1. (4^2x25^2 +32x125) / 2^3x5^2 = ..................?
2. (49^24x125^10x2^8-5^30x7^49x4^5) / 5^29x16^2x7^48 =.................?
Giúp mình với nhé xin cảm ơn.
K
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Những câu hỏi liên quan
29 tháng 10 2020
\(G=\frac{4^2.25^2+32.125}{2^3.5^2}\)
\(=\frac{\left(2^2\right)^2.\left(5^2\right)^2+2^5.5^3}{2^3.5^2}\)
\(=\frac{2^4.5^3.\left(5+2\right)}{2^3.5^2}\)
\(=2.5.7\)
\(=70\)
HP
1
KN
17 tháng 8 2020
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{48}-\frac{1}{49}\)
\(\Rightarrow1-A-\frac{1}{50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...-\frac{1}{48}+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow\frac{49}{50}-A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{48}+\frac{1}{49}+\frac{1}{50}\)
\(-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{50}\right)\)
\(\Rightarrow\frac{49}{50}-A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{25}\)
\(\Rightarrow\frac{49}{50}-A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
\(\Rightarrow A=\frac{49}{50}-\left(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+\frac{1}{29}+...+\frac{1}{50}\right)\)
Ta có :
\(\frac{1}{26}< \frac{1}{25};\frac{1}{27}< \frac{1}{25};\frac{1}{28}< \frac{1}{25};\frac{1}{29}< \frac{1}{25};\frac{1}{30}< \frac{1}{25};\)
\(\frac{1}{31}< \frac{1}{30};\frac{1}{32}< \frac{1}{30};..;\frac{1}{39}< \frac{1}{30};\frac{1}{40}< \frac{1}{30};\)
\(\frac{1}{41}< \frac{1}{40};\frac{1}{42}< \frac{1}{40};...;\frac{1}{49}< \frac{1}{40};\frac{1}{50}< \frac{1}{40}\)
\(\Rightarrow\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}< 5.\frac{1}{25}+10.\frac{1}{30}+10.\frac{1}{40}\)
\(\Rightarrow\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}\)
\(\Rightarrow A=\frac{49}{50}-\left(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+\frac{1}{29}+...+\frac{1}{50}\right)>\frac{49}{50}-\frac{4}{5}=\frac{9}{50}>\frac{10}{50}=\frac{1}{5}\)
\(\Rightarrow A>\frac{1}{5}\)( đpcm )
HP
0
NT
0
12 tháng 9 2016
\(\frac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{48}}\)
\(=\frac{\left(7^2\right)^{24}.\left(5^3\right)^{10}.2^8-5^{30}.7^{49}.\left(2^2\right)^5}{5^{29}.\left(2^4\right)^2.7^{48}}\)
\(=\frac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{48}}\)
\(=\frac{7^{48}.5^{30}.2^8.\left(1-7.2^2\right)}{5^{29}.2^8.7^{48}}\)
\(=5.\left(1-7.4\right)\)
\(=5.\left(1-28\right)\)
\(=5.\left(-27\right)=-135\)
ND
Nguyễn Đức Trí
VIP
10 tháng 7 2023
a) \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)
\(A=\dfrac{x-5+2x+10-2x-10}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)
b) \(A=-3\Rightarrow\dfrac{1}{x+5}=-3\)
\(\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}-5=\dfrac{-16}{3}\)
\(9x^2-42x+49=\left(3x-7\right)^2=\left(3.\dfrac{-16}{3}-7\right)^2=\left(-23\right)^2=529\) \(\left(x=\dfrac{-16}{3}\right)\)
TN
4
DA
10 tháng 8 2017
\(4\frac{2}{5}+2\frac{3}{7}-2\frac{2}{5}+5\frac{4}{7}\)
\(=\left(4\frac{2}{5}-2\frac{2}{5}\right)+\left(2\frac{3}{7}+5\frac{4}{7}\right)\)
\(=2+8\)
=10
10 tháng 8 2017
4 2/5 + 2 3/7 - 2 2/5 + 5 4/7
= 22/5 + 17/7 -12/5 + 39/7
= ( 22/5 - 12/5 ) + ( 17/7 + 39/7 )
= 2 + 8
= 10