tìm các số nguyên n biết
a. \(\left(\frac{1}{3}\right)^n=\frac{1}{18}\)
b. \(\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
c. \(\left(\frac{-3}{4}\right)^n=\frac{81}{256}\)
d. 27n : 3n=9
e. \(\frac{1}{2}.2^n+4.2^n=9.2^5\)
Tìm số nguyên N biết:
\(a,\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
\(b,\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
\(c,\left(\frac{-3}{4}\right)^n=\frac{81}{256}\)
\(d,\left(2x+3\right)^2=\frac{9}{121}^n\)
a) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\frac{1^4}{3^4}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow n=4\)
Vậy n = 4
b) \(\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\frac{-8^3}{7^3}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\left(\frac{-8}{7}\right)^3=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow n=3\)
Vậy n = 3
Tìm n∈Z biết :
a,27n/3n
b,\(\frac{25}{5^n}\)=5
c,\(\frac{81}{\left(-3\right)^n}=-243\)
d,\(\frac{1}{2}\cdot2^n+4\cdot2^n=9\cdot2^5\)
e,(\(\frac{1}{3}\))n=\(\frac{1}{81}\)
f,\(\left(\frac{-3}{4}\right)^n=\frac{81}{256}\)
g,\(\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
h,5-1*25n=125
k,3-1*3n+6*3n-1=7*36
a) Câu này thiếu đề nhé bạn.
b) \(\frac{25}{5^n}=5\)
\(\Rightarrow5^n=25:5\)
\(\Rightarrow5^n=5\)
\(\Rightarrow5^n=5^1\)
\(\Rightarrow n=1\)
Vậy \(n=1.\)
c) \(\frac{81}{\left(-3\right)^n}=-243\)
\(\Rightarrow\left(-3\right)^n=81:\left(-243\right)\)
\(\Rightarrow\left(-3\right)^n=-\frac{1}{3}\)
\(\Rightarrow\left(-3\right)^n=\left(-3\right)^{-1}\)
\(\Rightarrow n=-1\)
Vậy \(n=-1.\)
e) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow n=4\)
Vậy \(n=4.\)
f) \(\left(-\frac{3}{4}\right)^n=\frac{81}{256}\)
\(\Rightarrow\left(-\frac{3}{4}\right)^n=\left(-\frac{3}{4}\right)^4\)
\(\Rightarrow n=4\)
Vậy \(n=4.\)
Chúc bạn học tốt!
d) \(\frac{1}{2}.2^n+4.2^n=9.2^5\)
\(\Rightarrow2^n.\left(\frac{1}{2}+4\right)=288\)
\(\Rightarrow2^n.\frac{9}{2}=288\)
\(\Rightarrow2^n=288:\frac{9}{2}\)
\(\Rightarrow2^n=64\)
\(\Rightarrow2^n=2^6\)
\(\Rightarrow n=6\)
Vậy \(n=6.\)
g) \(-\frac{512}{343}=\left(-\frac{8}{7}\right)^n\)
\(\Rightarrow\left(-\frac{8}{7}\right)^n=\left(-\frac{8}{7}\right)^3\)
\(\Rightarrow n=3\)
Vậy \(n=3.\)
h) \(5^{-1}.25^n=125\)
\(\Rightarrow5^{-1}.5^{2n}=5^3\)
\(\Rightarrow5^{-1+2n}=5^3\)
\(\Rightarrow-1+2n=3\)
\(\Rightarrow2n=3+1\)
\(\Rightarrow2n=4\)
\(\Rightarrow n=4:2\)
\(\Rightarrow n=2\)
Vậy \(n=2.\)
k) \(3^{-1}.3^n+6.3^{n-1}=7.3^6\)
\(\Rightarrow3^{n-1}+6.3^{n-1}=7.3^6\)
\(\Rightarrow3^{n-1}.\left(1+6\right)=7.3^6\)
\(\Rightarrow3^{n-1}.7=7.3^6\)
\(\Rightarrow n-1=6\)
\(\Rightarrow n=6+1\)
\(\Rightarrow n=7\)
Vậy \(n=7.\)
Chúc bạn học tốt!
