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Ta có:\(A=3+3^2+3^3+...+3^{17}\)
\(3A=3\cdot\left(3+3^2+3^3+...+3^{17}\right)\)
\(3A=3^2+3^3+3^4+...+3^{18}\)
\(3A-A=\left(3^2+3^3+3^4+...+3^{18}\right)-\left(3+3^2+3^3+...+3^{17}\right)\)
\(2A=3^{18}-3\)
\(A=\frac{3^{18}-3}{2}\)
Vì \(3^{18}-3>3^{18}-4\)
\(\Rightarrow\frac{3^{18}-3}{2}>\frac{3^{18}-4}{2}\)
\(\Rightarrow A>B\)
A = 31 + 2 + 3 + 4 + 5 + 6 ... + 17
A = 3153
B = [ 318 - 4 ]
Ta thấy rõ ràng A sẽ lớn hơn B vì 153 > 18 ( chưa kể phải trừ thêm 4 ở biểu thức B )
A > B
Phạm Nguyễn Tất Đạt đúng nhưng hơi dài dòng quá !!
a)= \(2^3\left(17-14\right)\)
\(=8.3\)
\(=24\)
b)\(=17\left(85+15\right)-120\)
\(=17.100-120\)
\(=1700-120\)
\(=1580\)
c)\(=\left(25.4\right).\left(125.8\right).\left(2.5\right)\)
\(=100.1000.10\)
\(=1000000\)
d)\(=24.53+24.87-24.40\)
\(=24\left(53+87-40\right)\)
\(=24.100\)
\(=2400\)
e)\(=5.7.77-7.60-7.7.25-15.6.7\)
\(=7\left(5.77-60-7.25-15.6\right)\)
\(=7\left(385-60-175-90\right)\)
\(=7.60\)
\(=420\)
a) Ta có:
\(A=\frac{3^{10}\cdot11+3^{10}\cdot5}{3^0\cdot2^4}\)
\(A=\frac{3^{10}\left(11+5\right)}{1\cdot16}\)
\(A=\frac{3^{10}\cdot16}{16}=3^{10}\)
b) Ta có:
\(B=\frac{2^{10}\cdot13+2^{10}\cdot65}{28\cdot104}\)
\(B=\frac{2^{10}\cdot13\cdot\left(1+5\right)}{2^2\cdot7\cdot2^3\cdot13}\)
\(B=\frac{2^{10}\cdot6\cdot13}{2^5\cdot7\cdot13}=\frac{2^{11}\cdot3\cdot13}{2^5\cdot7\cdot13}\)
\(B=\frac{2^6\cdot3}{7}=\frac{192}{7}\)
a: \(=4\left(x^2+\dfrac{7}{4}x+\dfrac{13}{4}\right)\)
\(=4\left(x^2+2\cdot x\cdot\dfrac{7}{8}+\dfrac{49}{64}+\dfrac{159}{64}\right)\)
\(=4\left(x+\dfrac{7}{8}\right)^2+\dfrac{159}{16}>=\dfrac{159}{16}\)
Dấu '=' xảy ra khi x=-7/8
b: \(=x^2-8x+16-11\)
\(=\left(x-4\right)^2-11>=-11\)
Dấu '=' xảy ra khi x=4
a) \(\frac{1}{9}\cdot3^4\cdot3^n=3^7\)
\(\Leftrightarrow\frac{1}{3^2}\cdot3^4\cdot3^n=3^7\)
\(\Leftrightarrow3^{n+2}=3^7\)
\(\Rightarrow n+2=7\)
\(\Rightarrow n=5\)
b) \(\left(2n+1\right)^3=343\)
\(\Leftrightarrow2n+1=7\)
\(\Leftrightarrow2n=6\)
\(\Rightarrow n=3\)
c) \(2\cdot16>2^n>4\)
\(\Leftrightarrow2^5>2^n>2^2\)
\(\Rightarrow5>n>2\)
d) \(n^{45}=n\)
\(\Leftrightarrow n^{45}-n=0\)
\(\Leftrightarrow n\left(n^{44}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}n=0\\n^{44}-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}n=0\\n^{44}=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}n=0\\n=\pm1\end{cases}}\)
e) \(\left(7n-11\right)^3=2^5\cdot5^2+200\)
\(\Leftrightarrow\left(7n-11\right)^3=1000\)
\(\Leftrightarrow7n-11=10\)
\(\Leftrightarrow7n=21\)
\(\Rightarrow n=3\)
tìm số dư của
A= [22^6n+2(hai mũ hai mũ sáu n cộng hai) + 3]:7
B =[22^3n+1(hai mũ ba n cộng một)+3]:13
\(3^2+3^3+3^4+.......+3^{50}\)
Đặt \(A=3^2+3^3+3^4+.........+3^{50}\)
\(3.A=3.\left(3^2+3^3+3^4+........+3^{50}\right)\)
\(3A=3^3+3^4+3^5+........+3^{50}+3^{51}\)
\(3A-A=\left(3^3+3^4+3^5+.....+3^{51}\right)-\left(3^2+3^3+3^4+.......+3^{50}\right)\)
\(2A=3^{51}-3^2\)
\(A=\left(3^{51}-3^2\right):2\)
Vậy \(A=\frac{3^{51}-3^2}{2}\)
Nhớ k cho mình nhé! Thank you!!!
\(A=\left(-3\right)^2+5.\left(-2\right)^3+58:\left(-2\right).\)
\(A=9+5.\left(-8\right)+\left(-29\right)\)
\(A=9+\left(-40\right)+\left(-29\right)=\left(-31\right)+\left(-29\right)\)
\(A=-60\)