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1; 5.22 + (\(x\) + 3) = 52
5.4 + (\(x\) + 3) = 25
20 + (\(x\) + 3) = 25
\(x\) + 3 = 25 - 20
\(x+3\) = 5
\(x\) = 5 - 3
\(x\) = 2
Vậy \(x=2\)
2; 23 + (\(x\) - 32) = 53 - 43
8 + (\(x\) - 9) = 125 - 64
8 + (\(x\) - 9) = 61
\(x\) - 9 = 61 - 8
\(x\) - 9 = 53
\(x\) = 53 + 9
\(x\) = 62
Vậy \(x\) = 62
a) 5.2² + (x + 3) = 5²
5.4 + x + 3 = 25
20 + x + 3 = 25
x + 23 = 25
x = 25 - 23
x = 2
b) 2³ + (x - 3²) = 5³ - 4³
8 + (x - 9) = 125 - 64
8 + x - 9 = 61
x - 1 = 61
x = 61 + 1
x = 62
c) 4.(x - 5) - 2³ = 2⁴.3
4x - 20 - 8 = 16.3
4x - 28 = 48
4x = 48 + 28
4x = 76
x = 76 : 4
x = 19
d) 5.(x + 7) - 10 = 2³.5
5x + 35 - 10 = 8.5
5x + 25 = 40
5x = 40 - 25
5x = 15
x = 15 : 5
x = 3
e) 7² - 7.(13 - x) = 14
49 - 91 + 7x = 14
7x - 42 = 14
7x = 14 + 42
7x = 56
x = 56 : 7
x = 8
a) \(5\cdot2^2+\left(x+3\right)=5^2\)
\(\Rightarrow x+3=5^2-5\cdot2^2\)
\(\Rightarrow x+3=25-5\cdot4\)
\(\Rightarrow x+3=5\)
\(\Rightarrow x=5-3\)
\(\Rightarrow x=2\)
b) \(2^3+\left(x-3^2\right)=5^3-4^3\)
\(\Rightarrow8+\left(x-9\right)=125-64\)
\(\Rightarrow8+x-9=61\)
\(\Rightarrow x-1=61\)
\(\Rightarrow x=61+1\)
\(\Rightarrow x=62\)
c) \(4\left(x-5\right)-2^3=2^4\cdot3\)
\(\Rightarrow4\left(x-5\right)=2^4\cdot3+2^3\)
\(\Rightarrow4\cdot\left(x-5\right)=16\cdot3+8\)
\(\Rightarrow4\cdot\left(x-5\right)=56\)
\(\Rightarrow x-5=56:4\)
\(\Rightarrow x-5=14\)
\(\Rightarrow x=19\)
d) \(5\left(x+7\right)-10=2^3\cdot5\)
\(\Rightarrow5\left(x+7\right)=8\cdot5+10\)
\(\Rightarrow5\left(x+7\right)=40+10\)
\(\Rightarrow5\left(x+7\right)=50\)
\(\Rightarrow x+7=10\)
\(\Rightarrow x=10-7\)
\(\Rightarrow x=3\)
e) \(7^2-7\left(13-x\right)=14\)
\(\Rightarrow7\left(13-x\right)=7^2-14\)
\(\Rightarrow7\left(13-x\right)=49-14\)
\(\Rightarrow7\left(13-x\right)=35\)
\(\Rightarrow13-x=5\)
\(\Rightarrow x=13-5\)
\(\Rightarrow x=8\)
f) \(5x-5^2=10\)
\(\Rightarrow5x=10+5^2\)
\(\Rightarrow5x=10+25\)
\(\Rightarrow5x=35\)
\(\Rightarrow x=\dfrac{35}{5}\)
\(\Rightarrow x=7\)
g) \(9x-2\cdot3^2=3^4\)
\(\Rightarrow9x=3^4+2\cdot3^2\)
\(\Rightarrow9x=81+2\cdot9\)
\(\Rightarrow9x=99\)
\(\Rightarrow x=\dfrac{99}{9}\)
\(\Rightarrow x=11\)
h) \(10x+2^2\cdot5=10^2\)
\(\Rightarrow10x=10^2-2^2\cdot5\)
\(\Rightarrow10x=100-4\cdot5\)
\(\Rightarrow10x=80\)
\(\Rightarrow x=\dfrac{80}{10}\)
\(\Rightarrow x=8\)
i) \(125-5\left(4+x\right)=15\)
\(\Rightarrow5\left(4+x\right)=125-5\)
\(\Rightarrow5\left(4+x\right)=120\)
\(\Rightarrow4+x=\dfrac{120}{5}\)
\(\Rightarrow4+x=24\)
\(\Rightarrow x=24-4\)
\(\Rightarrow x=20\)
j) \(2^6+\left(5+x\right)=3^4\)
\(\Rightarrow5+x=3^4-2^6\)
\(\Rightarrow5+x=81-64\)
\(\Rightarrow5+x=17\)
\(\Rightarrow x=17-5\)
\(\Rightarrow x=12\)
a) 135 - 5(x + 4) = 35
<=> 135 - 5x - 20 = 35
<=> 115 - 5x = 35
<=> 5x = 115 - 35
<=> 5x = 80
<=> x = 16
b) 25 + 3 (x - 8) = 106
=> 3 ( x - 8) = 106 - 25
=> 3 ( x - 8) = 81
=> ( x - 8) = 81: 3
=> x - 8 = 27
=> x = 27 + 8
=> x = .......
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, 7\(x\) - \(x\) = 521 : 519 + 3.22.7
6\(x\) = 53 + 3.4.7
6\(x\) = 125 + 12.7
6\(x\) = 125 + 84
6\(x\) = 209
\(x\) = 209 : 6
\(x\) = \(\dfrac{209}{6}\)
b; 11\(x\) - 7\(x\) + 34 : 33 = 54 + 2\(x\)
4\(x\) + 3 = 625 + 2\(x\)
4\(x\) - 2\(x\) = 625 - 3
2\(x\) = 622
\(x\) = 622 : 2
\(x\) = 311
c; 75 - 5.(\(x-3\))3 = 700
5.(\(x\) - 3)3 = 700 - 75
5.(\(x\) - 3)3 = - 625
(\(x\) - 30)3 = - 625 : 5
(\(x\) - 30)3 = - 125
(\(x-3\))3 = (-5)3
\(x\) - 3 = - 5
\(x\) = - 5 + 3
\(x\) = -2
d, 3.(2\(x\) - 1)2 = 75
(2\(x\) - 1)2 = 75 : 3
(2\(x\) - 1)2 = 25
\(\left[{}\begin{matrix}2x-1=-5\\2x-1=5\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-5+1\\2x=5+1\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-4\\2x=6\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)