b)\(\frac{25}{5^n}\)=5
\(5^n\)=25/5
\(5^n\)=\(5^1\)
⇒ n = 1
c) \(\frac{81}{\left(-3\right)^n}\)=-243
\(\left(-3\right)^n\)=81/-243
\(\left(-3\right)^n\)=\(\frac{-1}{3}\)
\(\left(-3\right)^n\)=\(\left(-3\right)^{-1}\)
⇒n=-1
Tìm số tự nhiên n biết:
a) \(\left[\left(0,5\right)^3\right]^n\)=\(\frac{1}{64}\)
b) \(\frac{64}{\left(-2\right)^{n+1}}=4\)
c) \(\left(\frac{1}{3}\right)^{n+1}=\frac{1}{81}\)
d) \(\left(\frac{3}{4}\right)^n.\frac{1}{2}=\frac{81}{512}\)
\(a,\left[\left(0,5\right)^3\right]^n=\frac{1}{64}\Rightarrow\left(0,125\right)^n=0,125^2\Rightarrow n=2\)
\(b,\frac{64}{\left(-2\right)^{n+1}}=4\Rightarrow\left(-2\right)^{n+1}=\frac{64}{4}\Rightarrow\left(-2\right)^{n+1}=16\Rightarrow\left(-2\right)^{n+1}=\left(-2\right)^4\)
\(\Rightarrow n+1=4\Rightarrow n=3\)
\(c,\left(\frac{1}{3}\right)^{n+1}=\frac{1}{81}\Rightarrow\left(\frac{1}{3}\right)^{n+1}=\left(\frac{1}{3}\right)^4\Rightarrow n+1=4\Rightarrow n=3\)
\(d,\left(\frac{3}{4}\right)^n.\frac{1}{2}=\frac{81}{512}\Rightarrow\left(\frac{3}{4}\right)^n=\frac{81}{512}:\frac{1}{2}=\frac{81}{256}\Rightarrow\left(\frac{3}{4}\right)^n=\left(\frac{3}{4}\right)^4\Rightarrow n=4\)
a)\(27^x:3^x=9\)
b)\(\frac{125}{5^x}=25\)
c)\(\frac{-243}{\left(-3\right)}x=-245\)
d)\(\left(\frac{1}{3}\right)x=\frac{1}{81}\)
e)\(\frac{-512}{343}=\left(\frac{-8}{7}\right)^x\)
g)\(\left(\frac{-3}{4}\right)^x=\frac{81}{256}\)
a) x=1
b) x=1
c) x= -(245/81)
d) x= 1/27
e) x=3
g) x=4
1) Tính:
a) \(\frac{6^3-3.6^2+3^2}{-13}\)
b) \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
2) Tìm n \(\in\)Z:
a) 27n : 3n = 9
b) \(\frac{25}{5^n}=5\)
c) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
d) \(\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
ko bt làm thì xuống lớp 6 hocj đi
Bạn 12345678901 xuống lớp 1 học đạo đức làm người nhé bạn. Lịch sự tí đi
CMR:Với mọi số tự nhiên n \(\ne\)0 ta đều có:
a.\(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{\left(3n-1\right)\cdot\left(3n+2\right)}=\frac{n}{6n+4}\)
b.\(\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+\frac{5}{11\cdot15}+...+\frac{5}{\left(4n-1\right)\cdot\left(4n+3\right)}=\frac{5n}{4n+3}\)
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3n+2}\right]=\frac{1}{3}\left[\frac{3n+2}{2\left(3n+2\right)}-\frac{2}{2\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\cdot\frac{3n}{6n+4}=\frac{n}{6n+4}=VP\)
b) Ta có: \(\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{4n+3}{12n+9}-\frac{3}{12n+9}\right)\)
\(=\frac{5}{4}.\frac{4n}{12n+9}\)
\(=\frac{5n}{12n+9}\)
( sai đề )
Các bn giúp mk vs,tí nữa là phải đi hx rùi.1h15' mk quay lại.Nhanh nha
Tìm n:
a)\(\left(\frac{1}{5}\right)^{3n-1}=\frac{1}{25}\)
b)\(\left(\frac{4}{7}\right)^{n+2}=\frac{7}{4}\)
c)\(\left(\frac{2}{3}\right)^{-n+1}=\frac{3^3}{2^3}\)
d) \(\left(0,7\right)^{3n+1}=10^3:7^3\)
a)\(\left(\frac{1}{5}\right)^{3n-1}=\frac{1}{25}\)
\(\Leftrightarrow\left(\frac{1}{5}\right)^{3n-1}=\left(\frac{1}{5}\right)^2\)
\(\Leftrightarrow3n-1=2\)
\(\Leftrightarrow3n=3\)
\(\Leftrightarrow n=1\)
b)\(\left(\frac{4}{7}\right)^{n+2}=\frac{7}{4}\)
\(\Leftrightarrow\left(\frac{4}{7}\right)^{n+2}=\left(\frac{4}{7}\right)^{-1}\)
\(\Leftrightarrow n+2=-1\)
\(\Leftrightarrow n=-3\)
c)\(\left(\frac{2}{3}\right)^{-n+1}=\frac{3^3}{2^3}\)
\(\Leftrightarrow\left(\frac{2}{3}\right)^{-n+1}=\left(\frac{3}{2}\right)^3\)
\(\Leftrightarrow\left(\frac{2}{3}\right)^{-n+1}=\left(\frac{2}{3}\right)^{-3}\)
\(\Leftrightarrow-n+1=-3\)
\(\Leftrightarrow n=-4\)
c)\(\left(0,7\right)^{3n+1}=10^3:7^3\)
\(\Leftrightarrow\left(\frac{7}{10}\right)^{3n+1}=\left(\frac{10}{7}\right)^3\)
\(\Leftrightarrow\left(\frac{7}{10}\right)^{3n+1}=\left(\frac{7}{10}\right)^{-3}\)
\(\Leftrightarrow3n+1=-3\)
\(\Leftrightarrow3n=-4\)
\(\Leftrightarrow n=-\frac{4}{3}\)
Bài 1: a, Tìm số nguyên a để tích hai phân số \(\frac{-19}{5}\) và \(\frac{a}{a-1}\)là một số nguyên.
