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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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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. \(\dfrac{390^4}{130^4}=\dfrac{\left(130.3\right)^4}{130^4}=\dfrac{130^4.3^4}{130^4}=3^4=81\)
\(1,\)
\(a,x^2+x=0\)
\(x.\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=0-1=-1\end{matrix}\right.\)
Vậy \(x\in\left\{0;-1\right\}\)
\(b,\left(2x-3\right)^2=36\)
\(\left(2x-3\right)^2=6^2=\left(-6\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=6+3=9\\2x=-6+3=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{9}{2};\dfrac{-3}{2}\right\}\)
\(2,\)
\(a,64^8\)và \(16^2\)
Ta có \(64^8=\left(2^6\right)^8=2^{48}\)
\(16^2=\left(2^4\right)^2=2^8\)
Vì \(2^{48}>2^8\)
\(\Rightarrow64^8>16^2\)
\(b,\left(-5\right)^{30}\)và \(\left(-3\right)^{50}\)
\(\left(-5\right)^{30}=5^{30};\left(-3\right)^{50}=3^{50}\)
Vì \(5^3< 3^5\left(125< 243\right)\)
\(\Rightarrow\left(5^3\right)^{10}< \left(3^5\right)^{10}\)
\(\Rightarrow5^{30}< 3^{50}\)
\(\Rightarrow\left(-5\right)^{30}< \left(-3\right)^{50}\)
\(c,3^{4000}\) và \(9^{2000}\)
Ta có : \(9^{2000}=\left(3^2\right)^{2000}=3^{4000}\)
\(\Rightarrow3^{4000}=9^{2000}\)
\(d,2^{332}\)và \(3^{223}\)
Ta có : \(2^{332}< 2^{333}=\left(2^3\right)^{111}\)
\(3^{223}>3^{222}=\left(3^2\right)^{111}\)
Vì \(3^2>2^3\left(9>8\right)\)
\(\Rightarrow\left(2^3\right)^{111}< \left(3^2\right)^{111}\)
\(\Rightarrow2^{333}< 3^{222}\)
\(\Rightarrow2^{332}< 3^{223}\)
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) \(2^{91}\)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 \(3^{4000}=9^{2000}\)