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a,\(27^{11}=\left(3^3\right)^{11}=3^{33}\)
\(81^8=\left(3^4\right)^8=3^{32}\)
Vì \(3^{33}>3^{32}\Rightarrow27^{11}>81^8\)
b,\(3^{150}=\left(3^2\right)^{75}=9^{75}\\ 2^{225}=\left(2^3\right)^{75}=8^{75}\)
Vì \(9^{75}>8^{75}\Rightarrow3^{150}>2^{225}\)
c,Ta có :\(\dfrac{27^{50}}{240^{30}}\)
=\(\dfrac{27^{30}.27^{20}}{240^{30}}\)
=\(\dfrac{3^{30}.3^{30}.3^{30}.3^{20}.3^{20}.3^{20}}{3^{30}.80^{30}}\)
=\(\dfrac{3^{120}}{80^{30}}\)=\(\dfrac{\left(3^4\right)^{30}}{80^{30}}\)=\(\dfrac{81^{30}}{80^{30}}\)
\(\rightarrow\)\(\dfrac{81^{30}}{80^{30}}>1\rightarrow\dfrac{27^{50}}{240^{30}}>1\Rightarrow27^{50}>240^{30}\)
a)ta có 2711 và 818
= (33)11 và (34)8
=333 và 332
do 33>32
=>333>332hay2711>818
a) Ta có :
\(\hept{\begin{cases}27^{11}=\left(3^3\right)^{11}=3^{33}\\81^8=\left(3^4\right)^8=3^{32}\end{cases}}\)
Vì 333 > 332
=> 2711 > 818
b) Ta có:
\(\hept{\begin{cases}2^{225}=\left(2^3\right)^{75}=8^{75}\\3^{150}=\left(3^2\right)^{75}=9^{75}\end{cases}}\)
Vì 875 < 975
=> 2225 < 3150
Thôi còn lại bn tự làm nốt nha . Nhìn mà nản !!
a) \(\hept{\begin{cases}27^{11}=\left(3^3\right)^{11}=3^{33}\\81^8=\left(3^4\right)^8=3^{32}\end{cases}}\)
333 > 332 => 2711 > 818
b) \(\hept{\begin{cases}2^{225}=\left(2^3\right)^{75}=8^{75}\\3^{150}=\left(3^2\right)^{75}=9^{75}\end{cases}}\)
875 < 975 => 2225 < 3150
c) \(\hept{\begin{cases}2^{500}=\left(2^5\right)^{100}=32^{100}\\5^{200}=\left(5^2\right)^{100}=25^{100}\end{cases}}\)
32100 > 25100 => 2500 > 5200
d) \(\hept{\begin{cases}625^5=\left(5^4\right)^5=5^{20}\\125^7=\left(5^3\right)^7=5^{21}\end{cases}}\)
520 < 521 => 6255 < 1257
e) \(\hept{\begin{cases}5^{100}=\left(5^4\right)^{25}=625^{25}\\8^{75}=\left(8^3\right)^{25}=512^{25}\end{cases}}\)
62525 > 51225 => 5100 > 875
f) \(2^{16}=2^3\cdot2^{13}=8\cdot2^{13}\)
7 < 8 => 7.213 < 8.213 => 7.213 < 216
g) Ta có \(\frac{27^{50}}{240^{30}}=\frac{\left(3^3\right)^{50}}{3^{30}\cdot80^{30}}=\frac{3^{150}}{3^{30}\cdot80^{30}}=\frac{3^{120}}{80^{30}}=\frac{\left(3^4\right)^{30}}{80^{30}}=\frac{81^{30}}{80^{30}}\)
Vì 8130 > 8030 => 8130/8030 > 1 => 2750/24030 > 1 => 2750 > 24030
h) Ta có \(\hept{\begin{cases}63^9< 64^9=\left(2^6\right)^9=2^{54}\left(1\right)\\16^{14}=\left(2^4\right)^{14}=2^{56}< 17^{14}\left(2\right)\end{cases}}\)
Từ (1) và (2) => 639 < 254 < 256 < 1714
=> 639 < 1714
a.\(2^{225}=\left(2^3\right)^{75}=8^{75}\left(1\right)\)
\(3^{150}=\left(3^2\right)^{75}=9^{75}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow2^{225}< 3^{150}\)
