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\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\)
\(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\)
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)
\(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)
\(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
\(2^x=2\Rightarrow x=1\)
\(3^x=3^4\Rightarrow x=4\)
\(7^x=7^7\Rightarrow x=7\)
\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)
\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)
\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)
\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)
\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)
\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)
\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)
\(\left(-2\right)^{4x+2}=64\)
\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)
\(2x-5x=-4+1\)
\(-3x=-3\Rightarrow x=1\)
\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)
\(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)
\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)
\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).
hehe.
đánh tới què tay, hoa mắt lun r nekkk!!
a)
\((3x-7)^5=0\Rightarrow 3x-7=0\Rightarrow x=\frac{7}{3}\)
b)
\(\frac{1}{4}-(2x-1)^2=0\)
\(\Leftrightarrow (2x-1)^2=\frac{1}{4}=(\frac{1}{2})^2=(-\frac{1}{2})^2\)
\(\Rightarrow \left[\begin{matrix} 2x-1=\frac{1}{2}\\ 2x-1=\frac{-1}{2}\end{matrix}\right.\Rightarrow \Rightarrow \left[\begin{matrix} x=\frac{3}{4}\\ x=\frac{1}{4}\end{matrix}\right.\)
c)
\(\frac{1}{16}-(5-x)^3=\frac{31}{64}\)
\(\Leftrightarrow (5-x)^3=\frac{1}{16}-\frac{31}{64}=\frac{-27}{64}=(\frac{-3}{4})^3\)
\(\Leftrightarrow 5-x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{23}{4}\)
d)
\(2x=(3,8)^3:(-3,8)^2=(3,8)^3:(3,8)^2=3,8\)
\(\Rightarrow x=3,8:2=1,9\)
e)
\((\frac{27}{64})^9.x=(\frac{-3}{4})^{32}\)
\(\Leftrightarrow [(\frac{3}{4})^3]^9.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow (\frac{3}{4})^{27}.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow x=(\frac{3}{4})^{32}:(\frac{3}{4})^{27}=(\frac{3}{4})^5\)
f)
\(5^{(x+5)(x^2-4)}=1\)
\(\Leftrightarrow (x+5)(x^2-4)=0\)
\(\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2=4=2^2=(-2)^2\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=-5\\ x=\pm 2\end{matrix}\right.\)
g)
\((x-2,5)^2=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2\)
\(\Rightarrow \left[\begin{matrix} x-2,5=\frac{2}{3}\\ x-2,5=\frac{-2}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{19}{6}\\ x=\frac{11}{6}\end{matrix}\right.\)
h)
\((2x+\frac{1}{3})^3=\frac{8}{27}=(\frac{2}{3})^3\)
\(\Rightarrow 2x+\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)
3/ ta để ý thấy ở số mũ sẽ có thừa số 1000-103=0
nên số mũ chắc chắn bằng 0
mà số nào mũ 0 cũng bằng 1 nên A=1
5/ vì |2/3x-1/6|> hoặc = 0
nên A nhỏ nhất khi |2/3x-6|=0
=>A=-1/3
6/ =>14x=10y=>x=10/14y
23x:2y=23x-y=256=28
=>3x-y=8
=>3.10/4y-y=8
=>6,5y=8
=>y=16/13
=>x=10/14y=10/14.16/13=80/91
8/106-57=56.26-56.5=56(26-5)=59.56
có chứa thừa số 59 nên chia hết 59
4/ tính x
sau đó thế vào tinh y,z
Theo bài ra , ta có :
\(\left(2.5\right)^2=10^2\)
\(2^2.5^2=\left(2.5\right)^2=10^2\)
Vì \(10^2=10^2=100\)
Vậy \(\left(2.5\right)^2=2^2.5^2\)
b)
\(\left(\frac{1}{2}.\frac{3}{4}\right)^3=\left(\frac{1}{2}\right)^3.\left(\frac{3}{4}\right)^3\)
mà \(\left(\frac{1}{2}\right)^3.\left(\frac{3}{4}\right)^3\) là vế phải
Vậy \(\left(\frac{1}{2}.\frac{3}{4}\right)^3=\left(\frac{1}{2}\right)^3.\left(\frac{3}{4}\right)^3\)
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!
a) \(9\cdot3^3\cdot\frac{1}{81}\cdot3^2=3^2\cdot3^3\cdot\left(\frac{1}{3}\right)^43^2=3^7\cdot\frac{1}{3^4}=3^3\)
b) \(4\cdot2^5:\left(2^3\cdot\frac{1}{16}\right)=2^2\cdot2^5:\left(2^3\cdot\frac{1}{2^4}\right)=2^7:\frac{1}{2}=2^8\)
c) \(3^2\cdot2^5\cdot\left(\frac{2}{3}\right)^2=3^2\cdot2^5\cdot\frac{2^2}{3^2}=2^7\)
d) \(\left(\frac{1}{3}\right)^2\cdot\frac{1}{3}\cdot9^2=\frac{1}{3^2}\cdot\frac{1}{3}\cdot3^4=\frac{1}{3^3}\cdot3^4=3\)
a)9.33.\(\frac{1}{81}\).32
=32.33.34.\(\frac{1}{3^4}\).32
=311.\(\frac{1}{3^4}\)
=37
b)4.25:(\(2^3.\frac{1}{16}\))
=22.25:(\(2^3.\frac{1}{2^4}\))
=27:\(\frac{2^3}{2^4}\)
=27.\(\frac{2^4}{2^3}\)
=\(\frac{2^{11}}{2^3}\)
=28
c)32.25.\(\left(\frac{2}{3}\right)^2\)
=32.25.\(\frac{2^2}{3^2}\)
=\(\frac{3^2.2^5.2^2}{3^2}\)
=27
d)\(\left(\frac{1}{3}\right)^2.\frac{1}{3}.9^2\)
=\(\frac{1^2}{3^2}.\frac{1}{3}.\left(3^2\right)^2\)
=\(\frac{1^2}{3^2}.\frac{1}{3}.3^4\)
=\(\frac{1^2}{3^2}.\frac{3^4}{3}\)
=\(\frac{1^2}{3^2}.3^3\)
=3
Bài 1:
b) \(\left(3x-\frac{1}{5}\right)^3=\frac{64}{125}\)
=> \(\left(3x-\frac{1}{5}\right)^3=\left(\frac{4}{5}\right)^3\)
=> \(3x-\frac{1}{5}=\frac{4}{5}\)
=> \(3x=\frac{4}{5}+\frac{1}{5}\)
=> \(3x=1\)
=> \(x=1:3\)
=> \(x=\frac{1}{3}\)
Vậy \(x=\frac{1}{3}.\)
c) \(\left(3x-1\right)^3=-\frac{8}{27}\)
=> \(\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)
=> \(3x-1=-\frac{2}{3}\)
=> \(3x=\left(-\frac{2}{3}\right)+1\)
=> \(3x=\frac{1}{3}\)
=> \(x=\frac{1}{3}:3\)
=> \(x=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}.\)
Chúc bạn học tốt!
Bài 1: