12+(5+x)=20

5+x=8

    x=8/5

vãy=8/5

12 + ( 5 + x ) = 20              5.2² + ( x + 3 ) = 5²                         2³ + ( x + 3 ) = 5²                                 4³ - ( x - 2 ) = 5²

17 + x = 20                        5.4 + x + 3 = 25                               8 + x + 3 = 25                                  64 - x + 2 = 25

x = 20 - 17                         20 + 3 + x = 25                                11 + x = 25                                       66 - x = 25

x = 3                                  23 + x = 25                                       x = 25 - 11                                        x = 66 - 25

                                          x = 25 - 23                                        x = 14                                               x = 41

                                          x = 2

Đăng nhìu v bn :) Đáng quan ngại đây :)

1) \(2^x-15=17\)

\(\Leftrightarrow2^x=32=2^5\)

\(\Rightarrow x=5\)

2) \(\left(7x-11\right)^3=25\cdot5^2+200\)

\(\Leftrightarrow\left(7x-11\right)^3=825\)

\(\Leftrightarrow7x-11=\sqrt[3]{825}\)

\(\Leftrightarrow7x=11+\sqrt[3]{825}\)

\(\Rightarrow x=\frac{11+\sqrt[3]{825}}{7}\)

3) \(\left(x+1\right)^{100}-3\left(x+1\right)^{99}=0\)

\(\Leftrightarrow\left(x+1\right)^{99}\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^{99}=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

4) \(4x+5\left(x+3\right)=105\)

\(\Leftrightarrow9x+15=105\)

\(\Leftrightarrow9x=90\)

\(\Rightarrow x=10\)

5) \(5\cdot\left(x-2\right)+10\left(x+3\right)=170\)

\(\Leftrightarrow5\left[x-2+2\left(x+3\right)\right]=170\)

\(\Leftrightarrow3x+4=34\)

\(\Leftrightarrow3x=30\)

\(\Rightarrow x=10\)

a ) 4⁵ . 4 . 4⁷ = 4⁵ . 4¹ . 4⁷ = 4¹³

b ) 4 . 4³ . 4⁵ . 4⁶ = 4¹ . 4³ . 4⁵ . 4⁶ = 4¹⁵

c ) x . x³ . x⁴ = x¹ . x³ . x⁴ = x⁸

d ) x⁵ . x⁴ . x⁷ . x⁶ = x²²

Bài 2

a) x=5

b)x=1

\(x^4\cdot x^7\cdot...\cdot x^{100}\)

\(=x^{4+7+...+100}\)

\(=x^{52\cdot33}=x^{1716}\)

\(x^1\cdot x^2\cdot x^3\cdot...\cdot x^{2006}\)

Ta có : \(x^1\cdot x^2=x^{1+2}=x^3\)

Tương tự : \(x^1\cdot x^2\cdot x^3=x^{1+2+3}=x^6\)

Áp dụng vào bài toán :

\(x^1\cdot x^2\cdot x^3\cdot...\cdot x^{2006}=x^{1+2+3+...+2006}\)

\(\Rightarrow x^{1+2+3+...+2006}=x^{2013021}\)

a, 100 - 7 ( x - 5 ) = 31 + 3³

100 - 7 ( x - 5 ) = 31 + 27

100 - 7 ( x - 5 ) = 58

7 ( x - 5 ) = 100 - 58

7 ( x - 5 ) = 42

x - 5 = 42 : 7

x - 5 = 6

=> x = 6 +5

=> x = 11

Vậy x = 11

b, 12 ( x - 1 ) : 3 = 4³ + 2³

12 ( x - 1 ) : 3 = 64 + 8

12 ( x - 1 ) : 3 = 72

12 ( x - 1 ) = 72 . 3

12 ( x - 1 ) = 216

x - 1 = 216 : 12

x - 1 = 18

=> x = 18 + 1

=> x = 19

Vậy x = 19

c, 24 + 5x = 7⁵ : 7³

24 + 5x = 7²

24 + 5x = 49

5x = 49 - 24

5x = 25

=> x = 25 : 5

=> x = 5

Vậy x = 5

d, 5x - 206 = 2⁴ . 4

5x - 206 = 16 . 4

5x - 206 = 64

5x = 64 + 206

5x = 270

=> x = 270 : 5

=> x = 54

Vậy x = 54

e, 125 = x³

5³ = x³

=> x = 5

Vậy x = 5

g, 64 = x²

8² = x²

=> x = 8

Vậy x = 8

giúp mình với

bài 1) thực hiện phép tính

a) \(4^5-81:3^2=1024-9=1015\)

b) \(3^2.22-3^2.12=3^2.2\left(11-6\right)=18.5=90\)

c) \(2^3.15-\left[115-\left(12-5\right)\right]=120-\left(115-12+5\right)=120-115+12-5=12\)

d) \(3.3^2-19^{21}:19^{20}+2010^0=27-19+1=9\)

e)\(7^{25}:\left(7^{21}.46+7^{21}.3\right)=7^{25}:\left(7^{21}.49\right)=\frac{7^{21}.7^4}{7^{21}.49}=\frac{2401}{49}=49\)

bài 2)Tim x

a) 716 - (x - 143) = 695

<=> x - 143 = 716 - 695

<=> x - 143 = 21

<=> x = 21 + 143

<=> x = 164

vậy x = 164

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.