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a)\(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)
\(x=\left(\frac{3}{7}\right)^7\div\left(\frac{3}{7}\right)^5\)
\(x=\left(\frac{3}{7}\right)^2\)
\(x=\frac{9}{49}\)
Vậy...
b)\(\left(-\frac{1}{3}\right)^3.x=\left(\frac{1}{3}\right)^4\)
\(\left(-\frac{1}{3}\right)^3.x=\left(-\frac{1}{3}\right)^4\)
\(x=\left(-\frac{1}{3}\right)^4\div\left(\frac{-1}{3}\right)^3\)
\(x=-\frac{1}{3}\)
Vậy...
c)\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
=>\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
Vậy...
d)\(\left(x+\frac{1}{4}\right)^4=\left(\frac{2}{3}\right)^4\)
=>\(x+\frac{1}{4}=\frac{2}{3}\)
\(x=\frac{2}{3}-\frac{1}{4}\)
\(x=\frac{5}{12}\)
Vậy...
Phù, mãi mới xong, tk cho mk nha bn
a)\(\left(\frac{3}{5}\right)^5\times x=\left(\frac{3}{7}\right)^7\)
\(\Leftrightarrow\frac{3^5}{5^5}\times x=\frac{3^7}{7^7}\)
\(\Leftrightarrow x=\frac{3^7}{7^7}:\frac{3^5}{5^5}\)
\(\Leftrightarrow x=\frac{3^7\times5^5}{7^7\times3^5}\)
\(\Leftrightarrow x=\frac{3^2\times5^5}{7^7}\)
b)\(\left(\frac{-1}{3}\right)^3\times x=\frac{1}{81}\)
\(\Leftrightarrow\frac{\left(-1\right)^3}{3^3}\times x=\frac{1}{3^4}\)
\(\Leftrightarrow x=\frac{1}{3^4}:\frac{-1}{3^3}\)
\(\Leftrightarrow x=\frac{1\times3^3}{3^4\times\left(-1\right)}\)
\(\Leftrightarrow x=\frac{1}{-3}\)
c)\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)
\(\Leftrightarrow x=\frac{5}{6}\)
d)\(\Leftrightarrow\left(x+\frac{1}{2}\right)^4=\left(\frac{2}{3}\right)^4\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{2}{3}-\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{6}\)
Tìm x:
a)-23+0,5x=1,5
-8+0,5x = 1,5
0,5x = 1,5-(-8) = 9,5
x = 9,5:0,5 = 19
b)(−3)x81=−27
(-3)x:81 =-27
(-3)x = -27.81 = -2187
(-3)x = (-3)7
=> x=7
c)112.x−4=0,5
1,5.x = 0,5+4 = 4,5
x = 4,5:1,5 = 3
d)123:x4=6:0,3
\(\frac{5}{3}\):\(\frac{x}{4}\) = 20
\(\frac{x}{4}\) = \(\frac{5}{3}\):20 = \(\frac{1}{12}\)
=> x:4 = \(\frac{1}{12}\)
x = \(\frac{1}{12}\).4 = \(\frac{1}{3}\)
h) \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x+5^x.5^2=650\)
\(\Leftrightarrow5^x\left(1+25\right)=650\)
\(\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow x=2\)
haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=
bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây
1) Tính
a) 253 : 52 = (52)3 : 52 = 56 : 52 = 54 = 625
\(b)\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^6=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^6=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^9\) d) 9 . 32 . \(\dfrac{1}{81}\) . 32 = 32 . 32 . \(\dfrac{1}{3^4}\) . 32 = 9
2) Tìm x thuộc Q, biết:
a) 3x + 2 = 27
=> 3x + 2 = 33
x + 2 = 3
x = 3 - 2
x = 1
b) \(\left(\dfrac{1}{2}x-3\right)^4=81\)
\(\Rightarrow\left(\dfrac{1}{2}x-3\right)^4=3^4\)
\(\dfrac{1}{2}x-3=3^{ }\)
\(\dfrac{1}{2}x=3+3\)
\(\dfrac{1}{2}x=9\)
\(x=9:\dfrac{1}{2}\)
\(x=18\)
c) \(\left(x-\dfrac{1}{2}\right)^3=-27\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(-3\right)^3\)
\(x-\dfrac{1}{2}=-3\)
\(x=-3+\dfrac{1}{2}\)
\(x=\dfrac{-5}{2}\)
d) 5 . 5x + 1 = 125
5x + 1 = 125 : 5
5x + 1 = 25
5x + 1 = 52
x + 1 = 2
x = 2 - 1
x = 1.
a) \(\frac{\left(-3\right)^x}{81}=-27\)
\(\Rightarrow\left(-3\right)^x=\left(-27\right).81\)
\(\Rightarrow\left(-3\right)^x=-2187\)
\(\Rightarrow\left(-3\right)^x=\left(-3\right)^7\)
\(\Rightarrow x=7\)
Vậy \(x=7.\)
b) \(1\frac{1}{2}.x-4=0,5\)
\(\Rightarrow1\frac{1}{2}.x=0,5+4\)
\(\Rightarrow1\frac{1}{2}.x=\frac{9}{2}\)
\(\Rightarrow\frac{3}{2}.x=\frac{9}{2}\)
\(\Rightarrow x=\frac{9}{2}:\frac{3}{2}\)
\(\Rightarrow x=3\)
Vậy \(x=3.\)
Chúc bạn học tốt!
a) \(\left(\frac{1}{3}\right)^n=\frac{1}{27}\)
\(\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^3\)
\(\Rightarrow n=3\)
b) \(\left(\frac{3}{5}\right)^n=\frac{81}{625}\)
\(\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^4\)
\(\Rightarrow n=4\)
c) \(3^n\cdot2^n=36\)
\(\left(3\cdot2\right)^n=36\)
\(6^n=6^2\)
\(\Rightarrow n=2\)
d) \(\frac{2^n}{3^n}=\frac{8}{27}\)
\(\left(\frac{2}{3}\right)^n=\left(\frac{2}{3}\right)^3\)
\(\Rightarrow n=3\)