Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 2 :
\(\frac{-1}{39}+\frac{-1}{52}\) = \(\frac{-7}{156}\)
Chúc bạn học tốt
46 . 95 + 69 . 120 / 84 . 312 - 611
= ( 22 )6. ( 32 ) 5 + (2 . 3 ) 9 . 23 . 3 . 5 / ( 23 ) 4 . 312 - ( 2 . 3 ) 11
= 2^12.3^10+2^9.3^9.2^3.3.5 / 2^12.3^12-2^11.3^11
= 2^12.3^10+2^12.3^10.5 / 2^11.3^11.(2.3-1)
= 2^12.3^10.(1+5) / 2^11.3^11. (6-1)
= 2^12.3^10.6 / 2^11.3^11.5
= 2^13.3^11 / 2^11.3^11.5
= 2^2/5
=4/5
Giải toán trên mạng - Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học trực tuyến OLM
a: \(25S=25+5^4+...+5^{202}\)
=>24S=5^202-1
hay \(S=\dfrac{5^{202}-1}{24}\)
b:
4^30=2^30*2^30
=(2^3)^10*(2^2)^15>8^10*3^15=(8^10*3^10)*3^5>24^10*3
=>2^30+3^30+4^30>3*24^10
Đặt \(D=2+2^2+2^3+...+2^{100}\)
\(\Leftrightarrow2D=2^2+2^3+2^4+...+2^{101}\)
\(\Leftrightarrow2D-D=D=2^{101}-1\)
\(\Rightarrow C=2^{101}-1-2^{108}\)
\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)
\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{15}+2^9.3^9.8.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{15}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(6-1\right)}\)
\(=\frac{2^{10}.3^{10}\left(2^2.3^5+2^2.5\right)}{5.6^{11}}=\frac{6^{10}\left[4\left(243+5\right)\right]}{6^{10}.30}=\frac{4\left(243+5\right)}{30}\)
\(=\frac{992}{30}\)
\(S=1+5+5^2+...+5^{320}\)
\(5S=5.\left(1+5+5^2+...+5^{320}\right)\)
\(5S=5+5^2+5^3+...+5^{321}\)
\(5S-S=\left(5+5^2+5^3+...+5^{321}\right)-\left(1+5+5^2+...+5^{320}\right)\)
\(4S=5^{321}-1\)
\(S=\frac{5^{321}-1}{4}\)
=> 5S = 5 + 5^2 + . . . . . + 5^321
=> 5S - 1S = ( 5 + 5^2 + . . . . . +5^321 ) - ( 1 + 5 + . . . . . + 5^320 )
=> 5S - S = 5 + 5^2 + . . . . . +5^321 - 1 - 5 - . . . . . - 5^320
=> 4S = 5^321 - 1
=> S = \(\frac{5^{321}-1}{4}\)
VẬY . . .. . . . . . . .