Cho mk sửa lại câu b nè 
\(\frac{81}{\left(-3\right)^x}=-243\)
Những câu hỏi liên quan
Tìm x biết : \(\left(\frac{1}{3}\right)^x\left(\frac{1}{9}\right)^x\left(\frac{1}{27}\right)^x\left(\frac{1}{81}\right)^x\left(\frac{1}{243}\right)^x=\left(-\frac{1}{3}\right)^{30}\)
Tìm x biết:
\(\left(\frac{3}{5}\right)^x.\left(\frac{625}{81}\right)^3=\frac{243}{3125}\)
\(\left(\frac{3}{5}\right)^x.\left(\frac{625}{81}\right)^3=\frac{243}{3125}\)
=>\(\left(\frac{3}{5}\right)^x.\left(\frac{5}{3}\right)^{12}=\left(\frac{3}{5}\right)^5\)
=>\(\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^5:\left(\frac{5}{3}\right)^{12}\)
=>\(\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^{17}\)
=>x=17
Đúng 0
Bình luận (1)
\(\left(\frac{3}{5}\right)^x.\left(\frac{625}{81}\right)^3=\frac{243}{3125}\)
\(\Rightarrow\left(\frac{3}{5}\right)^x.\left(\frac{3}{5}\right)^{12}=\left(\frac{3}{5}\right)^5\)
\(\Rightarrow\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^5:\left(\frac{3}{5}\right)^{12}\)
\(\Rightarrow\left(\frac{3}{5}\right)^x=\left(\frac{5}{3}\right)^7\)
\(\Rightarrow x=-7\)
Đúng 1
Bình luận (0)
Tìm số tự nhiên x biết:
a, \(8< 2^x< \frac{2^9}{2^5}\)
b, \(27< 81^3:3^x< 243\)
c, \(\left(\frac{2}{5}\right)^x>\left(\frac{2}{5}\right)^3.\left(-\frac{2}{5}\right)^2\)
CÁC BN GIẢI HỘ MK NHA!!!! *-**.**=*
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
Đúng 0
Bình luận (0)
\(\left[x+\frac{1}{3}\right]+\left[x+\frac{1}{15}\right]+\left[x+\frac{1}{35}\right]+...+\left[x+\frac{1}{575}\right]\)
\(=11xx+\left[\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right]\)
\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{15}\right)+....+\left(x+\frac{1}{575}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(13x+\left(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(13x+\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(2x+\frac{12}{25}=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
Đặt \(A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(3A=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
\(3A-A=1-\frac{1}{3^5}=\frac{242}{243}=2A\)
=> \(A=\frac{121}{243}\)
=> \(2x+\frac{12}{25}=\frac{121}{243}\)
=> \(2x=\frac{121}{243}-\frac{12}{25}=\frac{109}{6075}\)
=> x = ......
Đúng 0
Bình luận (0)
Tìm \(x\)sao cho:
\(\left(x+\frac{1}{1\cdot3}\right)+\left(x+\frac{1}{3\cdot5}\right)+\left(x+\frac{1}{5\cdot7}\right)+...+\left(x+\frac{1}{23\cdot25}\right)=11\cdot x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
Chỉ biết \(x\) = \(\frac{109}{6075}\) thôi
Đúng 0
Bình luận (0)
tìm số nguyên x
a)\(27^n:3^n=9\)
b)\(\left(\frac{-1}{3}\right)^N=\frac{1}{81}\)c)\(\frac{25}{5^n}=5\)d)\(\frac{1}{2}\cdot2^n+4\cdot2^n=9\cdot2^5\)e)\(\frac{81}{\left(-3\right)^n}=-243\)
Bn nào giải đc câu nào thì giải nhé ko giải đc câu nào thì thôi
Tìm x:
\(\left(x+\frac{1}{1x3}\right)+\left(x+\frac{1}{3x5}\right)+......+\left(x+\frac{1}{23x25}\right)=11.x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+\frac{1}{243}\right)\)
Help me please! Thanks a lot!
Nhân 2 cả 2 vế lên:
\(\left(2x+\frac{2}{1x3}\right)+...+\left(2x+\frac{2}{23x25}\right)=22x+\frac{2}{3}+\frac{2}{9}+\frac{2}{81}+\frac{2}{243}\)2/243
\(24x+\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{23}-\frac{1}{25}\right)=22x+\frac{162+54+6+2}{243}\)
\(24x+\frac{24}{25}=22x+\frac{224}{243}\)
\(2x=\frac{224}{243}-\frac{24}{25}\)
\(2x=-\frac{232}{6025}\)
\(x=\frac{-116}{6075}\)
Đúng 0
Bình luận (0)
\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11.x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+\frac{1}{243}\right)\)
\(12x+\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\right)\right]=11.x+\left(\frac{81}{243}+\frac{27}{243}+\frac{3}{243}+\frac{1}{243}\right)\)
\(12x+\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{25}\right)\right]=11.x+\frac{112}{243}\)
\(12x+\left(\frac{1}{2}.\frac{24}{25}\right)=11.x+\frac{112}{243}\)
\(12x+\frac{12}{25}=11x+\frac{112}{243}\)
\(11x-12x=\frac{112}{243}-\frac{12}{25}\)
\(-1x=-\frac{116}{6075}\)
\(x=-\frac{116}{6075}\div\left(-1\right)\)
\(x=\frac{116}{6075}\)
Đúng 0
Bình luận (0)
so sánh:
a)\(\left(\frac{1}{80}\right)^7và\left(\frac{1}{243}\right)^6\)
b)\(\left(\frac{3}{8}\right)^5và\left(\frac{5}{243}\right)^3\)
giải chi tiết cho mk nha!!!