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a) \(10^{n+1}-6.10^n\)
\(=10^n.10-6.19^n\)
\(=10^n.\left(10-6\right)\)
\(=10^n.4\)
b) \(2^{n+3}+2^{n+2}-2^{n+1}+2^n\)
\(=2^n.2^3+2^n.2^2-2^n.2+2^n.1\)
\(=2^n.\left(2^3+2^2-2+1\right)\)
\(=2^n.11\)
c) \(90.10^k-10^{k+2}+10^{k+1}\)
\(=90.10^k-10^k.10^2+10^k.10\)
\(=10^k.\left(90-10^2+10\right)\)
\(=0\)
d) \(2,5.5^{n-3}.10+5^n-6.5^{n-1}\)
\(=\dfrac{2,5.5^n.10}{5^3}+5^n-\dfrac{6.5^n}{5}\)
\(=\dfrac{5^n}{5}+5^n-\dfrac{6.5^n}{5}\)
\(=\dfrac{5^n+5^{n+1}-6.5^n}{5}=\dfrac{5^n+5^n.5-6.5^n}{5}=\dfrac{5^n\left(1+5-6\right)}{5}=\dfrac{0}{5}=0\)
Bài 2:
a: =>50x+50=0
=>50x=-50
=>x=-1
b: \(\Leftrightarrow5^{2x-1}=5^3\)
=>2x-1=3
=>2x=4
=>x=2
c: \(\Leftrightarrow3^{x-1}+6\cdot3^{x-1}=7\cdot3^6\)
=>3^x-1=3^6
=>x-1=6
=>x=7
a, 5-1x 25n = 125 d, 25 < 5n:5 < 625
5-1 x 52n = 53 52 < 5n:5 < 54
=> -1+2n=3 => n=4
=>2n = 3--1
=>2n=4
=>n =2
a,\(5^{-1}\times25^n=125 \)
= \(\frac{1}{5}\times25^n=125\)
= \(25^n=125\div\frac{1}{5}\)
= \(25^n=625\)
= \(25^n=25^2\)
\(\Rightarrow n=2\)
8.2n +2n+1
=2n .(8+2)
=2n.10 chia hết cho 10
=> 8.2n +2n+1 chia hết cho 10
\(3^{n+3^{ }}-2.3^n+2^{n+5}-7.2^n\)
\(=3^n.\left(3^3-2\right)+2^n\left(2^5-7\right)\)
\(=3^n.25+2^n.25\)
=\(25.\left(3^n+2^n\right)\)chia hết cho 25
=>\(3^{n+3}-2.3^n+2^{n+5}-7.2^n\)
k cho mình nhé
a) 9.27n = 35
=> 32.33n = 35
=> 32 + 3n = 35
=> 2 + 3n = 5
=> 3n = 5 - 2
=> 3n = 3
=> n = 1
b) (23 : 4).2n = 4
=> 2.2n = 4
=> 2n = 4 : 2
=> 2n = 2
=> n = 1
c) 3-2.34 . 3n = 37
=> 3-2 + 4 + n = 37
=> 32 + n = 37
=> 2 + n = 7
=> n = 7 - 2 = 5
d) 2-1.2n + 4.2n = 9.25
=> (1/2 + 4).2n = 9.25
=> 9/2.2n = 9.25
=> 2n = 9.25 : 9/2
=> 2n = 26
=> n = 6
\(a,9\cdot27^n=3^5\)
\(\Rightarrow9\cdot27^n=243\)
\(\Rightarrow27^n=243:9=27\)
\(\Rightarrow27^n=27^1\)
\(\Rightarrow x=1\)
\(b,\left(2^3:4\right)\cdot2^n=4\)
\(\Rightarrow\left(8:4\right)\cdot2^n=4\)
\(\Rightarrow2\cdot2^n=4\)
\(\Rightarrow2^n=4:2=2\)
\(\Rightarrow n=1\)
\(c,3^{-2}\cdot3^4\cdot3^n=3^7\)
\(\Rightarrow3^2\cdot3^n=3^7\)
\(\Rightarrow3^n=3^7:3^2=3^5\)
\(\Rightarrow n=5\)
\(d,2^{-1}\cdot2^n+4\cdot2^n=9\cdot2^5\)
\(\Rightarrow2^n\cdot\left(2^{-1}+4\right)=9\cdot32\)
\(\Rightarrow2^n\cdot\frac{9}{2}=288\)
\(\Rightarrow2^n=288:\frac{9}{2}=64\)
\(\Rightarrow2^n=2^6\)
\(\Rightarrow n=6\)
Bài này là đê thi HSG khối 8 đó ko phải khối 7 đâu!
