3 mũ 1001 (5+2x) = 3 mũ 999
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 8
A = \(9999^{999^{99^9}}\)
Vì 999 không chia hết cho 2 nên \(999^{99^9}\) không chia hết cho 2
Vậy \(999^{99^9}\) = 2k + 1
A = 99992k+1
A = (99992)k.9999
A = \(\overline{...1}\)k. 9999
A = \(\overline{..1}\).9999
A = \(\overline{..9}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 8
B = vì 8 ⋮ 2 nên \(8^{7^{6^{5^{3^2}}}}\) ⋮ 2
Vậy B = 92k = (92)k = \(\overline{..1}\)k = \(\overline{..1}\)
2 tháng 8 2021
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: \(\left(x-5\right)^2=\left(1-3x\right)^2\)
\(\Leftrightarrow\left(x-5\right)^2-\left(3x-1\right)^2=0\)
\(\Leftrightarrow\left(x-5-3x+1\right)\left(x-5+3x-1\right)=0\)
\(\Leftrightarrow\left(-2x-4\right)\left(4x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)
2 tháng 8 2021
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) \(3+3^2+3^3+...+3^{2007}\)
\(=3\left(1+3+3^2\right)+...+3^{2005}\left(1+3+3^2\right)\)
\(=13\left(3+...+3^{2005}\right)⋮13\)
12 tháng 8 2016
a. \(12^2.3^2.2^3=2^4.3^2.3^2.2^3=2^7.3^4\)
b. \(8^3.3^2.6^3=2^9.3^2.2^3.3^3=2^{12}.3^5\)
c. \(5^{32}.5^2=5^{34}\)
d. \(100^6.2^3=\left(2^2.5^2\right)^6.2^3=2^8.5^8.2^3=2^{11}.5^8\)
e. \(100^2:10^2:5^2=\left(10.5.2\right)^2:10^2:5^2=2^2\)
f. \(121^3-11^2=11^6-11^2=11^2\left(11^4-1\right)\)
10 tháng 9 2020
a) \(\left(2x-5\right)^2-\left(2x+3\right)\left(2x-3\right)=10\Leftrightarrow\left(4x^2-20x+25\right)-\left(4x^2-9\right)-10=0\)
\(\Leftrightarrow-20x+24=0\Leftrightarrow x=\frac{6}{5}\)
b) \(\left(4x-1\right)\left(x+2\right)-\left(2x+3\right)^2-5\left(x-1\right)=9\Leftrightarrow-10x-15=0\)
\(\Leftrightarrow x=\frac{-3}{2}\)
c) \(\left(x+1\right)^3-\left(x-1\right)^3-2=6\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-8=0\)
\(\Leftrightarrow6x^2-6=0\Leftrightarrow x=\pm1\)
d) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x+1\right)\left(x^2-x+1\right)-3\left(-x-2\right)=5\)
\(\Leftrightarrow\left(x^3+8\right)-\left(x^3+1\right)+3x+6=5\Leftrightarrow3x+8=0\Leftrightarrow x=\frac{-8}{3}\)
10 tháng 10 2023
(5 mũ 2)mũ 3 : 5 mũ 2
5 mũ 2 x3 :5 mũ 2
5 mũ 6 : 5 mũ 2 =5 mũ 4=625
\(3^{1001}\cdot\left(5+2x\right)=3^{999}\)
\(\Rightarrow5+2x=\dfrac{3^{999}}{3^{1001}}\)
\(\Rightarrow5+2x=\dfrac{1}{3^2}\)
\(\Rightarrow5+2x=\dfrac{1}{9}\)
\(\Rightarrow2x=\dfrac{1}{9}-5\)
\(\Rightarrow2x=-\dfrac{44}{9}\)
\(\Rightarrow x=-\dfrac{44}{9}:2\)
\(\Rightarrow x=-\dfrac{22}{9}\)
\(3^{1001}.\left(5+2x\right)=3^{999}\)
\(\Rightarrow5+2x=\dfrac{3^{999}}{3^{1001}}\)
\(\Rightarrow5+2x=\dfrac{1}{3^2}\)
\(\Rightarrow5+2x=\dfrac{1}{9}\)
\(2x=\dfrac{1}{9}-5\)
\(x=-\dfrac{44}{9}:2\)
\(x=-\dfrac{22}{9}\)