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a, 20^2 - 6^ 2= 400 - 36 = 364
b,3^3 . 18 - 3^3 . 12 = 3^3 . ( 18-12) = 27 . 6 = 162
c, 39 . 213 + 87 . 39 = ( 213 + 87) . 39 = 300 . 39 = 110700
d, 80 - [ 130 - ( 12 - 4)^2] = 80 - [130 - (3^2) ] = 80 - 130 - 9 = -59
5^2 * 2^3 -5x = 5^99 : 5^97
25 * 8 - 5x = 5^2
200 - 5x = 25
5x = 200 - 25
5x =175
x = 175 : 5
x = 35
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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.
\(\left(5^2\times2^3-7^3\times2\right)\div2\times6-7\div2^5\)
\(=\left(25\times8-343\times2\right)\div2\times6-7\div32\)
\(=\left(200-686\right)\div2\times6-7\div32\)
\(=-486\div2\times6-7\div32\)
\(=-1458-0,21875\)
\(=-1458,21875\)
(7*x-11)^3 = 2^5 * 5^2 +200
(7*x-11)^3 = 32*25+200
(7*x-11)^3=800+200
(7*x-11)^3=1000
=>(7*x-11)^3=(1000)^3
7*x-11=1000
7x=1000+11
7x=1011
x=1011/7
(7.x-11)^3=2^5.5^2+200
(7.x-11)^3=32.25+200
(7.x-11)^3=800+200
(7.x-11)^3=1000
(7.x-11)^3=10^3
7x-11=10
7x=10+11
7x=21
x=21/7
x=3
`5^3` `-` `3x` `= 110`
`=>` `125` `- 3x` `= 110`
`=>` `3x` `=` `125` `- 110`
`=>` `3x` `= 15`
`=>` `x` `= 15` `: 3`
`=>` `x` `= 5`
Vậy `x` `= 15`
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