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NH
0
TG
3 tháng 8 2020
1) \(125^5:25^7\)
\(=\left(5^3\right)^5:\left(5^2\right)^7\)
\(=5^{15}:5^{14}\)
= 5
2) \(27^8:9^9\)
\(=\left(3^3\right)^8:\left(3^2\right)^9\)
\(=3^{24}:3^{18}\)
\(=3^6\)
3) \(36^5:6^8\)
\(=\left(6^2\right)^5:6^8\)
\(=6^{10}:6^8\)
\(=6^2\)
4) \(49^6:7^{10}\)
\(=\left(7^2\right)^6:7^{10}\)
\(=7^{12}:7^{10}=7^2\)
5) \(7^{20}:49^9\)
\(=7^{20}:\left(7^2\right)^9\)
\(=7^{20}:7^{18}=7^2\)
6) \(\frac{1}{2^{10}}:\frac{1}{8^3}\)
\(=\frac{1}{2^{10}}:\frac{1}{\left(2^3\right)^3}\)
\(=\frac{1}{2^{10}}:\frac{1}{2^9}=\frac{1}{2^{10}}.\frac{2^9}{1}=\frac{1}{2}\)
7) \(\left(-\frac{1}{2}\right)^{21}:\frac{1}{4^{10}}\)
\(=\frac{\left(-1\right)^{21}}{2^{21}}:\frac{1}{\left(2^2\right)^{10}}\)
\(=-\frac{1}{2^{21}}:\frac{1}{2^{20}}=-\frac{1}{2^{21}}.\frac{2^{20}}{1}\)
\(=-\frac{1}{2}\)
8) \(\frac{1}{16^5}:\left(-\frac{1}{2}\right)^{18}\)
\(=\frac{1}{\left(2^4\right)^5}:\frac{\left(-1\right)^{18}}{2^{18}}\)
\(=\frac{1}{2^{20}}:\frac{1}{2^{18}}\)
\(=\frac{1}{2^{20}}.\frac{2^{18}}{1}=\frac{1}{4}\)
9) \(\frac{1}{5^{30}}:\frac{1}{25^{14}}\)
\(=\frac{1}{5^{30}}:\frac{1}{\left(5^2\right)^{14}}\)
\(=\frac{1}{5^{30}}:\frac{1}{5^{28}}=\frac{1}{25}\)
TL
B={[5^10. 7^3 - 25^2 . 49^2] : [(125 . 7)^3+5^9 . 14^3]} - {[2^12-4^6 . 9^2] : [(2^2.3)^6+8^4 . 3^5]
1
9 tháng 7 2023
Sửa đề: \(5^9\cdot49^2\)
\(=\dfrac{5^{10}\cdot7^3-5^9\cdot7^4}{5^9\cdot7^3+5^9\cdot14^3}-\dfrac{2^{12}-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}\)
\(=\dfrac{5^9\cdot7^3\left(5-7\right)}{5^9\cdot7^3\left(1+8\right)}-\dfrac{2^{12}\left(1-3^4\right)}{2^{12}\left(3^6+3^5\right)}=\dfrac{-2}{9}+\dfrac{80}{972}\)
=-34/243
30 tháng 6 2015
\(D=\left(\frac{3}{7}\right)^{21}:\left(\left(\frac{3}{7}\right)^2\right)^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{2.6}=\left(\frac{3}{7}\right)^9\)
\(E=\left(-\frac{1}{3}\right)^{7+9}:\left(-\frac{1}{3}\right)^{5.3}+\left(-2\right)^{12+3}:\left(-2\right)^{15}=\left(-\frac{1}{3}\right)^{16}:\left(-\frac{1}{3}\right)^{15}+\left(-2\right)^{15}:\left(-2\right)^{15}=-\frac{1}{3}+1=\frac{2}{3}\)
15 tháng 5 2022
a: \(\Leftrightarrow\left(2x-2\right)\cdot\dfrac{3}{4}=-6-\dfrac{1}{3}=-\dfrac{19}{3}\)
\(\Leftrightarrow2x-2=-\dfrac{19}{3}:\dfrac{3}{4}=-\dfrac{76}{9}\)
=>2x=-58/9
hay x=-29/9
b: \(\Leftrightarrow\left(\dfrac{44}{7}x+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(\Leftrightarrow x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{5}{7}\)
\(\Leftrightarrow x\cdot\dfrac{44}{7}=-\dfrac{8}{7}\)
hay x=-2/11
e: \(\left(2x+\dfrac{3}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{3}{5}=\dfrac{3}{5}\\2x+\dfrac{3}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{5}\end{matrix}\right.\)
NB
24 tháng 8 2019
a)\(\left(5x+1\right)^2=\frac{36}{49}\\ \left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\\ \Rightarrow\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{matrix}\right.\)
vậy...
VM
24 tháng 8 2019
2.
a) \(\left(5x+1\right)^2=\frac{36}{49}\)
⇒ \(5x+1=\pm\frac{6}{7}\)
⇒ \(\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}5x=\frac{6}{7}-1=-\frac{1}{7}\\5x=\left(-\frac{6}{7}\right)-1=-\frac{13}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-\frac{1}{7}\right):5\\x=\left(-\frac{13}{7}\right):5\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{35};-\frac{13}{35}\right\}.\)
Chúc bạn học tốt!
NQ
19
4 tháng 9 2015
Ta có quy luật:
1+1=12 = 1
2+2=22 = 4
3+3 = 32 = 9
..................
.......................
9 + 9 = 92 = 81
\(\frac{9}{7}-\frac{2}{7}\times\frac{49}{6}+\frac{1}{3}=\frac{9}{7}-\frac{7}{3}+\frac{1}{3}=\frac{9}{7}-\left(\frac{7}{3}-\frac{1}{3}\right)=\frac{9}{7}-2=-\frac{5}{7}\)
=.= hok tốt!!