5 + ( x + 27 ) = 64

a) \(\Rightarrow x+27=59\Rightarrow x=32\)

b) \(\Rightarrow x-2=39\Rightarrow x=41\)

c) \(\Rightarrow x+5=-322\Rightarrow x=-327\)

d) \(\Rightarrow5x=35\Rightarrow x=7\)

e) \(\Rightarrow4\left(x-5\right)=56\Rightarrow x-5=14\Rightarrow x=19\)

f) \(\Rightarrow15+x=37\Rightarrow x=22\)

g) \(\Rightarrow7\left(13-x\right)=35\Rightarrow13-x=5\Rightarrow x=8\)

h) \(\Rightarrow10\left(x+1\right)=100\Rightarrow x+1=10\Rightarrow x=9\)

x2x2 là sao bn

`(5x+1)=36/49`

`<=> 5x = 36/49-1`

`<=> 5x = -13/49`.

`<=> x = -13/245.`

Vậy `x = -13/245`.

`b, x-2/9 = 2/3`.

`<=> x = 2/3 + 2/9`

`<=> x = 8/9`.

Vậy `x = 8/9`.

c: (8x-1)^(2x+1)=5^(2x+1)

=>8x-1=5

=>8x=6

=>x=3/4

d: Sửa đề: (x-3,5)^2+(y-1/10)^4=0

=>x-3,5=0 và y-0,1=0

=>x=3,5 và y=0,1

a) \(\left(x+1\right)^3-\left(x-1\right)^3-6\cdot\left(x-1\right)^2=10\)

\(\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x^2-2x+1\right)=10\)

\(\Rightarrow6x^2+2-6x^2+12x-6=10\)

\(\Rightarrow12x-4=10\)

\(\Rightarrow12x=14\)

\(\Rightarrow x=\dfrac{7}{6}\)

b) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)

\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)

\(\Rightarrow x^3-25x-x^3-8=42\)

\(\Rightarrow-25x-8=42\)

\(\Rightarrow-25x=50\)

\(\Rightarrow x=\dfrac{50}{-25}=-2\)

c) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)

\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)

\(\Rightarrow24x+25=49\)

\(\Rightarrow24x=24\)

\(\Rightarrow x=\dfrac{24}{24}=1\)

a) \(3\left(x-1\right)^2\cdot3x\left(x-5\right)=0\)

\(\Rightarrow9x\left(x-1\right)^2\left(x-5\right)=0\)

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

b) \(\left(x+3\right)^2-5x-15=0\)

\(\Rightarrow\left(x+3\right)^2-5\left(x+3\right)=0\)

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

\(\Rightarrow\left(x+3\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

c) \(2x^5-4x^3+2x=0\)

\(\Rightarrow2x\left(x^4-2x^2+1\right)=0\)

\(\Rightarrow2x\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2\right]=0\)

\(\Rightarrow2x\left(x^2-1\right)^2=0\)

\(\Rightarrow2x\left(x-1\right)^2\left(x+1\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

\(\text{#}Toru\)

a) \(\left(2x-5\right)^2=49\)    

     \(\left(2x-5\right)^2=\left(\pm7\right)^2\)   

      \(=>2x-5=7\)  hoặc \(2x-5=-7\)  

      \(\cdot2x-5=7\)                                        \(\cdot2x-5=-7\)   

       \(2x=5+7\)                                          \(2x=-7+5\)    

       \(2x=12\)                                                \(2x=-2\)   

         \(x=12:2\)                                             \(x=-2:2\)  

         \(x=6\)                                                     \(x=-1\)  

   Vậy x=6 hoặc x=-1

b/ \(\left(2x+5\right)^2-\left(1-2x\right)^2=10\)   

     \(4x^2+20x+25-\left(1-4x+4x^2\right)=10\)   

     \(4x^2+20x+25-1+4x-4x^2=10\)   

                  \(24x+24=10\)   

                   \(24x=10-24\)   

                    \(24x=-14\)    

                         \(x=\frac{-14}{24}\)

                          \(x=\frac{-7}{12}\)   

c/ \(\left(9-2x\right)^3=27\)   

     \(\left(9-2x\right)^3=3^3\)

      \(9-2x=3\)   

       \(2x=9-3\)

        \(2x=6\)

         \(x=6:2\)

         \(x=3\)

a/ (2x-5)^2=49

TH1: 2x-5=7

x=6

TH2: 2x-5=-7

x=-1

Ta có

(2x-1)^10 = 49^5

=> (2x-1)^10 = (7^2)^5 

=> (2x-1)^10 = 7^10

=> 2x-1 = 7

=> 2x= 8

=> x=4 

(2x-1)¹⁰=49⁵

=> (2x-1)¹⁰=(7²)⁵=[(-7)² ]⁵

=> (2x-1)¹⁰=7¹⁰=(-7)¹⁰

=> 2x-1=7           hoặc 2x-1=-7

=> 2x=8              hoặc 2x=-6

=> x=4               hoặc  x=-3 

\(a,2x+\dfrac{5}{2}=-\dfrac{3}{5}\)

\(2x=-\dfrac{3}{5}-\dfrac{5}{2}\)

\(2x=-\dfrac{31}{10}\)

\(x=-\dfrac{31}{10}:2\)

\(x=-\dfrac{31}{20}\)

\(b,\dfrac{1}{2}:x-\dfrac{5}{6}=-\dfrac{2}{3}\)

\(\dfrac{1}{2}:x=-\dfrac{2}{3}+\dfrac{5}{6}\)

\(\dfrac{1}{2}:x=\dfrac{1}{6}\)

\(x=\dfrac{1}{2}:\dfrac{1}{6}\)

\(x=3\)

\(c,\left(312-x\right):12,6=24,5\)

\(312-x=24,5\times12,6\)

\(312-x=308,7\)

\(x=312-308,7\)

`x=3,3`

a: =>2x=-3/5-5/2=-6/10-25/10=-31/10

=>x=-31/20

b: =>1/2:x=-2/3+5/6=5/6-4/6=1/6

=>x=1/2:1/6=3

c: =>312-x=308,7

=>x=3,3

`3(x-1)(x-5) =0`

`<=> (x-1) =0` hoặc `x-5 = 0`.

`<=> x =1` hoặc `x = 5`.

Vậy `x = 1` hoặc `x = 5.`

`b, 3x^2 + 7x = 10`.

`<=> 3x^2 + 7x - 10 = 0`

`<=> (3x+10)(x-1) =0`

`<=> 3x + 10 = 0` hoặc `x - 1=0`

`<=> x = -10/3` hoặc `x = 1.`

Vậy `x = -10/3` hoặc `x = 1.`