a) 71 - (33 + x) = 26

⇔ 71 - 33 - x = 26

⇔ 38 - x = 26

⇔ x = 38 - 26

⇔ x = 12

b) (x + 73) - 26 = 76

⇔ x + 73 - 26 = 76

⇔ x + 47 = 76

⇔ x = 76 - 47

⇔ x = 29

c) 45 - (x + 9) = 6

⇔ 45 - x - 9 = 6

⇔ 36 - x = 6

⇔ x = 36 - 6

⇔ x = 30

d) 89 - (73 - x) = 20

⇔ 89 - 73 + x = 20

⇔ 16 + x = 20

⇔ x = 20 - 16

⇔ x = 4

e) (x + 7) - 25 = 13

⇔ x + 7 - 25 = 13

⇔ x - 18 = 13

⇔ x = 13 + 18

⇔ x = 31

f) 198 - (x + 4) = 120

⇔ 198 - x - 4 = 120

⇔ 194 - x = 120

⇔ x = 194 - 120

⇔ x = 74

g) 140 : (x - 8) = 7

⇔ x - 8 = 140 : 7

⇔ x - 8 = 20

⇔ x = 20 + 8

⇔ x = 28

h) 4(x + 41) = 400

⇔ x + 41 = 400 : 4

⇔ x + 41 = 100

⇔ x = 100 - 41

⇔ x = 59

i) 11(x - 9) = 77

⇔ x - 9 = 77 : 11

⇔ x - 9 = 7

⇔ x = 7 + 9

⇔ x = 16

j) 5(x - 9) = 350

⇔ x - 9 = 350 : 5

⇔ x - 9 = 70

⇔ x = 70 + 9

⇔ x = 79

k) 2x - 49 = 5.3²

⇔ 2x - 49 = 45

⇔ 2x = 45 + 49

⇔ 2x = 94

⇔ x = 94 : 2

⇔ x = 47

l) 200 - (2x + 6) = 4³

⇔ 200 - 2x - 6 = 64

⇔ 194 - 2x = 64

⇔ 2x = 194 - 64

⇔ 2x = 130

⇔ x = 130 : 2

⇔ x = 65

a) 71 - (33 + x) = 26

⇔ 33+x=71-26

⇔ 33 + x = 45

⇔ x = 45 - 33

⇔ x = 12

Vậy x=12

b) (x + 73) - 26 = 76

⇔ (x + 73) = 76+26

⇔ x + 73 = 102

⇔ x = 102-73

⇔ x = 29

Vậy x=29

c) 45 - (x + 9) = 6

⇔ x + 9 = 45 - 6

⇔ x+9 = 39

⇔ x = 39 - 9

⇔ x = 30

Vậy x=30

d) 89 - (73 - x) = 20

⇔ 73 - x = 89 - 20

⇔ 73 - x = 69

⇔ x = 73 - 69

⇔ x = 4

Vậy x=4

e) (x + 7) - 25 = 13

⇔ x + 7 = 13 + 25

⇔ x + 7 = 38

⇔ x = 38 - 7

⇔ x = 31

Vậy x=31

f) 198 - (x + 4) = 120

⇔ x + 4 = 198 -120

⇔ x + 4 = 78

⇔ x = 78 - 4

⇔ x = 74

Vậy x=74

a: \(\Leftrightarrow73-x=67\)

hay x=6

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 \(x\in\left\{4;5;6\right\}\)

ko bt

Đc r

a: =>(2x+1/2)^2=1/4

=>2x+1/2=1/2 hoặc 2x+1/2=-1/2

=>x=-1/2 hoặc x=0

b: =>(x-1/5)^2=49

=>x-1/5=7 hoặc x-1/5=-7

=>x=-6,8 hoặc x=7,2

c: =>1,2x=12

=>x=10

d: =>3/4x+1/2x+1/2=-11/4

=>5/4x=-11/4-2/4=-13/4

=>x=-13/5

e: =>-0,25x+1,25x=0,2

=>x=0,2