Ta có:
A= \(5^n\left(5^n+1\right)-6^n\left(3^n+2^n\right)\)
\(=25^n+5^n-18^n-12^n\)
* \(=\left(25^n-18^n\right)-\left(12^n-5^n\right)\text{ do đó A chia hết cho 7}\)
* \(=\left(25^n-12^n\right)-\left(18^n-5^n\right)\text{ do đó A chia hết cho 13}\)
Do (7;13)=1 nên A chia hết cho 91
NOTE: mk đã lm theo cách lớp 7 đó! lớp 8 thì phải dùng đồng dư thức cơ! nhưng mk lâu rồi chưa lm lại ko biết có đúng ko mong bn kiểm tra rồi thông báo cho mk sớm nhất có thể nhé!!
1.
a.
\(\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{7}\right)\)
\(=\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\)
\(=\frac{35-21-15}{105}\)
\(=-\frac{1}{105}\)
b.
\(\frac{3}{5}-\left(\frac{3}{4}-\frac{1}{2}\right)\)
\(=\frac{3}{5}-\frac{3}{4}+\frac{1}{2}\)
\(=\frac{12-15+10}{20}\)
\(=\frac{7}{20}\)
c.
\(\frac{4}{7}-\left(\frac{2}{5}+\frac{1}{3}\right)\)
\(=\frac{4}{7}-\frac{2}{5}-\frac{1}{3}\)
\(=\frac{60-42-35}{105}\)
\(=-\frac{17}{105}\)
2.
a.
\(S=-\frac{1}{1\times2}-\frac{1}{2\times3}-\frac{1}{3\times4}-...-\frac{1}{\left(n-1\right)\times n}\)
\(S=-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{\left(n-1\right)\times n}\right)\)
\(S=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(S=-\left(1-\frac{1}{n}\right)\)
\(S=-1+\frac{1}{n}\)
b.
\(S=-\frac{4}{1\times5}-\frac{4}{5\times9}-\frac{4}{9\times13}-...-\frac{4}{\left(n-4\right)\times n}\)
\(S=-\left(\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{\left(n-4\right)\times n}\right)\)
\(S=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)
\(S=-\left(1-\frac{1}{n}\right)\)
\(S=-1+\frac{1}{n}\)
Chúc bạn học tốt![]()
c, \(\frac{-32}{-2^n}=4\)
\(\Rightarrow-2^n=-32:4\)
\(\Rightarrow-2^n=-8\)
\(\Rightarrow-2^n=-2^3\Rightarrow n=3\)
d, \(\frac{8}{2^n}=2\)
\(\Rightarrow2^n=8:2\)
\(\Rightarrow2^n=4\)
\(\Rightarrow2^n=2^2\Rightarrow n=2\)
e, \(\frac{25^3}{5^n}=25\)
\(\Rightarrow5^n=25^3:25\)
\(\Rightarrow5^n=25^2\)
\(\Rightarrow5^n=5^4\Rightarrow n=4\)
i , \(8^{10}:2^n=4^5\)
\(\Rightarrow2^n=8^{10}:4^5\)
\(\Rightarrow2^n=\left(2^3\right)^{10}:\left(2^2\right)^5\)
\(\Rightarrow2^n=2^{30}:2^{10}\)
\(\Rightarrow2^n=2^{20}\Rightarrow n=20\)
k, \(2^n.81^4=27^{10}\)
\(\Rightarrow2^n=27^{10}:81^4\)
\(\Rightarrow2^n=\left(3^3\right)^{10}:\left(3^4\right)^4\)
\(\Rightarrow2^n=3^{30}:3^{16}\)
\(\Rightarrow2^n=3^{14}\)
\(\Rightarrow2^n=4782969\)Không chia hết cho 2 nên ko có Gt n thỏa mãn
\(5^{n+1}+5^n=6\cdot125\)
`-> 5^n*5+5^n=750`
`-> 5^n(5+1)=750`
`-> 5^n*6=750`
`-> 5^n = 125`
`-> 5^n = 5^3`
`-> n=3`
\(5^{n+1}+5^n=6.125\)
\(\Leftrightarrow5^n.5+5^n=750\)
\(\Leftrightarrow5^n\left(5+1\right)=750\)
\(\Leftrightarrow5^n=125\)
\(\Rightarrow n=3\)
Lớp 7 chưa học phương trình tương đương. Không được sử dụng tùy tiện dấu? Từ đầu là tương đương đến cuối là suy ra?