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\(a,\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\dfrac{1}{2}\)
\(b,\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
\(\Rightarrow x=-\dfrac{1}{4}\)
\(c,\left(2x+3\right)^2=\dfrac{9}{121}\)
\(\Rightarrow\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)
\(\Rightarrow2x+3=\dfrac{3}{11}\)
\(\Rightarrow2x=-\dfrac{30}{11}\)
\(\Rightarrow x=-\dfrac{15}{11}\)
\(d,\left(2x-1\right)^3=-\dfrac{8}{27}\)
\(\Rightarrow\left(2x-1\right)^3=\left(-\dfrac{2}{3}\right)^3\)
\(\Rightarrow2x-1=-\dfrac{2}{3}\)
\(\Rightarrow2x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{6}\)
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\Leftrightarrow x=\dfrac{-1}{4}\)
\(\left(2x+3\right)^2=\dfrac{9}{121}\Leftrightarrow\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\Leftrightarrow2x+3=\dfrac{3}{11}\Leftrightarrow x=\dfrac{-15}{11}\)
\(\left(2x-1\right)^3=-8\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\Leftrightarrow2x-1=-2\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\)
a) Ta có :
\(2^{225}=\left(2^3\right)^{75}=8^{75}\)
\(3^{150}=\left(3^2\right)^{75}=9^{75}\)
Mà 8^75 < 9^75 => 2^225<3^150
b) Ta có
2^91=(2^13)^7=8192^7
3^35=(3^5)^7=243^7
mà 8192^7<243^7=> 2^91<3^35
c) 3^4000=(3^2)^2000=9^2000
d) 2^332 < 2^333=2^3^111=8^111
3^223>3^222=9^111
=>2^332<3^223
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a) \(2^{91}\)và \(5^{35}\)
Ta có :
\(2^{91}=\left(2^{13}\right)^7=8192^7\)
\(5^{35}=\left(5^5\right)^7=3125^7\)
Vì \(8192^7>3125^7\)nên \(2^{91}>5^{35}\)
b) \(3^{4000}\)và \(9^{2000}\)
Ta có :
\(3^{4000}=\left(3^4\right)^{1000}=81^{1000}\)
\(9^{2000}=\left(9^2\right)^{1000}=81^{1000}\)
Vì \(81^{1000}=81^{1000}\)nên \(3^{4000}=9^{2000}\)
\(2^{91}\)và \(5^{35}\)
Ta có :
\(2^{91}=\left(2^{13}\right)^7=8192^7\)
\(5^{35}=\left(5^5\right)^7=3125^7\)
Vì \(8192>3125\)nên \(2^{91}>5^{35}\)
\(3^{4000}\)và \(9^{2000}\)
Ta có :
\(3^{4000}=\left(3^4\right)^{1000}=81^{1000}\)
\(9^{2000}=\left(9^2\right)^{1000}=81^{1000}\)
Vì \(81=81\)nên \(3^{4000}=9^{2000}\)
Giải:
a) \(\dfrac{120^3}{40^3}=\left(\dfrac{120}{40}\right)^3=30^3=2700\)
b) \(\dfrac{390^4}{130^4}=\left(\dfrac{390}{130}\right)^4=30^4=810000\)
c) \(\dfrac{3^2}{\left(0,375\right)^2}=\left(\dfrac{3}{0,375}\right)^2=8^2=64\)
Đáp số: a) 2700; b) 810000; c) 64.
Chúc bạn học tốt!!!
