a) \(-28-7|-3x+15|=-70\)

\(\Rightarrow7|-3x+15|=42\)

\(\Rightarrow|-3x+15|=6\)

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

\(\Rightarrow|3|.|5-x|=6\)

\(\Rightarrow3|5-x|=6\)

\(\Rightarrow|5-x|=2\)

\(\Rightarrow\orbr{\begin{cases}5-x=2\\5-x=-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=7\end{cases}}\)

Vậy \(x\in\left\{3;7\right\}\)

b) \(|18-2|-x+5||=12\)

\(\Rightarrow\orbr{\begin{cases}18-2|-x+5|=12\\18-2|-x+5|=-12\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2|5-x|=6\\2|5-x|=30\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}|5-x|=3\left(1\right)\\|5-x|=15\left(2\right)\end{cases}}\)

Từ \(\left(1\right):\Rightarrow\orbr{\begin{cases}5-x=3\\5-x=-3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=8\end{cases}}\)

Từ \(\left(2\right):\Rightarrow\orbr{\begin{cases}5-x=15\\5-x=-15\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-10\\x=20\end{cases}}\)

Vậy \(x\in\left\{2;8;-10;20\right\}\)

c) \(12-2\left(-x+3\right)^2=-38\)

\(\Rightarrow2\left(3-x\right)^2=50\)

\(\Rightarrow\left(3-x\right)^2=100\)

\(\Rightarrow\orbr{\begin{cases}3-x=10\\3-x=-10\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-7\\x=13\end{cases}}\)

Vậy \(x\in\left\{-7;13\right\}\)

d) \(-20+3\left(2x+1\right)^3=-101\)

\(\Rightarrow3\left(2x+1\right)^3=-81\)

\(\Rightarrow\left(2x+1\right)^3=-27\)

\(\Rightarrow2x+1=-3\)

\(\Rightarrow2x=-4\)

\(\Rightarrow x=-2\)

Vậy \(x=-2\)

Trả lời:

a, -28 - 7| -3x + 15 | = -70

=> 7| -3x + 15 | = 42

=> | -3x + 15 | = 6

=> -3x + 15 = 6 hoặc -3x + 15 = -6

=>   -3x = -9                    -3x = -21

=>  x = 3                            x = 7

Vậy x = 3; x = 7

b, | 18 - 2 | -x + 5 || = 12

=> 18 - 2| -x + 5 | = 12 hoặc 18 -  2| -x + 5 | = -12

=> 2 | -x + 5 | = 6   hoặc    2 | -x + 5 | = 30

=> | -x + 5 | = 3    hoặc      | -x + 5 | = 15

=>  -x + 5 = 3 hoặc -x + 5 = -3 hoặc -x + 5 = 15 hoặc -x + 5 = -15

=>  x = 2                     x = 8                  x = -10               x = 20

Vậy x \(\in\){ 2; 8; -10; 20 }

c, 12 - 2.( -x + 3 )² = -38

=> 2.( -x + 3 )² = 50

=> ( -x + 3 )² = 25

=> -x + 3 = 5 hoặc -x + 3 = -5

=>   x = -2                x = 8

Vậy x = -2; x = 8

d, -20 + 3.( 2x + 1 )³ = -101

=> 3.( 2x + 1)³ = -81

=> ( 2x + 1 )³ = -27

=> 2x + 1 = -3 

=> 2x = -4

=> x = -2

Vậy x = -2

=> x = 1

Bài 1: 

a) Ta có: \(\left(2x-1\right)^{20}=\left(2x-1\right)^{18}\)

\(\Leftrightarrow\left(2x-1\right)^{20}-\left(2x-1\right)^{18}=0\)

\(\Leftrightarrow\left(2x-1\right)^{18}\left[\left(2x-1\right)^2-1\right]=0\)

\(\Leftrightarrow\left(2x-1\right)^{18}\cdot\left(2x-2\right)\cdot2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

b) Ta có: \(\left(2x-3\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

c) Ta có: \(\left(x-5\right)^2=\left(1-3x\right)^2\)

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

\(\Leftrightarrow\left(x-5-3x+1\right)\left(x-5+3x-1\right)=0\)

\(\Leftrightarrow\left(-2x-4\right)\left(4x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)

Bài 2: 

a) \(15^{20}-15^{19}=15^{19}\left(15-1\right)=15^{19}\cdot14⋮14\)

b) \(3^{20}+3^{21}+3^{22}=3^{20}\left(1+3+3^2\right)=3^{20}\cdot13⋮13\)

c) \(3+3^2+3^3+...+3^{2007}\)

\(=3\left(1+3+3^2\right)+...+3^{2005}\left(1+3+3^2\right)\)

\(=13\left(3+...+3^{2005}\right)⋮13\)

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

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)

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

=> x+2=2x

=> 2=2x-x

=>x=2

tương tự

chịu

Chịu 

Chịu thôi🙂