vừa nãy mk cũng ko vào đc 

2x-3x +5 = 2(3x-3)+17

\(\Leftrightarrow-x+5=6x-6+17\)

\(\Leftrightarrow-x-6x=-6+17-5\)

\(\Leftrightarrow-7x=6\)

\(\Leftrightarrow=-\frac{6}{7}\)

2x-3x+5=2(3x-3)+17

2x-3x+5=6x-6+17

2x-3x-6x=-6+17-5

x(2-3-6)=6

x(-7)=6

x=-6/7

kb nha

''học tốt''

ĐKXĐ: ...

\(\Leftrightarrow3x-1-x\sqrt{3x-1}+x\sqrt{x+1}-\sqrt{\left(x+1\right)\left(3x-1\right)}=0\)

\(\Leftrightarrow\sqrt{3x-1}\left(\sqrt{3x-1}-x\right)-\sqrt{x+1}\left(\sqrt{3x-1}-x\right)=0\)

\(\Leftrightarrow\left(\sqrt{3x-1}-\sqrt{x+1}\right)\left(\sqrt{3x-1}-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x-1}=\sqrt{x+1}\\\sqrt{3x-1}=x\end{matrix}\right.\)

\(\Leftrightarrow...\)

ĐKXĐ: x \(\ge\)\(\dfrac{1}{3}\)

pt\(\Leftrightarrow\)x(\(\sqrt{x+1}-\sqrt{3x-1}\))+\(\sqrt{3x-1}\left(\sqrt{3x-1}-\sqrt{x+1}\right)\)=0

  \(\Leftrightarrow\)(\(\sqrt{x+1}-\sqrt{3x-1}\))(1-\(\sqrt{3x-1}\))=0

  \(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x+1}=\sqrt{3x-1}\\1=\sqrt{3x-1}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{2}{3}\end{matrix}\right.\)(t/m x \(\ge\)\(\dfrac{1}{3}\))

Vậy.....................

\(x\left(3-\sqrt{3x-1}\right)=\sqrt{3x^2+2x-1}-x\sqrt{x+1}+1\)(Đk x≥\(\dfrac{1}{3}\))

ta có:\(x\left(3-\sqrt{3x-1}\right)\)

=\(3x-x\sqrt{3x-1}\)

=\(3x-1-x\sqrt{3x-1}+1\)

=\(\sqrt{3x-1}\left(\sqrt{3x-1}-x\right)+1\)

Ta có \(\sqrt{3x^2+2x-1}-x\sqrt{x+1}+1\)

=\(\sqrt{x^2+2x+1-2+2x^2}-x\sqrt{x+1}+1\)

=\(\sqrt{\left(x+1\right)\left(3x-1\right)}-x\sqrt{x+1}+1\)

=\(\sqrt{x+1}\left(\sqrt{3x-1}-x\right)+1\)

ta có \(x\left(3-\sqrt{3x-1}\right)=\sqrt{3x^2+2x-1}-x\sqrt{x+1}+1\)

⇔\(\sqrt{3x-1}\left(\sqrt{3x-1}-x\right)+1\)=\(\sqrt{x+1}\left(\sqrt{3x-1}-x\right)+1\)

⇔\(\sqrt{3x-1}\left(\sqrt{3x-1}-x\right)=\sqrt{x+1}\left(\sqrt{3x-1}-x\right)\)

⇔​​\(\sqrt{3x-1}=\sqrt{x+1}\)

⇔​\(3x-1=x+1\)

⇔\(2x=2\)

⇔x=1(N)

​Vậy x=1

-3-2x+17=12-3x

-2x+3x=12-17+3

x=-2

b.43-2x-2=-48+x

  -2x-x = -48+2-43

   -3x=-89

    x=-89/-3

c  -253-2x=102+19-x

        -2x+x=102+19+253

         -x=374

             x=-374

ma doi

Nguyễn Khánh Phương

Bài 1 :

a) 149 - ( 35 : x + 3 ) x 17 = 13

              ( 35 : x + 3 ) x 17 = 149 - 13 

               ( 35 : x + 3 ) x 17 = 136

                ( 35 : x + 3 )       = 136 : 17

                  ( 35 : x + 3 )     = 8

                    35 - x             = 8 - 3 

                    35 - x             = 5

                            x = 35 - 5

                            x = 30

b, 121 : 11 − ( 4x + 5 )  : 3 = 4
11 − 4x + 5 : 3 = 4
4x + 5 : 3 = 11 − 4
4x + 5 : 3 = 7
4x + 5 = 7 x 3
4x + 5 = 21
4x = 21 − 5
4x = 16

x = 16 : 4

  x = 4

a,

149-(35:x+3)x17=13

(35:x+3)x17=149-13=136

35:x+3=136:17=8

35:x=8-3=5

x=35:5=7

b,

121:11-(4x+5):3=4

11-(4x+5):3=4

(4x+5):3=11-4=7

4x+5=7x3=21

4x=21-5=16

x=16:4=4

c,

mình không biết

d,\(x^{2017}=x^{2018}\)

=>x=1 hoặc x=0

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

\(\Rightarrow3\left(x-1\right)-3\left(x+2\right)=2\left(x+2\right)-2x\left(2+3x\right)\)

\(\Rightarrow3\left(x-1-x-2\right)=2\left(x+2\right)-2\left(2x+3x^2\right)\)

\(\Rightarrow3\left(-3\right)=2\left(x+2-2x-3x^2\right)\)

\(\Rightarrow-9=2\left(2-x-3x^2\right)\)

\(\Rightarrow2-x-3x^2=-4,5\)

\(\Rightarrow x-3x^2=6,5\)(hình như sai đề)

Sai đề nha mn

Ko cần phải giải đâu,lên giải giúp mk câu trên nhá 

ukm mk cũng thấy nó kì 

ĐKXĐ \(X\ne-3\)

\(=\frac{4X+26}{X+3}=\frac{4X+12+14}{X+3}=4+\frac{14}{X+3}.\)

để bt trên nguyên thì \(x+3\inƯ\left(14\right)\in\left\{\mp1;\mp2;\mp7;\mp14\right\}\)

đến đây bn tự giải ha nếu có x=-3 thì loại còn lấy hêt ha