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a, \(125^3:5^7=\left(5^3\right)^3:5^7=5^9:5^7=5^2\)
b, \(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{4}{49}\right)^5:\left(\dfrac{8}{343}\right)^2\)
= \(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{2^2}{7^2}\right)^5:\left(\dfrac{2^3}{7^3}\right)^2\)
= \(\left(\dfrac{2}{7}\right)^{18}:\left[\left(\dfrac{2}{7}\right)^2\right]^5:\left[\left(\dfrac{2}{7}\right)^3\right]^2\)
=\(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{2}{7}\right)^{10}:\left(\dfrac{2}{7}\right)^6\)
= \(\left(\dfrac{2}{7}\right)^{18-10-6}=\left(\dfrac{2}{7}\right)^2\)
c, \(3-\left(\dfrac{-7}{9}\right)^0+\left(\dfrac{1}{3}\right)^5.3^5\)
= 3 - 1 +\(\left[\left(\dfrac{1}{3}\right)^5.3^5\right]\)
= 2 + 1=3
d, \(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{\left(9.5\right)^{10}.5^{20}}{\left(25.3\right)^{15}}=\dfrac{\left(3^2\right)^{10}.5^{10}.5^{20}}{\left(5^2\right)^{15}.3^{15}}\)
= \(\dfrac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5\)
a.\(3^{x-1}=243\)
\(3^x:3^1=243\)
\(3^x=729\)
\(\Leftrightarrow3^6=729\)
\(\Leftrightarrow x=6\)
b.\(\left(\dfrac{2}{3}\right)^{x+1}=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x.\left(\dfrac{2}{3}\right)=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x=3\)
Câu b tính đến đây rồi không mò đc x nữa.
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)\(\left(\dfrac{1}{2}\right)^n=\dfrac{1}{32}\)
=>\(\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{2}\right)^5\)
=>n=5
b)\(\left(\dfrac{343}{125}\right)=\left(\dfrac{7}{5}\right)^n\)
=>\(\left(\dfrac{7}{5}\right)^3=\left(\dfrac{7}{5}\right)^n\)
=>n=3
c)\(\dfrac{16}{2^n}=2\)
=>2n=\(\dfrac{16}{2}\)
=>2n=8
=>2n=23
=>n=3
d)\(\dfrac{\left(-3\right)^n}{81}=-27\)
=>(-3)n=-27.81
=>(-3)n=-2187
=>(-3)n=(-3)7
=>n=7
e)8n:2n=4
=>(23)n:2n=4
=>23n:2n=4
=>23n-n=4
=>22n=4
=>22n=22
=>2n=2
=>n=1
f)32.3n=35
=>3n=35:32
=>3n=35-2
=>3n=33
=>n=3
g) (22:4).2n=4
=>1.2n=22
=>n=2
h)3-2.34.3n=37
=>\(\left(\dfrac{1}{3}\right)^2\).34.3n=37
=>32.3n=37
=>32+n=37
=>2+n=7
=>n=5
cứ phan tích cho hết đi là đc 9^6. 9^10 = (3^2)^6...................
