(2.n+1)^2=625
Tìm số tự nhiên n, biết : a) 5^n +1 = 625; b) 7^n= 7^2. 7^4; c) 7. 2^3n-1=224
`5^(n + 1) = 625`
`=> 5^(n + 1) = 5^4`
`=> n + 1 = 4`
`=> n = 4 -1`
`=> n = 3`
`7^n = 7^2 . 7^4`
`=> 7^n = 7^(2 + 4)`
`=> 7^n = 7^6`
`=> n = 6`
`7. 2^(3n - 1) = 224`
`=>2^(3n-1) = 224 : 7`
`=> 2^(3n-1) = 32`
`=> 2^(3n -1) = 2^5`
`=> 3n - 1 = 5`
`=> 3n = 6`
`=> n = 2`
a: =>5^(n+1)=5^4
=>n+1=4
=>n=3
b: =>7^n=7^6
=>n=6
c: =>2^(3n-1)=32
=>3n-1=5
=>3n=6
=>n=2
a) \(5^{n+1}=625\)
\(\Rightarrow5^{n+1}=5^4\)
\(\Rightarrow n+1=4\)
\(\Rightarrow n=4-1\)
\(\Rightarrow n=3\)
b) \(7^n=7^2\cdot7^4\)
\(\Rightarrow7^n=7^{2+4}\)
\(\Rightarrow7^n=7^6\)
\(\Rightarrow n=6\)
c) \(7\cdot2^{3n-1}=224\)
\(\Rightarrow2^{3n-1}=224:7\)
\(\Rightarrow2^{3n-1}=32\)
\(\Rightarrow2^{3n-1}=2^5\)
\(\Rightarrow3n-1=5\)
\(\Rightarrow3n=5+1\)
\(\Rightarrow3n=6\)
\(\Rightarrow n=\dfrac{6}{3}\)
\(\Rightarrow n=2\)
(625-1 mũ 3).(625-2 mũ 3).(625-3 mũ 3)....(625-2015 mũ 3)
tim x thuoc N biet :a,(2 nhan x cong 1 ) binh phuong 2 =625
( 2 x X + 1 )2 = 625
( 2 x X + 1 ) 2 = 252
2 x X + 1 = 25
2 x X = 24
x = 12
Tìm số mũ n biết
(-1/2)^n=1/64
81/625=(3/5)^n
3^n/32=2
(-3)^3×(-3)^n=243
Tìm tất cả các số nguyên n sao cho
\(\sqrt{\dfrac{25}{2}+\sqrt{\dfrac{625}{4}-n}}+\sqrt{\dfrac{25}{2}-\sqrt{\dfrac{625}{4}-n}}\)
ĐK: \(n\le\dfrac{625}{4}\le156\) (vì \(n\in Z\) )
Đặt \(a=\sqrt{\dfrac{25}{2}+\sqrt{\dfrac{625}{4}-n}}+\sqrt{\dfrac{25}{2}-\sqrt{\dfrac{625}{4}-n}}\) \(\left(a\ge0,a\in Z\right)\)
\(\Rightarrow a^2=25+2\sqrt{\dfrac{625}{4}-\dfrac{625}{4}+n}\)
\(\Rightarrow a^2=25+2\sqrt{n}\) (1)
Để \(a\in Z\Rightarrow a^2\in Z\Rightarrow\sqrt{n}\in Z^+\)
Vì \(2\sqrt{n}⋮2\) mà 25 không chia hết cho 2
\(\Rightarrow a^2\) không chia hết cho 2
\(\Rightarrow\) a không chia hết cho 2
Đặt \(a=2k+1\left(k>0,k\in Z\right)\)
\(\left(1\right)\Rightarrow\left(2k+1\right)^2=25+2\sqrt{n}\)
\(\Rightarrow2\sqrt{n}=4k^2+4k-24\)
\(\Rightarrow\sqrt{n}=2k^2+2k-12\)
Vì \(\sqrt{n}\ge0\Rightarrow2k^2+2k-12\ge0\)
\(\Rightarrow\left(k+3\right)\left(k-2\right)\ge0\)
