=-\dfrac{15}{2}\)

d.\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)

\(\left|2x-\dfrac{5}{12}\right|=3\dfrac{1}{4}:\dfrac{39}{16}\)

\(\left|2x-\dfrac{5}{12}\right|=\dfrac{4}{3}\)

\(\rightarrow\left[{}\begin{matrix}2x-\dfrac{5}{12}=\dfrac{4}{3}\\2x-\dfrac{4}{12}=-\dfrac{4}{3}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{4}\\2x=-\dfrac{11}{12}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{11}{24}\end{matrix}\right.\)

A, \(\dfrac{4}{9}+x=\dfrac{5}{3}\)

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

\(x\)\(=\dfrac{11}{9}\)

B,\(\dfrac{3}{4}.x=\dfrac{-1}{2}\)

\(x=\dfrac{-1}{2}:\dfrac{3}{4}\)

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

a)

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

=> \(x = \frac{5}{3}-\frac{4}{9}\)

=> \(x = \) \(\frac{11}{9}\)

Vậy \(x = \dfrac{11}{9}\)

b) 

\(\dfrac{3}{4} .x = \dfrac{-1}{2}\)

=> \(x = \dfrac{-1}{2} : \dfrac{3}{4}\)

=> \(x = \dfrac{-2}{3}\)

Vậy \(x = \dfrac{-2}{3}\)

c)

\( \dfrac{3}{7}+ \dfrac{5}{7}:x = \dfrac{1}{3}\)

=> \(\dfrac{5}{7}:x = \dfrac{1}{3}-\) \( \dfrac{3}{7}\)

=> \(\dfrac{5}{7}:x = \dfrac{-2}{21}\)

=> \(x = \dfrac{5}{7}:\dfrac{-2}{21}\)

=> \(x = \dfrac{-15}{2}\)

Vậy \(x = \dfrac{-15}{2}\)

d) 

\(3\dfrac{1}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)

=> \(\dfrac{13}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)

=> \( |2x - \dfrac{5}{12} | =\dfrac{13}{4} : \dfrac{39}{16}\)

=> \(|2x-\dfrac{5}{12} |= \dfrac{4}{3}\)

=> \(\left[\begin{matrix} 2x - \dfrac{5}{12} = \dfrac{4}{3}\\ 2x - \dfrac{5}{12} = \dfrac{4}{3}\end{matrix}\right.\)

=> \(\left[\begin{matrix} 2x = \dfrac{-4}{3}+\dfrac{5}{12}\\ 2x = \dfrac{-4}{3}+\dfrac{5}{12} \end{matrix}\right.\)

=> \(\left[\begin{matrix} 2x = \dfrac{7}{4}\\ 2x = \dfrac{-11}{12} \end{matrix}\right.\)

=> \(\left[\begin{matrix} x = \dfrac{7}{8}\\ x = \dfrac{-11}{24} \end{matrix}\right.\)

Vậy \(x \in \) { \(\dfrac{7}{8} ; \dfrac{-11}{24}\) }

Bài 2: a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)

\(\Leftrightarrow\left(x-3\right).7=\left(x+5\right).5\)

\(\Leftrightarrow7x-21=5x+25\)

\(\Leftrightarrow7x-5x=21+25\)

\(\Leftrightarrow2x=46\)

\(\Rightarrow x=46:2=23\)

b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=63\)

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

\(\Leftrightarrow x^2=64\)

\(\Rightarrow x^2=\left(\pm8\right)^2\)

\(\Rightarrow x=8\) hoặc \(x=-8\)

2)a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)

\(\Leftrightarrow7\left(x-3\right)=5\left(x+5\right)\)

\(7x-21=5x+25\)

\(7x-5x+25=21\)

\(2x+25=21\)

\(2x=-4\Rightarrow x=-2\)

b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

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

\(63=x\left(x-1\right)+1\left(x-1\right)\)

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

\(x^2=63+1=64\)

\(x=\left\{\pm8\right\}\)

c) \(\dfrac{x+4}{20}=\dfrac{2}{x+4}\)

\(\Leftrightarrow\left(x+4\right)\left(x+4\right)=2.20=40\)

\(x\left(x+4\right)+4\left(x+4\right)=40\)

\(x^2+4x+4x+16=40\)

\(x^2+8x=40-16=24\)

\(x\left(x+8\right)=24\)

\(x\in\left\{\varnothing\right\}\)

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

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=\left(x-1\right)\left(x+3\right)\)

\(x\left(x-2\right)+2\left(x-2\right)=x\left(x+3\right)-1\left(x+3\right)\)

\(x^2-2x+2x-4=x^2+3x-x-3\)

\(\)\(x^2-4=x^2+2x-3\)

\(\Leftrightarrow x^2-x^2-2x+3=4\)

\(-2x+3=4\)

\(-2x=1\)

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

bài 1) a) ta có : \(\dfrac{x}{10}=\dfrac{y}{15}\) \(\Leftrightarrow\) \(10y=15x\) \(\Leftrightarrow\) \(10y=15.\dfrac{7}{2}=\dfrac{105}{2}\)

\(\Leftrightarrow y=\dfrac{105}{2}:10=\dfrac{21}{4}\)

vậy \(x+2y-3z=20\Leftrightarrow\) \(\dfrac{7}{2}+2.\dfrac{21}{4}-3z=20\)

\(\Leftrightarrow\) \(-3z=20-14=6\) \(\Leftrightarrow\) \(z=-2\)

vậy \(x=\dfrac{7}{2}\) ; \(y=\dfrac{21}{4}\) ; \(z=-2\)

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

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

Tìm x trong tỉ lệ thức

a)