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a) \(9\cdot3^3\cdot\frac{1}{81}\cdot3^2=3^2\cdot3^3\cdot\left(\frac{1}{3}\right)^43^2=3^7\cdot\frac{1}{3^4}=3^3\)
b) \(4\cdot2^5:\left(2^3\cdot\frac{1}{16}\right)=2^2\cdot2^5:\left(2^3\cdot\frac{1}{2^4}\right)=2^7:\frac{1}{2}=2^8\)
c) \(3^2\cdot2^5\cdot\left(\frac{2}{3}\right)^2=3^2\cdot2^5\cdot\frac{2^2}{3^2}=2^7\)
d) \(\left(\frac{1}{3}\right)^2\cdot\frac{1}{3}\cdot9^2=\frac{1}{3^2}\cdot\frac{1}{3}\cdot3^4=\frac{1}{3^3}\cdot3^4=3\)
a)9.33.\(\frac{1}{81}\).32
=32.33.34.\(\frac{1}{3^4}\).32
=311.\(\frac{1}{3^4}\)
=37
b)4.25:(\(2^3.\frac{1}{16}\))
=22.25:(\(2^3.\frac{1}{2^4}\))
=27:\(\frac{2^3}{2^4}\)
=27.\(\frac{2^4}{2^3}\)
=\(\frac{2^{11}}{2^3}\)
=28
c)32.25.\(\left(\frac{2}{3}\right)^2\)
=32.25.\(\frac{2^2}{3^2}\)
=\(\frac{3^2.2^5.2^2}{3^2}\)
=27
d)\(\left(\frac{1}{3}\right)^2.\frac{1}{3}.9^2\)
=\(\frac{1^2}{3^2}.\frac{1}{3}.\left(3^2\right)^2\)
=\(\frac{1^2}{3^2}.\frac{1}{3}.3^4\)
=\(\frac{1^2}{3^2}.\frac{3^4}{3}\)
=\(\frac{1^2}{3^2}.3^3\)
=3
a) \(2^{2014}\) và \(3^{1343}\)
Ta có:
\(2^{2014}=(2^3)^{\frac{2014}{3}}=8^{\frac{2014}{3}}< 9^{\frac{2014}{3}}\)
\(3^{1343}=(3^2)^{\frac{1343}{2}}=9^{\frac{1343}{2}}> 9^{\frac{2014}{3}}\)
\(\rightarrow 2^{2014}< 3^{1343}\)
b) \(31^{11}\) và \(17^{44}\)
Có: \(17^{44}=(17^4)^{11}> (17.2)^{11}>31^{11}\)
c)
\(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\)
\(\Rightarrow 2A=1+\frac{1}{2^1}+\frac{1}{2^2}+..+\frac{1}{2^{49}}\)
Lấy vế sau trừ vế trước thu được:
\(2A-A=1-\frac{1}{2^{50}}< 1\)
\(\Leftrightarrow A< 1\)
d) \(B=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(\Rightarrow 3B=1+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
Lấy vế sau trừ vế trước:
\(\Rightarrow 3B-B=1-\frac{1}{3^{100}}< 1\)
\(\Leftrightarrow 2B< 1\Rightarrow B< \frac{1}{2}\)
Bài 1:
1. \(x:-\left(-\frac{1}{2}\right)=-\frac{1}{2}\)
⇒ \(x:\frac{1}{2}=-\frac{1}{2}\)
⇒ \(x=\left(-\frac{1}{2}\right).\frac{1}{2}\)
⇒ \(x=-\frac{1}{4}\)
Vậy \(x=-\frac{1}{4}.\)
3. \(\frac{16}{2^n}=2\)
⇒ \(2^n=16:2\)
⇒ \(2^n=8\)
⇒ \(2^n=2^3\)
⇒ \(n=3\)
Vậy \(n=3.\)
4. \(\frac{-3^n}{81}=-27\)
⇒ \(\left(-3\right)^n=\left(-27\right).81\)
⇒ \(\left(-3\right)^n=-2187\)
⇒ \(\left(-3\right)^n=\left(-3\right)^7\)
⇒ \(n=7\)
Vậy \(n=7.\)
Chúc bạn học tốt!
Câu 1
4 p/s cộng thêm 1,p/s cuối trừ 4 rồi nhóm vs nhau
d/s la x= - 329
Câu 2
NHân vs 7 thành 7S rồi rút gọn là đc
Câu 1 :
a) \(\Leftrightarrow\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Leftrightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Rightarrow\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Dễ thấy \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\ne0\) \(\Rightarrow x+329=0\Rightarrow x=-329\)
a: \(A=\left(5xy-2xy+1.3xy\right)+3x-2y-3.5y^2\)
\(=4.3xy+3x-2y-3.5y^2\)
b: \(B=\left(\dfrac{1}{2}ab^2-\dfrac{1}{2}ab^2-\dfrac{7}{8}ab^2\right)+\left(\dfrac{3}{4}a^2b-\dfrac{3}{8}a^2b\right)\)
\(=-\dfrac{7}{8}ab^2+\dfrac{3}{8}a^2b\)
c: \(C=\left(2a^2b+5a^2b\right)+\left(-8b^2-3b^2\right)+\left(5c^2+4c^2\right)\)
\(=7a^2b-11b^2+9c^2\)
b) \(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2018}\right)\)
\(=\frac{2-1}{2}.\frac{3-1}{3}.\frac{4-1}{4}....\frac{2018-1}{2018}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2017}{2018}=\frac{1.2.3...2017}{2.3.4...2018}=\frac{1}{2018}\)
c) Giữa các biểu thức là dấu nhân hay dấu cộng vậy bạn?
d)
\(D=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(D=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
e) \(E=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)
\(2E=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(2E=\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+....+\frac{99-97}{97.99}\)
\(2E=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(\Rightarrow E=\frac{16}{99}\)
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
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
S = 1 + 3 + 32 +..+399
=> S = (1 + 3) + ... + (3^98 + 3^99)
=> S = (1 + 3) + ... + 3^98.(1 + 3)
=> S = 4 + ... + 3^98.4
=> S = 4.(1 +... + 3^98) chia hết cho 4 (Đpcm)