ở dưới làm r

3 . ( 5x + 15 ) + x - 11 = 98

=>3(5x+15)+x=109

=>15x+45+x=109

=>16x=64

=>x=4

a) 13-[7-(x+8)]=11         

    [7-(x+8)]=13-11

    [7-(x+8)]=2

    (x+8)=7-2

    (x+8)=5

     x=5-8

     x=-3

hơi khó anh mai ơi !

hơi bị khó... chờ mình ghi lại để hỏi cô!!!

dấu này"/" là chia hay phần

A = ( 4/4 + 2/3 ) - ( 51/3 - 6/5 ) - ( 6 - 7/4 + 3/2 )

Sau đó quy đồng rồi trừ cả là đc 

B tương tự 

C=13/15 

D cx thế . Bạn tự vận dụng đi . Xl vì ko giải đc . Mik đang gấp

+)  4x - 12 -11 - x - 58 = 0
     3x - 81 = 0
     3x = 81
     x = 27
+)  5x - 2 + 3x -15 - x = 0
      7x - 17 = 0
      7x = 17
       x = 17/7
+)  4x - 44 - 6x -15 + x = 0
      -x -59 = 0
       -x = 59
        x = -59
     
 

5x-2+3(x-5)=x

5x-2+3x-15=x

5x+3x-x=15+2

7x=17

  x=17/7

vậy x=17/7

4(x-3)-(11+x)=58                   5x-2+3(x-5)=x                   4(x-11)-2(3x)=15-x

4x-12-11-x=58                       5x-2+3x-15-x=0                 4x-44-6x-15+x=0

3x-23=58                               7x-17=0                              -x-59=0

3x=81                                     7x=17                                  -x=59

x=27                                         x=-17/7                               x=-59

b) PT \(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow\left(15x-35\right)\left(5x+3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

 Vậy \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)

c) PT \(\Leftrightarrow\left(2-3x\right)\left(x-11\right)+\left(2-3x\right)\left(2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(-9-4x\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{9}{4}\end{matrix}\right.\)

  Vậy \(S=\left\{\dfrac{2}{3};-\dfrac{9}{4}\right\}\)

a)(x-1)(5x+3)=(3x-8)(x-1)

\(\Leftrightarrow\)(x-1)(5x+3)-(3x-8)(x-1)=0

\(\Leftrightarrow\left(x-1\right)\left(5x-3-3x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\)

\(\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{1;\dfrac{5}{2}\right\}\)

a) Ta có: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

\(\Leftrightarrow5x^2+3x-5x-3=3x^2-3x-8x+8\)

\(\Leftrightarrow5x^2-2x-3=3x^2-11x+8\)

\(\Leftrightarrow5x^2-2x-3-3x^2+11x-8=0\)

\(\Leftrightarrow2x^2+9x-11=0\)

\(\Leftrightarrow2x^2+11x-2x-11=0\)

\(\Leftrightarrow x\left(2x+11\right)-\left(2x+11\right)=0\)

\(\Leftrightarrow\left(2x+11\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+11=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-11\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{11}{2}\\x=1\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{11}{2};1\right\}\)

b) Ta có: \(3x\left(25x+15\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow3x\cdot5\cdot\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow\left(5x+3\right)\left(15x-35\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+3=0\\15x-35=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\15x=35\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=\dfrac{7}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)

c) Ta có: \(\left(2-3x\right)\left(x-11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\Leftrightarrow2x-22-3x^2+33x=6x-15x^2-4+10x\)

\(\Leftrightarrow-3x^2+35x-22=-15x^2+16x-4\)

\(\Leftrightarrow-3x^2+35x-22+15x^2-16x+4=0\)

\(\Leftrightarrow12x^2+19x-18=0\)

\(\Leftrightarrow12x^2+27x-8x-18=0\)

\(\Leftrightarrow3x\left(4x+9\right)-2\left(4x+9\right)=0\)

\(\Leftrightarrow\left(4x+9\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+9=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-9\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{4}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{9}{4};\dfrac{2}{3}\right\}\)

4.(x-3)-(11+x)=58

4x-3-11-x=58

4x-x=58+3+11

3x=72

  x=72:3

  x=24

vậy x=24

4(x-11)-2(3x)=15-x

4x-44-6x=15-x

4x-6x+x=15+44

-x=59

 x=-59

vậy x=-59

b) ghi thiếu đề rui

\(4.\left(x-3\right)-\left(11+x\right)=58\)

 \(4x-12-11-x=58\)

 \(3x-23=58\)

\(x=27\)

           câu b) tương tự nha

c)   \(4x-44-6x=15-x\)

\(-44-2x=15-x\)

   \(-44=15-x+2x\)

      \(-44=15+x\)

     \(x=-59\)