tự làm đi
1. Tính:
a. \(\dfrac{9^6.9^{10}}{3^{32}}=\dfrac{\left(3^2\right)^6.\left(3^2\right)^{10}}{3^{32}}=\dfrac{3^{12}.3^{20}}{3^{32}}=\dfrac{3^{32}}{3^{32}}=1\)
b. \(\dfrac{25^8.25^{10}}{5^{34}}=\dfrac{\left(5^2\right)^8.\left(5^2\right)^{10}}{5^{34}}=\dfrac{5^{16}.5^{20}}{5^{34}}=\dfrac{5^{36}}{5^{34}}=5^{36}:5^{34}=5^2=25\)
c. \(\dfrac{7^{56}}{49^9.49^{20}}=\dfrac{7^{56}}{\left(7^2\right)^9.\left(7^2\right)^{20}}=\dfrac{7^{56}}{7^{18}.7^{40}}=\dfrac{7^{56}}{7^{58}}=7^{56}:7^{58}=\dfrac{7^{56}}{7^{56}.7^2}=\dfrac{1}{7^2}=\dfrac{1}{49}\)
d. \(\dfrac{4^2.4^3}{2^{10}}=\dfrac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}=\dfrac{2^4.3^6}{2^{10}}=\dfrac{2^{10}}{2^{10}}=1\)
e. \(\dfrac{2^{17}.25^5}{10^8.8^3}=\dfrac{2^{17}.\left(5^2\right)^5}{\left(2.5\right)^8.\left(2^3\right)^3}=\dfrac{2^{17}.5^{10}}{2^8.5^8.2^9}=\dfrac{2^{17}.5^{10}}{2^{17}.5^8}=\dfrac{5^{10}}{5^8}=5^{10}:5^8=5^2=25\)
f. \(\dfrac{3^{15}.25^4}{15^6.27^3}=\dfrac{3^{15}.\left(5^2\right)^4}{\left(3.5\right)^6.\left(3^3\right)^3}=\dfrac{3^{15}.5^8}{5^6.3^6.3^9}=\dfrac{3^{15}.5^8}{5^6.3^6.3^9}=\dfrac{5^8}{5^6}=5^8:5^6=5^2=25\)
2. Tính lũy thừa âm:
a. 3-2 = \(\dfrac{1}{3^2}\) = \(\dfrac{1}{9}\)
b. 2-3 = \(\dfrac{1}{2^3}\) = \(\dfrac{1}{8}\)
3. Tính :
a. \(\dfrac{\left(0,8\right)^4}{\left(0,4\right)^3}=\dfrac{\left(0,8\right)^3.0,8}{\left(0,4\right)^3}=\left(\dfrac{0,8}{0,4}\right)^3.0,8=2^3.0,8=8.0,8=6,4\)
b. \(\dfrac{\left(0,8\right)^3}{\left(0,4\right)^4}=\dfrac{\left(0,8\right)^3}{\left(0,4\right)^3:0,4}=\left(\dfrac{0,8}{0,4}\right)^3.\dfrac{1}{0,4}=2^3.2,5=8.2,5=20\)
c. \(\dfrac{\left(0,6\right)^5}{\left(0,2\right)^6}=\dfrac{\left(0,6\right)^5}{\left(0,2\right)^5.\left(0,2\right)}=\left(\dfrac{\left(0,6\right)}{\left(0,2\right)}\right)^5.\dfrac{1}{0,2}=3^5.\dfrac{1}{0,2}=\dfrac{3^5}{0,2}=1215\)
P/s : Chế Mai Ngọc Trâm thử tham khảo thử đi!!!
1. Tìm n, biết:
a) \(\dfrac{-32}{\left(-2\right)^n}=4\)
\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)
\(\Rightarrow\left(-2\right)^n.\left(-2\right)^2=\left(-2\right)^5\)
(-2)n + 2 = (-2)5
n + 2 = 5
n = 5 - 2
n = 3.
b) \(\dfrac{8}{2^n}=2\)
\(\Rightarrow\dfrac{2^3}{2^n}=2\)
\(\Rightarrow\) 2n . 2 = 23
n + 1 = 3
n = 3 - 1
n = 2.
c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)
2n - 1 = 3
2n = 3 + 1
2n = 4
n = 4 : 2
n = 2.
