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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 \(3^{4000}=9^{2000}\)
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
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. \(\dfrac{390^4}{130^4}=\dfrac{\left(130.3\right)^4}{130^4}=\dfrac{130^4.3^4}{130^4}=3^4=81\)
a) Ta có: \(2^{225}=2^{3.75}=\left(2^3\right)^{75}=8^{75}\)
\(3^{150}=3^{2.75}=\left(3^2\right)^{75}=9^{75}\)
\(\Rightarrow8^{75}< 9^{75}\)\(\Rightarrow2^{225}< 3^{150}\)
b) Ta có : \(2^{91}=2^{7.13}=\left(2^{13}\right)^7=8192^7\)
\(5^{35}=5^{5.7}=\left(5^5\right)^7=3125^7\)
\(\Rightarrow8192^7>3125^7\)\(\Rightarrow2^{91}>3^{35}\)
c) Ta có: \(99^{20}=99^{2.10}=\left(99^2\right)^{10}=\left(99.99\right)^{10}\)
\(9999^{10}=\left(99.101\right)^{10}\)
Vì 99<101 \(\Rightarrow\left(99.99\right)^{10}< \left(99.101\right)^{10}\)\(\Rightarrow99^{20}< 9999^{10}\)
a) ta có :\(2^{24}=\left(2^2\right)^{12}=4^{12}\)
\(3^{36}=\left(3^2\right)^{12}=9^{12}\)
Vì \(4^{12}< 9^{12}\left(4< 9\right)\)
Nên bạn tự kết luận
b) ta có : \(10^{20}=\left(10^2\right)^{10}=100^{10}\)
Vì \(100^{10}>90^{10}\left(100>90\right)\)
Nên bạn tự kết luận
c) ta có : \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8^{111}< 9^{111}\left(8< 9\right)\)
Nên bạn tự kết luận
224=(22)12=412
336=(33)12=2712
Tự so sánh nhé
phần sau tương tự
\(a,\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\dfrac{1}{2}\)
\(b,\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
\(\Rightarrow x=-\dfrac{1}{4}\)
\(c,\left(2x+3\right)^2=\dfrac{9}{121}\)
\(\Rightarrow\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)
\(\Rightarrow2x+3=\dfrac{3}{11}\)
\(\Rightarrow2x=-\dfrac{30}{11}\)
\(\Rightarrow x=-\dfrac{15}{11}\)
\(d,\left(2x-1\right)^3=-\dfrac{8}{27}\)
\(\Rightarrow\left(2x-1\right)^3=\left(-\dfrac{2}{3}\right)^3\)
\(\Rightarrow2x-1=-\dfrac{2}{3}\)
\(\Rightarrow2x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{6}\)
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\Leftrightarrow x=\dfrac{-1}{4}\)
\(\left(2x+3\right)^2=\dfrac{9}{121}\Leftrightarrow\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\Leftrightarrow2x+3=\dfrac{3}{11}\Leftrightarrow x=\dfrac{-15}{11}\)
\(\left(2x-1\right)^3=-8\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\Leftrightarrow2x-1=-2\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\)
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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