bài 4 : c1 \(3^{4000}\)và \(9^{2000}\)
\(\Leftrightarrow9^{2000}\Leftrightarrow\left(3^2\right)^2^{000}\Leftrightarrow3^{4000}\)
vì \(3^{4000}=3^{4000}\Leftrightarrow3^{4000}=9^{2000}\)
c2
ta có
\(3^{4000}=\left(3^4\right)^{1000}=81^{1000}\)
\(9^{2000}=\left(9^2\right)^{1000}=81^{1000}\)
vì \(81^{1000}=81^{1000}\Leftrightarrow3^{4000}=9^{2000}\)
bài 5
\(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)
vì \(8^{111}< 9^{111}\Leftrightarrow2^{332}< 3^{223}\)
3) M = 22010 - (22009 + 22008 + .... + 21 + 20)
Đặt N = 22009 + 22008 + .... + 21 + 20
=> 2N = 22010 + 22009 + .... + 22 + 21
=> 2N - N = (22010 + 22009 + .... + 22 + 21) - (22009 + 22008 + .... + 21 + 20)
=> N = 22010 - 1
Khi đó M = 22010 - (22010 - 1) = 1
4) C1 Ta có 34000 = (34)1000 = 811000 = (92)1000 = 92000
34000 = 92000
C2 Ta có : 34000 = (34)1000 = 811000 (1)
Lại có 92000 = (92)1000 = 811000 (2)
Từ (1) (2) => 34000 = 92000
5 Ta có 2332 < 2333 = (23)111 = 8111 < 9111 = (32)111 = 3222 < 3223
=> 2332 < 3223
2) Ta có n150 < 5225
=> (n5)75 < (53)75
=> n5 < 53
=> n5 < 125
Vì n là số nguyên lớn nhất => n = 2
a) $\dfrac{120^3}{40^3}=(\dfrac{120}{40})^3=3^3=27$
b) $\dfrac{390^4}{130^4}=(\dfrac{390}{130})^4=3^4=81$
c) $\dfrac{3^2}{(0,375)^2}=(3:0,375)^2=(3:\dfrac{3}{8})^2=8^2=64$
1.Tính
(0,25)4.1024=(1/4)4.1024=4
2.So sánh
291=(213)7=81927
535=(55)7=31257
Mà 8192>3125=> 81927>31257
=> 291>535
3. Tìm giá trị biểu thức
a) \(\dfrac{45^{10^{ }}.5^{20^{ }}}{75^{15}}=\dfrac{\left(3^{2^{ }}.5\right)^{10^{ }}.5^{20}}{^{ }\left(3.5^2\right)^{15}}=\dfrac{3^{20}.5^{30}}{3^{15}.5^{30}}=3^5=243\)
b)\(\dfrac{\left(0,8\right)^5}{\left(0,4\right)^6}=\dfrac{\left(2.0,4\right)^5}{0,4.0,4^5}=\dfrac{2^{5^{ }}.0,4^5}{0,4.0,4^5}=\dfrac{2^5}{0,4}=80\)
c)\(\dfrac{2^{15}.9^4}{6^6.8^3}=\dfrac{2^{15^{ }}.3^8}{3^6.2^6.2^9}=\dfrac{2^{15}.3^8}{3^6.2^{15}}=3^2=9\)
Tic hộ tui đi !!! chúc bn hok tôts 
1. sai dấu nhé
2.a, \(\frac{45^{10}.5^{20}}{75^{15}}=\frac{\left(3^2.5\right)^{10}.5^{20}}{\left(5^2.3\right)^{15}}=\frac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5=243\)
b, \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(\frac{4}{5}\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\cdot2\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\right)^5\cdot2^5}{\left(\frac{2}{5}\right)^5\cdot\frac{2}{5}}=2^5\div\frac{2}{5}=32\cdot\frac{5}{2}=80\)
c, \(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^2}{2^{15}}=3^2=9\)
, bạn nào làm đc mik tick cho
Câu 1 :
\(\text{a) }B=\dfrac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\\ B=\dfrac{\left(2^2\right)^6\cdot\left(3^2\right)^5+\left(2\cdot3\right)^9\cdot\left(2^3\cdot3\cdot5\right)}{\left(2^3\right)^4\cdot3^{12}-6^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-\left(2\cdot3\right)^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\left(6-1\right)}\\ B=\dfrac{2\cdot6}{3\cdot5}\\ B=\dfrac{4}{5}\\ \)
