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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.
\(4^8.2^{20}=2^{16}.2^{20}=2^{36}\)
\(9^{12}.27^5.81^4=3^{24}.3^{15}.3^{16}=3^{55}\)
mk chỉnh đề
\(64^3.4^5.16^2=4^9.4^5.4^4=4^{18}\)
\(25^{20}.125^4=5^{40}.5^{12}=5^{52}\)
\(x^7.x^4.x^3=x^{14}\)
\(A=5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9\)
\(A=5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9\)
\(A=5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}\)
\(A=5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}\)
\(A=2^{29}\cdot3^{18}\cdot\left(5\cdot2^1\cdot1-1\cdot3^2\right)\)
\(A=2^{29}\cdot3^{18}\cdot\left(5-9\right)\)
\(A=-2^2\cdot2^{29}\cdot3^{18}\)
\(A=-2^{31}\cdot3^{18}\)
_______________
\(B=5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6\)
\(B=5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6\)
\(B=5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}\)
\(B=2^{28}\cdot3^{18}\cdot\left(5\cdot1\cdot3-7\cdot2\cdot1\right)\)
\(B=2^{28}\cdot3^{18}\cdot\left(15-14\right)\)
\(B=2^{28}\cdot3^{18}\)
Ta có: \(A:B\)
\(=\left(-2^{31}\cdot3^{18}\right):\left(2^{28}\cdot3^{18}\right)\)
\(=\left(-2^{31}:2^{28}\right)\cdot\left(3^{18}:3^{18}\right)\)
\(=-2^3\cdot1\)
\(=-8\)
a. 3x+3x+1+3x+2=1053
=> 3x.(1+3+32)=1053
=> 3x.(1+3+9)=1053
=> 3x.13=1053
=> 3x=1053:13
=> 3x=81
=> 3x=34
=> x=4
b. 5x.519=520.511
=> 519+x=520+11
=> 19+x=20+11
=> 19+x=31
=> x=31-19
=> x=12
\(3^x+3^{x+1}+3^{x+2}=1053\)
\(3^x\left(1+3+3^2\right)=1053\)
\(3^x\cdot13=1053\)
\(3^x=1053:13=81\)
\(3^x=3^4\)
=> x = 4
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) \(3+3^2+3^3+...+3^{2007}\)
\(=3\left(1+3+3^2\right)+...+3^{2005}\left(1+3+3^2\right)\)
\(=13\left(3+...+3^{2005}\right)⋮13\)
a: Ta có: \(60-3\left(x-2\right)=51\)
\(\Leftrightarrow3\left(x-2\right)=9\)
\(\Leftrightarrow x-2=3\)
hay x=5
b: Ta có: \(4x-20=2^5:2^3\)
\(\Leftrightarrow4x=24\)
hay x=6
\(\left(3-x\right)^{10}:\left(3-x\right)^{20}=1\)
\(\Rightarrow\frac{1}{\left(3-x\right)^{10}}=1\)
\(\Rightarrow\left(3-x\right)^{10}=1\)
\(\Rightarrow3-x=\orbr{\begin{cases}1\\-1\end{cases}}\)
\(\Rightarrow x=\orbr{\begin{cases}2\\4\end{cases}}\)
a) \(a^2\cdot a^3\cdot a^7\cdot b^2\cdot b\)
\(=\left(a^2\cdot a^3\cdot a^7\right)\cdot\left(b^2\cdot b\right)\)
\(=a^{12}\cdot b^3\)
b) \(b^6\cdot b\cdot c^7\cdot c^8\)
\(=\left(b^6\cdot b\right)\cdot\left(c^7\cdot c^8\right)\)
\(=b^7\cdot c^{15}\)
c) \(a^8\cdot a^9\cdot a\cdot c\cdot c^{20}\)
\(=\left(a^8\cdot a^9\cdot a\right)\cdot\left(c\cdot c^{20}\right)\)
\(=a^{18}\cdot c^{21}\)
d) \(a^2\cdot a^3\cdot b^4\cdot c\cdot c^3\)
\(=\left(a^2\cdot a^3\right)\cdot b^4\cdot\left(c\cdot c^3\right)\)
\(=a^5\cdot b^4\cdot c^4\)
a) Kiểm tra lại nhé
b) \(b^6.b^7.c^8\)
\(=b^{6+7}.c^8=b^{13}.c^8\)
c) \(a^8.a^9.a.c.c^{20}\)
\(=a^{8+9+1}.c^{1+20}\)
\(=a^{18}.c^{21}\)
d) \(a^2.a^3.b^4.c.c^3\)
\(=a^{2+3}.b^4.c^{1+3}\)
\(=a^5.b^4.c^4\)
\(#WendyDang\)
\(x^{20}=x\)
=> \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
x20 = x
<=> x20 - x = 0
<=> x . ( x19 - 1 ) = 0
<=> x = 0 hoặc x19 - 1 = 0
<=> x = 0 hoặc x19 = 1
<=> x = 0 hoặc x = 1
Cre : olm '-'