b.\(2^{91}=2^{7.13}=\left(2^{13}\right)^7=8192^7\left(1\right)\)
\(5^{35}=5^{7.5}=\left(5^5\right)^7=3125^7\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow5^{35}< 2^{91}\)
c.\(9999^{10}=\left(99.101\right)^{10}=99^{10}.101^{10}>99^{10}.99^{10}=99^{20}\)
\(\Rightarrow9999^{10}>99^{20}\)
Bài 2:
\(b.\)\(75^{20}=\left(3.5^2\right)^{20}=\left(3^{20}.5^{10}\right).5^{30}=\left[̣\left(3^2\right)^{10}.5^{10}\right].5^{30}=45^{10}.5^{30}\)
\(\Rightarrowđpcm\)
Bài 3:
a,\(25^4.2^8=\left(5^2\right)^4.2^8=5^8.2^8=10^8\)
b. \(\frac{27^2}{25^3}=\frac{\left(3^3\right)^2}{\left(5^2\right)^3}=\frac{3^6}{5^6}=\left(\frac{3}{5}\right)^3\)\(:v\)
\(27^2.25^3=3^6.5^6=15^6\)( đề phòng you viết sai đề )
Câu 6 :
Vì bình phương một số luôn lớn hơn hoặc bằng 0
Mà tổng của chúng bằng 0
\(\Rightarrow2x+3=3x-2=0\)
\(\Leftrightarrow2x-3x=-2-3\)
\(\Leftrightarrow-x=-5\)
\(\Leftrightarrow x=5\left(\text{Thỏa mãn}\right)\)
Vậy có số hữu tỉ x thỏa mãn
\(\hept{\begin{cases}\left(2x+3\right)^2\ge0\\\left(3x-2\right)^2\ge0\end{cases}\Rightarrow\left(2x+3\right)^2+\left(3x-2\right)^2\ge0}\)
dấu = xảy ra khi: \(\hept{\begin{cases}\left(2x+3\right)^2=0\\\left(3x-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{3}{2}\\x=\frac{2}{3}\end{cases}}}\)
=> ko có giá trị x nào t/m để \(\left(2x+3\right)^2+\left(3x-2\right)^2=0\)
p/s: Trần Thanh Phương sai rồi
vì 1/27^11 = 1/(3^3)^11 = 1/3^33
1/81^8= 1/(9^2)^8 = 1/9^16 = 1/(3^2)^16 = 1/3^32
=> <
vì 1/3^99= 1/(3)^33.3=1/99^3 = 1/ (33.3)^3 = 1/33^9
1/11^21=1/(11)^3.7=1/33^7
=> " > "
nhớ ****
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)27^11=(3^3)^11=3^33
81^8=(3^4)8=3^32
vì 3^33>3^32 nên 27^11>81^8
b)ko biết làm chỉ biết 3^150>2^225
c)27^50=27^5x10=(27^5)^10=14348907^10
240^30=240^3x10=(240^3)^10=13824000^10
suy ra 27^50>240^30
a) Ta có: \(27^{11}=\left(3^3\right)^{^{11}}=3^{3.11}=3^{33}\)
\(81^8=\left(3^4\right)^{^8}=3^{4.8}=3^{32}\)
Vì \(3^{33}>3^{32}\)
nên \(27^{11}>81^8\)
b) Ta có: \(3^{150}=3^{2.75}=\left(3^2\right)^{^{75}}=9^{75}\)
\(2^{225}=2^{3.75}=\left(2^3\right)^{^{75}}=8^{75}\)
vì \(9^{75}>8^{75}\)
nên \(3^{150}>2^{225}\)
c) Ta có:
\(\frac{27^{50}}{240^{30}}=\frac{27^{30}.27^{20}}{240^{30}}=\frac{3^{30}.3^{30}.3^{30}.3^{20}.3^{20}.2^{20}}{3^{30}.80^{30}}\)
\(=\frac{3^{120}}{80^{30}}=\frac{\left(3^4\right)^{^{30}}}{80^{30}}=\frac{81^{30}}{80^{30}}\)
Vì \(\frac{81^{30}}{80^{30}}>1\)\(\Rightarrow\frac{27^{50}}{240^{30}}>1\)\(\Rightarrow27^{50}>240^{30}\)