\(\text{b) }C=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\\ C=\dfrac{5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9}{5\cdot2^9\cdot\left(2\cdot3\right)^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{2^{29}\cdot3^{18}\left(10-9\right)}{2^{28}\cdot3^{18}\left(15-14\right)}\\ C=\dfrac{2^{29}\cdot3^{18}}{2^{28}\cdot3^{18}}\\ C=2\\ \)
\(\text{c) }D=\dfrac{49^{24}\cdot125^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot4^5}{5^{29}\cdot16^2\cdot7^{48}}\\ D=\dfrac{\left(7^2\right)^{24}\cdot\left(5^3\right)^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot\left(2^2\right)^5}{5^{29}\cdot\left(2^4\right)^2\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{5^{29}\cdot2^8\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1-28\right)}{5^{29}\cdot2^8\cdot7^{48}}\\ D=5\cdot\left(-27\right)\\ D=-135\)
Câu 2 :
\(\text{a) }9^{x+1}-5\cdot3^{2x}=324\\ \Leftrightarrow9^x\cdot9-5\cdot9^x=81\cdot4\\ \Leftrightarrow9^x\left(9-5\right)=9^2\cdot4\\ \Leftrightarrow9^x\cdot4=9^2\cdot4\\ \Leftrightarrow9^x=9^2\\ \Leftrightarrow x=2\\ \text{Vậy }x=2\\ \)
Sorry . Mình chỉ biết đến đây thôi
Bài 1 :
a) Ta có :
\(2^{225}=\left(2^3\right)^{75}=8^{75}\)
\(3^{150}=\left(3^2\right)^{75}=9^{75}\)
Vì \(8^{75}< 9^{75}\Leftrightarrow2^{225}< 3^{150}\)
b) Ta có :
\(2^{91}=\left(2^{13}\right)^7=8192^7\)
\(5^{35}=\left(5^5\right)^7=3125^7\)
Vì \(8192^7>3125^7\Leftrightarrow2^{91}>5^{35}\)
c)Ta có :
\(3^{4000}=\left(3^4\right)^{1000}=81^{1000}\)
\(9^{2000}=\left(9^2\right)^{1000}=81^{1000}\)
Vì \(81^{1000}=81^{1000}\Leftrightarrow3^{4000}=9^{2000}\)
d) Ta có :
\(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{223}< 3^{222}=\left(3^2\right)^{111}=9^{111}\)
Mà \(8^{111}< 9^{111}\Leftrightarrow2^{332}< 3^{223}\)
Bài 2 :
a) \(\dfrac{120^3}{40^3}=\left(\dfrac{120}{4}\right)^3=3^3=27\)
b) \(\dfrac{390^4}{130^4}=\left(\dfrac{390}{130}\right)^4=3^4=81\)
c) \(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{\left(3^2.5\right)^{10}.5^{20}}{\left(3.5^2\right)^{15}}=\dfrac{3^{20}.5^{10}.5^{20}}{3^{15}.5^{30}}=3^5=243\)
Bài 1:
a.Ta có :
\(2^{225}=\left(2^3\right)^{75}=8^{75}\)
\(3^{150}=\left(3^2\right)^{75}=9^{75}\)
Vì \(8^{75}< 9^{75}\) nên \(2^{225}< 3^{150}\)
b. Ta có :
\(2^{91}=\left(2^{13}\right)^7=8192^7\)
\(5^{35}=\left(5^5\right)^7=3125^7\)
Vì \(8192^7>3125^7\) nên \(2^{91}>5^{35}\)
c. Ta có :
\(3^{4000}=\left(3^2\right)^{2000}=9^{2000}\)
Vì \(9^{2000}=9^{2000}\) nên \(3^{4000}=9^{2000}\)
Bài 2:
a. \(\dfrac{120^3}{30^3}=\dfrac{\left(30.4\right)^3}{30^3}=\dfrac{30^3.4^3}{30^3}=4^3=64\)
b. \(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{\left(5.3^2\right)^{10}.5^{20}}{\left(3.5^2\right)^{15}}=\dfrac{5^{10}.3^{20}.5^{20}}{3^{15}.5^{30}}=\dfrac{5^{30}.3^{20}}{3^{15}.5^{30}}=3^5=243\)
c. \(\dfrac{390^4}{130^4}=\dfrac{\left(130.3\right)^4}{130^4}=\dfrac{130^4.3^4}{130^4}=3^4=81\)