b, Tìm số nguyên a để \(\frac{5}{4}\): \(\frac{a}{a+1}\)được thương là một số nguyên.
c,Tìm phân số dương \(\frac{a}{b}\)nhỏ nhất sao cho khi chia \(\frac{a}{b}cho\frac{7}{9}\)hoặc khi chia cho \(\frac{5}{12}\)được mỗi thương là một số tự nhiên
Bài 2:a,Với giá trị nào của x thì ta có:
1,A= \(\left(x-\frac{3}{4}\right)\left(x+\frac{1}{2}\right)\)là số dương 2,B=\(\frac{x-0,5}{x+1}\)là số âm.
b,Cho phân số \(\frac{a}{b}\left(b\ne0\right)\).Tìm phân số \(\frac{c}{d}\left(c\ne0,d\ne0\right)\)sao cho \(\frac{a}{b}:\frac{c}{d}=\frac{a}{b}.\frac{c}{d}\)
c, Tìm các cặp số nguyên (x,y) để: \(B=\frac{1}{x-y}:\frac{x+2}{2\left(x-y\right)}\)là số nguyên.
Bài 3: a, Tính : A=\(\left(-2\right)\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{n}\right)\left(n\in N,n\ne0\right)\)
B=\(\frac{4\frac{1}{4}}{11\frac{1}{3}.5\frac{1}{4}}\) C= \(\frac{-1:1\frac{1}{15}}{3\frac{1}{8}:6\frac{2}{3}}:\frac{4\frac{7}{8}:13}{5:1\frac{7}{8}}\) D=\(-\frac{7}{4}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
E=\(\frac{1}{2}:\left(-1\frac{1}{2}\right):1\frac{1}{3}:\left(-1\frac{1}{4}\right):1\frac{1}{5}:\left(-1\frac{1}{6}\right):...:\left(-1\frac{1}{100}\right)\) F=\(4+\frac{1}{1+\frac{1}{1+\frac{2}{1+\frac{3}{4}}}}\)
nhiều thế ??????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????
CMR:Với mọi số tự nhiên n \(\ne\)0 ta đều có:
a.\(\frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+...+\frac{1}{\left(3n-1\right)\times\left(3n+2\right)}=\frac{1}{6n+4}\)
b.\(\frac{5}{3\times7}+\frac{5}{7\times11}+\frac{5}{11\times15}+...+\frac{5}{\left(4n-1\right)\times\left(4n+3\right)}=\frac{5n}{4n+3}\)
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\)
\(=\frac{1}{2}-\frac{1}{3n+2}=\frac{3n+2}{2\cdot\left(3n+2\right)}-\frac{2}{2\cdot\left(3n+2\right)}\)
\(=\frac{3n+2-2}{6n+4}=\frac{3n}{6n+4}=VP\)
b)\(VT=\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{4}\left[\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right]\)
\(=\frac{5}{4}\cdot\left[\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right]\)
\(=\frac{5}{4}\cdot\left[\frac{1}{3}-\frac{1}{4n+3}\right]=\frac{5}{4}\cdot\left[\frac{4n+3}{3\left(4n+3\right)}-\frac{3}{3\left(4n+3\right)}\right]\)
\(=\frac{5}{4}\cdot\left[\frac{4n+3-3}{12n+9}\right]\)\(=\frac{5}{4}\cdot\frac{4n}{12n+9}=\frac{5n}{12n+9}\)