2. Tính:
a) \(\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left[\left(\dfrac{1}{2}\right)^2\right]^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{2}\right)^4\)
\(=\left(\dfrac{1}{2}\right)^7\)
\(=\dfrac{1}{128}\)
b) 273 : 93
= (33)3 : (32)3
= 39 : 36
= 33
= 27
c) 1252 : 253
= (53)2 : (52)3
= 56 : 56
= 1
d) \(\dfrac{27^2.8^5}{6^6.32^3}\)
\(=\dfrac{\left(3^3\right)^2.\left(2^3\right)^5}{6^6.\left(2^5\right)^3}\)
\(=\dfrac{3^6.2^{15}}{6^6.2^{15}}\)
\(=\dfrac{3^6}{6^6}\)
\(=\dfrac{1}{64}.\)
B2 :
b) 27\(^3\): 9\(^3\)= (27:9)\(^3\)= 3\(^3\)
c) 125\(^2\): 25\(^3\)= 15625 : 15625 = 1
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
a) \(A=2^{24}=\left(2^3\right)^8=8^8.\)(1)
\(B=3^{16}=\left(3^2\right)^8=9^8\)(2)
Từ (1) và (2) \(\Rightarrow A< B\)
Vậy \(A< B.\)
b) \(B=\left(0,3\right)^{30}=\left(0,3^2\right)^{15}=0,09^{15}\)(1)
\(A=\left(0,1\right)^{15}\)(2)
Từ (1) và (2) \(\Rightarrow A>B\)
Vậy \(A>B.\)
c) \(A=\left(\frac{-1}{4}\right)^8=\left(\frac{1}{4}\right)^8=\left[\left(\frac{1}{2}\right)^2\right]^8=\left(\frac{1}{2}\right)^{16}\)(1)
\(B=\left(\frac{1}{8}\right)^5=\left[\left(\frac{1}{2}\right)^3\right]^5=\left(\frac{1}{2}\right)^{15}\)(2)
Từ (1) và (2) \(\Rightarrow A>B\)
Vậy \(A>B.\)
d) \(A=102^7=102^6.102\)(1)
\(B=9^{13}=9^{12}.9=\left(9^2\right)^6.9=81^6.9\)(2)'
Từ (1) và (2) \(\Rightarrow A>B\)
Vậy \(A>B.\)
e) \(8A=8\frac{8^{18}+1}{8^{19}+1}=\frac{8^{19}+8}{8^{19}+1}=1+\frac{7}{8^{19}+1}\)(1)
\(8B=8\frac{8^{23}+1}{8^{24+1}}=\frac{8^{24}+8}{8^{24}+1}=1+\frac{7}{8^{24}+1}\)(2)
Từ (1) và (2) \(\Rightarrow8A>8B\Rightarrow A>B\)
Vậy \(A>B.\)
f) \(A=\frac{5^5}{5+5^2+5^3+5^4}=\frac{5^4}{1+5+5^2+5^3}=\frac{625}{156}>\frac{468}{156}=3.\)(1)
\(B=\frac{3^5}{3+3^2+3^3+3^4}=\frac{3^4}{1+3+3^2+3^3}=\frac{81}{40}< \frac{120}{40}=3.\)(2)
Từ (1) và (2) \(\Rightarrow A>B\)
Vậy \(A>B.\)
a, ta có A=2^24=64^4
B=3^16=81^4
Vì 64^4<81^4
Vậy 2^24<3^36
b, ta có A=0,1^15
B=0,3^30=0,09^15
Vì 0,1^15< 0,09^15
Vậy 0,1^15<0,3^30
a)\(VT=\left(-\dfrac{1}{8}\right)^{100}=\dfrac{1}{8^{100}}=\dfrac{1}{\left(2^3\right)^{100}}=\dfrac{1}{2^{300}}\)
\(VP=\left(-\dfrac{1}{4}\right)^{200}=\dfrac{1}{\left(2^2\right)^{200}}=\dfrac{1}{2^{400}}\)
\(\Rightarrow VT>VP\)
b) \(VT=4^{100}=\left(2^2\right)^{100}=2^{200}< 2^{202}=VP\)
c) \(VT=5^{2000}=\left(5^2\right)^{1000}=25^{1000}>10^{1000}=VP\)
d) \(VT=31^5< 32^5=\left(2^5\right)^5=2^{25}\)
\(VP=17^7>16^7=\left(2^4\right)^7=2^{28}\)
\(VP>VT\)