a: Ta có: \(60-3\left(x-2\right)=51\)

\(\Leftrightarrow3\left(x-2\right)=9\)

\(\Leftrightarrow x-2=3\)

hay x=5

b: Ta có: \(4x-20=2^5:2^3\)

\(\Leftrightarrow4x=24\)

hay x=6

c, 10 + 2 . x = 4⁵ : 4³

    10 + 2 . x = 4²

    10 + 2 . x = 16

            2 . x = 16 - 10

            2 . x = 6

                 x = 6 : 2

                 x = 3

             Vậy x = 3.

Đây nha bạn!!!

Chúc bạn học tốt!!!

may cai bai day ma cung khong biet oc cho

a) 89-(73-x)=20

=>73-x=89-20

=>73-x=69

=>x=73-69

=>x=4

Dài quá bn ơi, vs lại mấy cái này cx ko khó lắm đâu (nếu ko lm đc đăng mỗi lần một ít thui chứ dài quá mk đọc cn lười chứ đừng ns là làm)

a. 3^x+3^x+1+3^x+2=1053

=> 3^x.(1+3+3²)=1053

=> 3^x.(1+3+9)=1053

=> 3^x.13=1053

=> 3^x=1053:13

=> 3^x=81

=> 3^x=3⁴

=> x=4

b. 5^x.5¹⁹=5²⁰.5¹¹

=> 5^19+x=5²⁰⁺¹¹

=> 19+x=20+11

=> 19+x=31

=> x=31-19

=> x=12

\(3^x+3^{x+1}+3^{x+2}=1053\)

\(3^x\left(1+3+3^2\right)=1053\)

\(3^x\cdot13=1053\)

\(3^x=1053:13=81\)

\(3^x=3^4\)

=> x = 4 

A)x=4

B)x=12 100% đó

a) 2² ( x + 3² ) - 5 = 55

=>4(x+9)-5=55

=>4(x+9)=60

=>x+9=15

=>x=6

b)7 ( x + 5² ) - 20 = 190

=>7(x+25)=210

=>x+25=30

=>x=5

c)6 ( x + 2³ ) + 40 = 100

=>6(x+8)+40=100

=>6(x+8)=60

=>x+8=10

=>x=2

a)6

b)5

c)2

6;5;2

Bài 5 : 

f, bạn xem lại đề hay là tìm x chứa tham số a ? 

g, \(x^2+3x-\left(2x+6\right)=0\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)

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

h, \(5x+20-x^2-4x=0\Leftrightarrow5\left(x+4\right)-x\left(x+4\right)=0\)

\(\Leftrightarrow\left(5-x\right)\left(x+4\right)=0\Leftrightarrow x=-4;x=5\)

m, \(x^3-5x^2-x+5=0\Leftrightarrow x^2\left(x-5\right)-\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-5\right)=0\Leftrightarrow x=\pm1;x=5\)

n, \(x\left(x-3\right)-7x+21=0\Leftrightarrow x\left(x-3\right)-7\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-3\right)=0\Leftrightarrow x=3;x=7\)

x=7 nha

1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅

3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1

5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)

6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅

7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅

8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1

9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)

\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)

\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

Câu 3, 4 tương tự nhé.

\(5.\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Leftrightarrow\left(x-5\right)^{2018}-\left(x-5\right)^{2016}=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left[\left(x-5\right)^2-1\right]=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left(x-5-1\right)\left(x-5+1\right)=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left(x-6\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^{2016}=0\\x-6=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x=6\\x=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)

Vậy \(x\in\left\{4;5;6\right\}\)

b ) 20 + 5x = 5⁵ : 5³

20 + 5x = 5²

20 + 5x = 25

5x = 25 - 20

5x = 5

x = 5 : 5 

x = 1

c ) 5x - 201 = 2⁴.4

5x - 201 = 16.4

5x - 201 = 64

5x = 64 + 201

5x = 265

x = 265 : 5

x = 53

1,a) 695- [200+ (11- 1²)]

    = 695- [200+ (11- 1)]

    = 695- [200+ 10]

    = 695- 210

    = 485

  b) (5¹⁹: 5¹⁷+ 3): 7

   = (5²+ 3): 7

   = (25+ 3): 7

   = 28: 7

   = 4

  c) 129- 5[29- (6- 1²)]

   = 129- 5[29- (6- 1)]

   = 129- 5[29- 5]

   = 129- 5. 24

   = 129- 120

   = 9

3,a) 2x- 49= 5. 3²

       2x- 49= 5. 9

      2x- 49= 45

      2x     = 45+ 49

      2x     = 94

      x       = 94: 2

      x       = 47

  c) 2^x- 15= 17

     2^x       = 17+ 15

     2^x        = 32

     2^x      = 2⁵

 => x       = 5

Câu 3b bạn tự làm nhé, xin loiosxn vì không giúp được cả bài.

CHÚC BẠN HỌC GIỎI !!!

MÌNH TÌM RA CÁCH LÀM CÂU 3b RỒI !!!

5x+ 2x= 45+ 20: 15

5x+ 2x= 45+ \(\frac{4}{3}\)

5x+ 2x= \(\frac{139}{3}\)

(5+ 2)x=\(\frac{139}{3}\)

7x       =\(\frac{139}{3}\)

x         =\(\frac{139}{3}\): 7

x         =\(\frac{139}{21}\)

CHÚC BẠN HỌC GIỎI !!!