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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)\(\left(\frac{-1}{3}\right)^3\cdot x=\frac{1}{81}\) \(< =>\frac{-1}{27}x=\frac{1}{81}\)\(< =>x=\frac{-1}{3}\)
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 = \(\frac{1}{3}\)
Ta có:
\(\left(\frac{1}{81}\right)^x=\left(\frac{1}{3}\right)^{4x}\)
\(\left(\frac{1}{27}\right)^4=\left(\frac{1}{3}\right)^{12}\)
=> 4x=12
=> x=3
\(\left(\frac{1}{81}\right)^x=\left(\frac{1}{27}\right)^4\)
\(\Leftrightarrow\left(\frac{1}{3^4}\right)^x=\left(\frac{1}{3^3}\right)^4\)
\(\Leftrightarrow\left(\frac{1}{3}\right)^{4x}=\left(\frac{1}{3}\right)^{12}\)
\(\Leftrightarrow4x=12\)
\(\Leftrightarrow x=3\)
Bài 2:
a) \(\frac{x}{-27}=\frac{-3}{x}\Leftrightarrow-\frac{x}{27}=-\frac{3}{x}\Leftrightarrow-x.x=\left(-27\right).\left(-3\right)\Leftrightarrow-x^2=-81\Leftrightarrow\orbr{\begin{cases}x=9\\x=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=9\\x=-9\end{cases}}\)
b) \(\frac{-9}{x}=\frac{-x}{\frac{4}{49}}\Leftrightarrow-\frac{9}{x}=-\frac{49x}{4}\Leftrightarrow-9.4=-x.49x\Leftrightarrow-36=-49x^2\Leftrightarrow\orbr{\begin{cases}x=\frac{6}{7}\\x=-\frac{6}{7}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{6}{7}\\x=-\frac{6}{7}\end{cases}}\)
1/vì (1,782x-2-1,78x):1,78x=0
nên 1,78x2-2-1,78x=0
=>1,782x-2=1,78x
=>2x-2=x
2x=x+2
=>x=2
2/vì cơ số bằng nhau nên ta có
x-2=1;-1;0
ta có: x-2=1 => x=3
x-2=-1 => x=1
x-2=0 => x=2
3/ta có
(x+2)3=33 =>x+2=3 =>x=1
mik mệt rồi bạn cứ gải tiếp đi
1. \(\frac{x^7}{81}=27\Leftrightarrow x^7=2187\)
\(\Leftrightarrow x^7=3^7\Leftrightarrow x=3\)
2. \(\left(x^4\right)^2=\frac{x^{12}}{x^5}\Leftrightarrow x^8=x^7\)
\(\Leftrightarrow x^8-x^7=0\Leftrightarrow x^7\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy,...
3.\(x^{10}=25x^8\Leftrightarrow x^{10}-25x^8=0\)
\(\Leftrightarrow x^8\left(x^2-25\right)=0\Leftrightarrow x^8\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^8=0\\x+5=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\\x=5\end{matrix}\right.\)
4. \(\left(3x-1\right)^3=\frac{-8}{27}\Leftrightarrow\left(3x-1\right)^3=\left(\frac{-2}{3}\right)^3\)
\(\Leftrightarrow3x-1=\frac{-2}{3}\Leftrightarrow3x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{9}\)
Bài 1
A= \(\frac{81^{10}.3^{17}}{27^{10}.9^{13}}\)
= \(\frac{\left(3^4\right)^{10}.3^{17}}{\left(3^3\right)^{10}.\left(3^2\right)^{13}}\)
= \(\frac{3^{40}.3^{17}}{3^{30}.3^{26}}\)
= \(\frac{3^{57}}{3^{56}}\)= 3
\(\frac{x}{7}\)= \(\frac{-12}{49}\)
=> 49x = (-12) x 7
=> 49x = -84
=> x= \(\frac{-12}{7}\)
( 1/81 )x=(1/27)4
=> [(1/3)4 ]x=[(1/3)3 ]4
=> (1/3)x.4=(1/3)3.4
=> x.4=3.4
=> x=3