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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.
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/ a) \(2.3.12.12.3=2.3.2^2.3.2^2.3.3=2^5.3^4\)
b) \(3.5.27.125=3.5.3^3.5^3=3^4.5^4=\left(3.5\right)^4\)
2/ a) \(\left(27^3\right)^4=27^{3.4}=27^{12}\)
Vậy \(\left(27^3\right)^4=27^{12}\)
b) \(5^{36}=\left(5^6\right)^6\) và \(11^{24}=\left(11^4\right)^6\)
Do đó \(5^6=15625\) và \(11^4=14641\)
Vì 15625>14641 nên\(\left(5^6\right)^6>\left(11^4\right)^6hay5^{36}>11^{24}.\)
3/ a) \(x^3=125=>x=5\)
b) \(\left(3x-14\right)^3=2^5.5^2+200\)
\(\left(3x-14\right)^3=1000\)
\(3x-14=10^3\)
\(3x=10^3+14\)
\(3x=1014\)
\(x=\frac{1014}{3}=338\)
c) \(\left(2x-1\right)^4=81\)
\(\left(2x-1\right)^4=3^4\)
\(2x-1=3\)
\(2x=3+1\)
\(x=\frac{4}{2}=2\)
d) \(5x+3^4=2^2.7^2\)
\(5x+3^4=\left(2.7\right)^2=14^2\)
\(5x+81=196\)
\(5x=196-81\)
\(5x=115\)
\(x=\frac{115}{5}=23\)
e) \(4^x=1024=>x=5\).