Vì \(k>0\Rightarrow k\ge2\) (2)
Mặt khác: \(n\le156\Rightarrow\sqrt{n}\le\sqrt{156}\) mà \(\sqrt{n}\in Z\)
\(\Rightarrow\sqrt{n}\le12\Rightarrow2k^2+2k-12\le12\)
\(\Rightarrow\left(k-3\right)\left(k+4\right)\le0\)
Vì \(k>0\Rightarrow0< k\le3\) (3)
Từ (2) và (3)\(\Rightarrow\left[{}\begin{matrix}k=2\\k=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}n=0\\n=144\end{matrix}\right.\) (t/m)
Vậy n=0, n=144
Nguyễn Việt Lâm Uyen Vuuyen Trần Trung Nguyên JakiNatsumi Vương Đại Nguyên bullet sivel Nguyễn Thanh Hằng KHUÊ VŨ @Nk>↑@ mấy best toán chỉ e với
\([(-\dfrac{1}{5})^n]^2=\dfrac{1}{625}\) Tìm n \(\in N\)
\(\left[\left(-\dfrac{1}{5}\right)^n\right]^2=\dfrac{1}{625}\) \(\Leftrightarrow\left(-\dfrac{1}{5}\right)^n=\pm\sqrt{\dfrac{1}{625}}=\pm\dfrac{1}{25}\)
\(th1:\left(-\dfrac{1}{5}\right)^n=\dfrac{1}{25}=\left(-\dfrac{1}{5}\right)^2\Rightarrow n=2\)
\(th2:\left(-\dfrac{1}{5}\right)^n=\dfrac{-1}{25}\) (không có giá trị nào của n thỏa mãn)
vậy \(n=2\)
\(\left[\left(-\dfrac{1}{5}\right)^n\right]^2=\dfrac{1}{625}\)
\(\Rightarrow\left(\dfrac{1}{5}\right)^{2n}=\left(\dfrac{1}{5}\right)^4\)
Vì \(\dfrac{1}{5}\ne-1;\dfrac{1}{5}\ne0;\dfrac{1}{5}\ne1\) nên \(2n=4\Rightarrow n=2\)
Vậy.............
Chúc bạn học tốt!!!
5 * = 625
2 *x 16=128
3* :9 =27
(2n +1) mũ 3 =27
(n -2 ) mũ 2 =(n-2) mũ 4
\(5^x=625\)
\(\Rightarrow5^x=5^4\)
\(\Rightarrow x=4\)
\(3^x:9=27\)
\(3^x=27.3\)
\(3^x=81\)
\(3^x=3^4\)
\(\Rightarrow x=4\)
\(\left(2n+1\right)^3=27\)
\(\left(2n+1\right)^3=3^3\)
\(\Rightarrow2n+1=3\)
\(2n=3-1\)
\(2n=2\)
\(n=2:2\)
\(n=1\)
1. tim n
a,5n=625
b.12n=144
c.(n-2)2=(n-2)4
a. \(5^n=625\)
\(\Rightarrow5^n=5^4\)
\(\Rightarrow n=4\)
Vậy.....
b. \(12^n=144\)
\(\Rightarrow12^n=12^2\)
\(\Rightarrow n=2\)
Vậy........
a) 5n = 625
\(\Rightarrow\) 5n = 54
\(\Rightarrow\) n = 4
a,5n=625
5n=54
=>n=4
b,12n=144
12n=122
=>n=2
Tìm tất cả các số nguyên n sao cho biểu thức \(\sqrt{\frac{25}{2}+\sqrt{\frac{625}{4}-n}}+\sqrt{\frac{25}{2}-\sqrt{\frac{625}{4}-n}}\) có giá trị nguyên.