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\(2\left(x-1\right)+3\left(3x-2\right)=x-4\)
\(2x-2+9x-6=x-4\)
\(2x+9x-x-2-6=-4\)
\(10x-2-6=-4\)
\(10x-2=2\)
\(10x=4\)
\(x=\frac{2}{5}\)
Vậy \(x=\frac{2}{5}\)
\(3\left(4-x\right)-2\left(x-1\right)=x+20\)
\(12-3x-2x+2=x+20\)
\(12-5x+2=x+20\)
\(12-5x-x+2=20\)
\(12-6x+2=20\)
\(12-6x=18\)
\(6x=-6\)
\(x=-1\)
Vậy \(x=-1.\)
\(4\left(2x+7\right)-3\left(3x-2\right)=24\)
\(8x+28-9x+6=24\)
\(8x-9x+28+6=24\)
\(-x+34=24\)
\(-x=-10\)
\(x=10\)
Vậy \(x=10\)
\(3\left(x-2\right)+2x=10\)
\(3x-6+2x=10\)
\(3x+2x-6=10\)
\(5x=16\)
\(x=\frac{16}{5}\)
Vậy \(x=\frac{16}{5}\)
2(x-1)+3(3x-2)=x-4
=>2x-2=9x-6-x+4=0
=>10x-4=0
=>x=\(\frac{2}{5}\)

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.

a)(x+1)+(x+2)+(x+3)+...+(x+20)=250
x+1+x+2+x+3+...+x+20 = 250
20x+(1+2+3+...+20) = 250
20x + 210 = 250
20x = 250 - 210
20x = 40
x = 40 : 20
x = 2
b)(x+2)+(x+4)+(x+6)+...+(x+20)=220
x+2+x+4+x+6+...+x+20=220
10x + (2+4+6+...+20)=220
10x + 110 = 220
10x = 220 - 110
10x = 110
x = 110 : 10
x = 11
(x + 1) + (x + 2) + ... + (x + 20) = 250
=> 20x + (1 + 2 + 3 + ... + 20) = 250
=> 20x + \(\frac{20.21}{2}\)= 250
=> 20x + 210 = 250
=> 20x = 40
=> x = 2
(x + 2) + (x + 4) + ... + (x + 20) = 220
=> 10x + (2 + 4 + ... + 20) = 220
=> 10x + \(\frac{\left[\left(20-2\right):2+1\right].\left(20+2\right)}{2}\)= 220
=> 10x + 110 = 220
=> 10x = 110
=> x = 11

3 ( x + 2 ) - 5 (x - 1 ) = - x + 10
=> 3x + 6 - 5x + 5 = - x + 10
=> 3x - 5x + x = 10 - 5 - 6
=> - x = -1
=> x = 1
- 4 ( 2x - 1 ) - 3 ( x + 2 ) = - 20
=> - 8x + 4 - 3x - 6 = - 20
=> - 8x - 3x = -20 - 4 + 6
=> - 11x = - 18
=> x = \(\frac{18}{11}\)
- 2 ( 3x + 2 ) - 2 ( 4x - 1 ) = 26
=> - 6x - 4 - 8x + 2 = 26
=> - 6x - 8x = 26 - 2 + 4
=> - 14x = 28
=> x = -2
\(\left(x+2\right)^2=4\)
=> \(\orbr{\begin{cases}x+2=2\\x+2=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}}\)
\(\left(x-3\right)^2=25\)
=> \(\orbr{\begin{cases}x-3=5\\x-3=-5\end{cases}\Rightarrow\orbr{\begin{cases}x=8\\x=-2\end{cases}}}\)
- 3 ( x + 2 ) = 0
=> x + 2 = 0
=> x = -2
4 ( x - 1 ) = 0
=> x - 1 = 0
=> x = 1

A = 1 x 2 + 2 x 3 + 3 x 4 + ... + 19 x 20
=> 3A = 1 x 2 x 3 + 2 x 3 x ( 4 - 1 ) + 3 x 4 x ( 5 - 2 ) + ... + 19 x 20 x ( 21 - 18 )
=> 3A = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + .... + 19 x 20 x 21 - 18 x 19 x 20
=> 3A = 19 x 20 x 21
=> A = 19 x 20 x 7
Vậy A = 19 x 20 x 7
\(3\left(4-x\right)-2\left(x-1\right)=x+20\)
\(\Rightarrow12-3x-2x+2=x+20\)
\(\Rightarrow14-5x=x+20\)
\(\Rightarrow14-6x=20\)
\(\Rightarrow-6=6x\)
\(\Rightarrow x=-1\)