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help 

Giải:

a)-3/10-(-1/5)+x)=-3/2

             -1/5+x   =-3/10-(-3/2)

              -1/5+x   =6/5

                       x   =6/5-(-1/5)

                       x   =7/5

b)-(-x+3/4)-12/8.(-32/15)=-(-1/2)

           x-3/4+16/5          =1/2

           x-3/4                =1/2-16/5

           x                           =-27/10

           x                           =-27/10+3/4

           x                           =-39/20

c)x-3/x+5=4/3

=>(x-3).3=4.(x+5)

     3x-9   =4x+20

     3x-4x =20+9

     -1x     =29

        x     =-29

Câu b cậu nên tính lại cho kĩ nhé, ấn máy tính dễ nhầm lắm đấy!

Mk phải ấn: -(39/20+3/4)-12/8.-32/15=1/2

Vì x là số âm mà đằng trước x là dấu ''-'' nên -(-39/20)=39/20 ; -(-1/2)=1/2

Chúc bạn học tốt!

a) Ta có: \(-3\dfrac{1}{4}\cdot x-75\%+\dfrac{3x}{2}=-1.2:\dfrac{-9}{10}-1\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{-13x}{4}-\dfrac{3}{4}+\dfrac{3x}{2}=\dfrac{-6}{5}\cdot\dfrac{10}{-9}-\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{-13x-3+6x}{4}=\dfrac{4}{3}-\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{-7x-3}{4}=\dfrac{1}{12}\)

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

\(\Leftrightarrow-7x=\dfrac{10}{3}\)

hay \(x=-\dfrac{10}{21}\)

b) Ta có: \(\dfrac{5}{3}+\dfrac{5}{15}+\dfrac{5}{35}+...+\dfrac{5}{x\left(x+2\right)}=2\dfrac{8}{17}\)

\(\Leftrightarrow\dfrac{5}{2}\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{x\left(x+2\right)}\right)=2\dfrac{8}{17}\)

\(\Leftrightarrow\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=2+\dfrac{8}{17}\)

\(\Leftrightarrow\left(1-\dfrac{1}{x+2}\right)=\dfrac{42}{17}:\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{x+1}{x+2}=\dfrac{42}{17}\cdot\dfrac{2}{5}=\dfrac{84}{85}\)

\(\Leftrightarrow85x+85=84x+168\)

\(\Leftrightarrow x=83\)

b: \(\Leftrightarrow4x^2-8x+4=x^2+2x+1+3\left(x^2+x-6\right)\)

\(\Leftrightarrow3x^2-10x+3=3x^2+3x-18\)

=>-13x=-21

hay x=21/13

c: \(\Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\)

=>x-100=0

hay x=100

\(\dfrac{x-1}{15}+\dfrac{x-2}{10}+\dfrac{x-3}{5}=1\)

\(\Leftrightarrow2\left(x-1\right)+3\left(x-2\right)+6\left(x-3\right)=30\)

\(\Leftrightarrow2x-2+3x-6+6x-18=30\)

\(\Leftrightarrow11x=56\Rightarrow x=\dfrac{56}{11}\)

Ta có: \(\dfrac{x-1}{15}+\dfrac{x-2}{10}+\dfrac{x-3}{5}=1\)

\(\Leftrightarrow\dfrac{2\left(x-1\right)}{30}+\dfrac{3\left(x-2\right)}{30}+\dfrac{6\left(x-3\right)}{30}=\dfrac{30}{30}\)

\(\Leftrightarrow2x-2+3x-6+6x-18=30\)

\(\Leftrightarrow12x=56\)

hay \(x=\dfrac{14}{3}\)

Vậy: \(x=\dfrac{14}{3}\)

a)4/5+x=2/3

x=2/3-4/5

x=-2/15

b)-5/6-x=2/3

x=-5/6-2/3

x=-3/2

c)1/2x+3/4=-3/10

1/2x=-3/10-3/4

1/2x=-21/20

x=-21/20:1/2

x=-21/10

d)x/3-1/2=1/5

x/3=1/5+1/2

x/3=7/10

10x/30=21/30

10x=21

x=21:10

x=21/10

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(#yH\)

\(#NBaoNgoc\)

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=10\\\dfrac{1}{x}-\dfrac{2}{y}=15\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=10\\\dfrac{2}{x}-\dfrac{4}{y}=30\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}-\dfrac{2}{x}+\dfrac{4}{y}=10-30\\\dfrac{2}{x}+\dfrac{3}{y}=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{y}=-20\\\dfrac{2}{x}+\dfrac{3}{y}=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-7}{20}\\\dfrac{2}{x}+\dfrac{3}{-\dfrac{7}{20}}=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-7}{20}\left(tm\right)\\x=\dfrac{7}{65}\left(tm\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=10.\\\dfrac{1}{x}-\dfrac{2}{y}=15.\end{matrix}\right.\) \(\left(x\ne0;y\ne0\right).\)

\(\Leftrightarrow\left\{{}\begin{matrix}2.\dfrac{1}{x}+3.\dfrac{1}{y}=10.\\\dfrac{1}{x}-2.\dfrac{1}{y}=15.\end{matrix}\right.\)

Đặt \(a=\dfrac{1}{x};b=\dfrac{1}{y}\left(a\ne0;b\ne0\right).\)

\(\Rightarrow\left\{{}\begin{matrix}2a+3b=10.\\a-2b=15.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2a+3b=10.\\2a-4b=30.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7b=-20.\\a-2b=15.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=-\dfrac{20}{7}.\\a=\dfrac{65}{7}.\end{matrix}\right.\) \(\left(TM\right).\)

\(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{20}{7}=\dfrac{1}{y}.\\\dfrac{65}{7}=\dfrac{1}{x}.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{7}{20}.\\x=\dfrac{7}{65}.\end{matrix}\right.\) \(\left(TM\right).\)

Vậy hệ phương trình có nghiệm duy nhất là \(\left(x;y\right)=\left(-\dfrac{7}{20};\dfrac{7}{65}\right).\)