e) \(\left(x+3\right)^3=\left(2x\right)^3\)

\(\Rightarrow x+3=2x\)

\(\Rightarrow2x-x=3\)

\(\Rightarrow x=3\)

f) \(\left(5-x\right)^5=32\)

\(\Rightarrow\left(5-x\right)^5=2^5\)

\(\Rightarrow5-x=2\)

\(\Rightarrow x=5-2\)

\(\Rightarrow x=3\)

g) \(\left(5x-6\right)^3=64\)

\(\Rightarrow\left(5x-6\right)^3=4^3\)

\(\Rightarrow5x-6=4\)

\(\Rightarrow5x=4+6\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=\dfrac{10}{2}\)

\(\Rightarrow x=5\)

h) \(5\cdot9^x=405\)

\(\Rightarrow9^x=\dfrac{405}{5}\)

\(\Rightarrow9^x=81\)

\(\Rightarrow9^x=9^2\)

\(\Rightarrow x=2\)

i) \(11^5:11^{n-2}=11^5\)

\(\Rightarrow11^{n-2}=11^5:11^5\)

\(\Rightarrow11^{n-2}=1\)

\(\Rightarrow11^{n-2}=11^0\)

\(\Rightarrow n-2=0\)

\(\Rightarrow n=2\)

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

\(\Rightarrow3x=2x+1\)

\(\Rightarrow3x-2x=1\)

\(\Rightarrow x=1\)

\(-x+3=2x+9\\ \Rightarrow-x-2x=9-3\\ \Rightarrow-3x=6\\ \Rightarrow x=-\dfrac{6}{3}\\ \Rightarrow x=-2\)

Vậy \(x=-2\)

\(-x-5=-2x+11\\ \Rightarrow-x+2x=11+5\\ \Rightarrow x=16\)

Vậy \(x=16\)

b: =>15-x=-10

hay x=25

a: =>-2x+17=9

=>-2x=-8

hay x=4

d: \(\Leftrightarrow9x^2=81\)

hay \(x\in\left\{3;-3\right\}\)

e: \(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\3-x=0\end{matrix}\right.\Leftrightarrow x\in\left\{2;3\right\}\)

Lằng nhằng vậy😅

đợi xíu

Giúp mình với mình cần gấp lắm

\(2x+11=3\left(x-9\right)\)

\(\Leftrightarrow2x+11=3x-27\)

\(\Leftrightarrow x=11+27=38\)

Ta có: 2x+11=3(x-9)

\(\Leftrightarrow3x-27=2x+11\)

\(\Leftrightarrow3x-2x=11+27\)

hay x=38

Vậy: x=38

a; -2\(x\) - 3.(\(x-17\)) = 34 - 2.( - \(x\) + 25)

    - 2\(x\) - 3\(x\) + 51 = 34 + 2\(x\) - 50

       2\(x\) + 2\(x\) + 3\(x\) = - 34 + 50 + 51

            7\(x\)            = 67

               \(x\)           =  67 : 7

                \(x\)         =  \(\dfrac{67}{7}\)

Vậy \(x\) = \(\dfrac{67}{7}\) 

b; 17\(x\) + 3.(- 16\(x\) - 37) = 2\(x\) + 43 - 4\(x\)

    17\(x\) - 48\(x\)  - 111 =  2\(x\) - 4\(x\) + 43

    - 31\(x\) - 2\(x\) + 4\(x\) = 111 + 43

        - \(x\) x (31 + 2 - 4) = 154

        - \(x\) x (33 - 4) =  154

         - \(x\) x 29 = 154

         - \(x\)         = 154 : (-29)

           \(x\)        = - \(\dfrac{154}{29}\)

