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\(\frac{1}{2}.2^n+4.2^n=9.2^5\Rightarrow2^n\left(\frac{1}{2}+4\right)=288\Rightarrow2^n.\frac{9}{2}=288\Rightarrow2^{n-2}.9=288\Rightarrow2^{n-2}=32\)(dấu "=>" số 3 bn sửa thành 2n-1.9=288=>2n-1=32 nha)
=>2n-1=25=>n-1=5=>n=5+1=6
vậy......
~~~~~~~~~~~~~~~
a) 4. ( 1.1/4)2 + [(3/4)2 : (5/4)3] : (3/2)3
= 4.1/16 + [9/16 : 125/64] : 27/8
= \(\frac{1}{4}+\frac{9}{16}:\frac{125}{64}:\frac{27}{8}=\frac{1}{4}+\frac{36}{125}:\frac{27}{8}\)
= \(\frac{1}{4}+\frac{36}{125}.\frac{8}{27}\)
=\(\frac{1}{4}+\frac{32}{375}=\frac{375}{1500}+\frac{128}{1500}=\frac{503}{1500}\)
b] = 2^3 + 3 x 1 - 1 + ( 2^2 x 2 ) x 2^3
= 2^3 + 3 - 1 + 2^3 x 2^3
= 2^3 + 2 + 2^6 = 74
a] = 4 x ( 1/4 )2 + ( 32/42 : 53/43 ) : 27/8
= 4 x 1/16 + ( 32 x 4/53 ) x 8/27
= 1/4 + 36/53 x 8/27 = 1/4 + 4/125 x 8/3 = 503/1500 sấp sỉ 0,335333
3/ ta để ý thấy ở số mũ sẽ có thừa số 1000-103=0
nên số mũ chắc chắn bằng 0
mà số nào mũ 0 cũng bằng 1 nên A=1
5/ vì |2/3x-1/6|> hoặc = 0
nên A nhỏ nhất khi |2/3x-6|=0
=>A=-1/3
6/ =>14x=10y=>x=10/14y
23x:2y=23x-y=256=28
=>3x-y=8
=>3.10/4y-y=8
=>6,5y=8
=>y=16/13
=>x=10/14y=10/14.16/13=80/91
8/106-57=56.26-56.5=56(26-5)=59.56
có chứa thừa số 59 nên chia hết 59
4/ tính x
sau đó thế vào tinh y,z
c: \(=\dfrac{7}{23}\cdot\left(\dfrac{-4}{3}-\dfrac{5}{2}\right)=\dfrac{7}{23}\cdot\dfrac{-8-15}{6}\)
\(=\dfrac{7}{23}\cdot\dfrac{-23}{6}=-\dfrac{7}{6}\)
d: \(=\dfrac{5}{7}\left(23+\dfrac{1}{4}-13-\dfrac{1}{4}\right)=\dfrac{5}{7}\cdot10=\dfrac{50}{7}\)
e: \(=\dfrac{2^5\cdot3^3\cdot5^3}{2^3\cdot3^3\cdot2^2\cdot5^2}=5\)
i: \(=\dfrac{1}{3^{10}}\cdot3^{50}-\dfrac{2^{10}}{3^{10}}:\dfrac{4^5}{3^{10}}\)
\(=3^{40}-1\)
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
2) -12:\(\left(-\dfrac{5}{6}\right)^2\)=\(-12:\dfrac{25}{36}=-12\cdot\dfrac{36}{25}=-\dfrac{432}{25}\)
s) \(-\dfrac{1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)=-\dfrac{1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
= \(-\dfrac{1}{12}-\dfrac{55}{24}=-\dfrac{2}{24}-\dfrac{55}{24}=-\dfrac{57}{24}=-\dfrac{19}{8}\)
t) \(-1,75-\left(-\dfrac{1}{9}-2\dfrac{1}{18}\right)=-1,75-\left(-\dfrac{2}{18}-\dfrac{37}{18}\right)\)
= -1,75-(\(-\dfrac{13}{6}\)) = \(-\dfrac{7}{4}+\dfrac{13}{6}=\dfrac{5}{12}\)
c) \(\left(\sqrt{\dfrac{1}{9}}-0,5\right)^3+\dfrac{-1}{3}=\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^3-\dfrac{1}{3}\)
= \(\left(-\dfrac{1}{6}\right)^3-\dfrac{1}{3}=\dfrac{-1}{216}-\dfrac{1}{3}=-\dfrac{73}{216}\)
d) \(\left(\dfrac{1}{2}-\sqrt{\dfrac{4}{25}}\right)^2-2\dfrac{1}{2}=\left(\dfrac{1}{2}-\dfrac{2}{5}\right)^2-\dfrac{5}{2}\)
= \(\left(\dfrac{1}{10}\right)^2-\dfrac{5}{2}=\dfrac{1}{100}-\dfrac{250}{100}=-\dfrac{249}{100}=-2,49\)
a. ta có : \(9^{10}=3^{20}< 3^{21}=27^7\text{ vậy }9^{10}< 27^7\)
b. ta có \(81^3=3^{12}< 4^{12}=64^4\text{ Vậy }81^3< 64^4\)
c. ta có \(\left(-8\right)^{12}=8^{12}=2^{36}=2^{16}.2^{20}=2^{16}.4^{10}>2^{16}.3^{10}>2^{16}.3^8=12^8=\left(-12\right)^8\)
Vậy \(\left(-8\right)^{12}>\left(-12\right)^8\)
d. ta có \(\left(-0.66\right)^{99}< 0< \left(-0.99\right)^{66}\text{ nên }\left(-0.66\right)^{99}< \left(-0.99\right)^{66}\)
e. ta có \(0.216^5=\left(0.6^3\right)^5=0.6^{15}>0.6^{16}=\left(0.6^2\right)^8=0.36^8\text{ Vậy }0.216^5>0.36^8\)