\(\left(\frac{3}{4}x-\frac{9}{16}\right)\cdot\left(1,5+\frac{-3}{5}:x\right)=0\\ \Rightarrow\left[{}\begin{matrix}\frac{3}{4}x-\frac{9}{16}=0\\1,5+\frac{-3}{5}:x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{3}{4}x=\frac{9}{16}\\\frac{-3}{5}:x=-1,5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{3}{4}\\x=\frac{2}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{3}{4};\frac{2}{5}\right\}\)

\(Q=\left(\frac{x^2+1}{x+1}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)=\left(\frac{x^2+1-\left(x+1\right)}{x+1}\right)\left(\frac{4x-2\left(x-1\right)}{x\left(x-1\right)}\right)\)

    \(=\left(\frac{x^2+1-x-1}{x+1}\right)\left(\frac{4x-2x+2}{x\left(x-1\right)}\right)=\left(\frac{x^2-x}{x+1}\right)\left(\frac{2\left(x+1\right)}{x\left(x-1\right)}\right)=\frac{2x\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}=2\)

Vậy Q = 2

Hình như đề là rút gọn thì phải.

Giải

\(Q=\left(\frac{x^2+1}{x+1}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)\)

\(=\left(\frac{x^2}{x}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)=\left(x-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)\)

\(=\frac{4\left(x-1\right)}{x-1}-\frac{2\left(x-1\right)}{x}=4-\frac{2x-2}{x}\)

Chết, nhầm rồi =((((

\(a,5\frac{4}{7}:x=13\Leftrightarrow x=\frac{39}{7}:13\Leftrightarrow x=\frac{39}{7}.\frac{1}{13}=\frac{3}{7}\)

\(b,\left(2,8x-32\right):\frac{2}{3}=-90\)

\(\Leftrightarrow2,8x-32=-90.\frac{2}{3}=-60\)

\(\Leftrightarrow2,8x=-60+32=-28\)

\(\Leftrightarrow x=\frac{-28}{2,8}=-10\)

d, \(7x=3,2+3x\Leftrightarrow7x-3x=3,2\Leftrightarrow4x=3,2\Leftrightarrow x=3,2:4=3,2.\frac{1}{4}=\frac{4}{5}\)

Câu c bị sai đề :\(\frac{19}{10}-1-\frac{2}{5}=\frac{1}{2}\ne1\)bạn nha.

mình lộn \(\left(\frac{19}{10}-1-\frac{2}{5}\right)+\frac{4}{5}=\frac{13}{10}\ne1\)ms đúng nha

a) \(\frac{39}{7}:x=13\)

     \(x=\frac{39}{7}:13\)

      \(x=\frac{3}{7}\)

b) \(2,8x-32=-90.\frac{2}{3}\)

  \(2,8x-32=-60\)  

    \(2,8x=-60+32\)

     \(2,8x=-28\)

     \(x=-28:2,8\)

     \(x=-10\)

c)  mik nghĩ b nên xem lại đề có đúng là có 2 dấu = ko

d) \(7x-3x=3,2\)( quy tắc chuyển vế )

    \(4x=3,2\)

    \(x=3,2:4\)

    \(x=0,8\)

tk nha

                                      \(\frac{3}{5}x-\frac{13}{9}:\left(\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}\right)=-10\)

                                 <=> \(\frac{3}{5}x-\frac{13}{9}:\left(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}\right)=-10\)

                                 <=> \(\frac{3}{5}x-\frac{13}{9}:13:\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)=-10\)

                                 <=> \(\frac{3}{5}x-\frac{1}{9}:\left(\frac{21}{315}+\frac{9}{315}+\frac{5}{315}\right)=-10\)

                                 <=> \(\frac{3}{5}x-\frac{1}{9}:\frac{35}{315}=-10\)

                                 <=> \(\frac{3}{5}x-\frac{1}{9}:\frac{1}{9}=-10\)

                                 <=> \(\frac{3}{5}x-1=-10\)

                                 <=> \(\frac{3}{5}x=-9\)

                                 <=> \(x=-15\)

Vậy x = -15.

\(\frac{3}{5}x-1\frac{4}{9}:\left(\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}\right)=-10\)

\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}=-10\right)\)

\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left[\frac{13}{2}\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\right)\right]=-10\)

\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left[\frac{13}{2}\left(\frac{2}{3.5}+\frac{2}{.57}+\frac{2}{7.9}\right)\right]=-10\)

\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left[\frac{13}{2}\left(\frac{1}{3}-\frac{1}{9}\right)\right]=-10\)

\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left(\frac{13}{2}.\frac{2}{9}\right)=-10\)

\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\frac{26}{18}=-10\)

\(\Leftrightarrow\frac{3}{5}x-1=-10\)

\(\Leftrightarrow\frac{3}{5}x=-10+1\)

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

\(\Rightarrow x=-9:\frac{3}{5}\)

\(\Rightarrow x=-15\)

Vậy \(x=-15\)

\(\frac{x}{2}+\frac{3}{2}\sqrt{x^2-4x+4}-2=\frac{x}{2}+\frac{3}{2}\left|x-2\right|-2\)

Với x >= 2 thì \(\frac{x}{2}+\frac{3\left(x-2\right)}{2}=\frac{3x+x-6}{2}=\frac{4x-6}{2}=2x-3\)

Với x < 2 thì \(\frac{x}{2}+\frac{3\left(2-x\right)}{2}=\frac{x+6-3x}{2}=\frac{6-2x}{2}=3-x\)

a) \(\frac{x-3}{3}-1=\frac{x}{-4}\)

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

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

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

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

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

\(\Leftrightarrow7x=24\)

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

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

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

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

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

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

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

c) \(\frac{x}{38}-\frac{1}{19}=\frac{-1}{2}\)

\(\Leftrightarrow\frac{x}{38}-\frac{2}{38}=\frac{-1}{2}\)

\(\Leftrightarrow\frac{x-2}{38}=\frac{-1}{2}\)

\(\Leftrightarrow2\left(x-2\right)=-38\)

\(\Leftrightarrow2x-4=-38\)

\(\Leftrightarrow2x=-38+4\)

\(\Leftrightarrow2x=-34\)

\(\Leftrightarrow x=-34\div2\)

\(\Leftrightarrow x=-17\)

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

\(\Rightarrow\left(-x\right).x=\left(-9\right).4\)

\(\Rightarrow-x^2=-36\)

\(\Rightarrow x^2=36\)

\(\Rightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)

vậy \(x=\pm6\)