          Vậy \(x=-\dfrac{154}{29}\) 

c; {-3\(x\) + 2.[45 - \(x\) - 3(3\(x\) + 7) - 2\(x\)] + 4\(x\) } = 55

    {-3\(x\) + 2.[45 - \(x\) - 9\(x\) - 21 - 2\(x\)] + 4\(x\)} = 55

    {- 3\(x\) + 2.[(45 - 21) - (\(x+9x\)+2\(x\))] + 4\(x\) } = 55

     { (- 3\(x\) + 4\(x\)) + 2.[24 - 12\(x\)] } = 55

                   \(x\) + 48 - 24\(x\) = 55

                   \(x-24x\)         = 55 - 48

                     - 23\(x\)           = 7

                            \(x\)           = 7 : - 23

                             \(x=-\dfrac{7}{23}\)

Vậy \(x=-\dfrac{7}{23}\) 

Tìm x : 

a) | x + 12x | = 2x

=> \(\orbr{\begin{cases}13x=2x\\13x=-2x\end{cases}}\)

=>  \(\orbr{\begin{cases}11x=0\\15x=0\end{cases}}\)

=>  \(x=0\)

b) 3x − |x + 1| = 1

=> |x + 1|  = 3x -1

=>\(\orbr{\begin{cases}x+1=3x-1\\x+1=1-3x\end{cases}}\)

=>  \(\orbr{\begin{cases}2x=2\\4x=0\end{cases}}\)

=>  \(\orbr{\begin{cases}x=1\\x=0\end{cases}}\)

c) |2x + 3| = x + 1

=> \(\orbr{\begin{cases}2x+3=x+1\\2x+3=-x-1\end{cases}}\)

=>  \(\orbr{\begin{cases}x=-2\\3x=-4\end{cases}}\)

=> \(\orbr{\begin{cases}x=-2\\x=-\frac{4}{3}\end{cases}}\)

b) 3x - |x + 1| = 1

<=> |x + 1| = 3x - 1 (1)

ĐK : \(x\ge\frac{1}{3}\)

Khi đó (1) <=> \(\orbr{\begin{cases}x+1=3x-1\\x+1=-3x+1\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=2\\4x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(\text{loại}\right)\\x=1\end{cases}}\)

Vậy x = 1

c) ĐK : x + 1\(\ge0\Rightarrow x\ge-1\)

Khi đó |2x + 3| = x + 1

<=> \(\orbr{\begin{cases}2x+3=x+1\\2x+3=-x-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{4}{3}\end{cases}}\left(\text{loại}\right)\)

Vậy \(x\in\varnothing\)

d) ||x + 9| + 11| = 2x + 11 (1)

ĐK : \(2x+11\ge0\Rightarrow x\ge-\frac{5}{2}\)

Khi đó (1) <=> \(\orbr{\begin{cases}\left|x+9\right|+11=2x+11\\\left|x+9\right|+11=-2x-11\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left|x+9\right|=2x\\\left|x+9\right|=-2x-22\end{cases}}\)

Khi |x + 9| = 2x (x \(\ge0\))

<=> \(\orbr{\begin{cases}x+9=2x\\x+9=-2x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\left(tm\right)\\x=-3\left(\text{loại}\right)\end{cases}}\)

Khi |x + 9| = -2x - 22 ( \(-\frac{5}{2}\le x\le-11\))

<=> \(\orbr{\begin{cases}x+9=-2x-22\\x+9=2x+22\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{31}{3}\\x=-13\end{cases}}\left(\text{loại}\right)}\)

Vậy x = 9 

đúng. Mik học qua lớp anh rồi lên mik biết

a: =>-2x=9-17=-8

hay x=4

b: =>15-x=-10

hay x=25

d: \(\Leftrightarrow9x^2=81\)

hay \(x\in\left\{3;-3\right\}\)

e: \(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\3-x=0\end{matrix}\right.\Leftrightarrow x\in\left\{2;3\right\}\)

2x + 11 = 3(x - 9)

=> 2x + 11 = 3x - 27

=> 2x - 3x = -11 - 27

=> -x = -38

=> x = 38

=>3x-2x=27+11

=>x=38

x= 38

còn giải thì có người giải rùi mk rảnh