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S
19 tháng 3 2020
a) \(\left(-\frac{3}{4}\right)^{3x-1}=\frac{-27}{64}\)
\(\Leftrightarrow\left(-\frac{3}{4}\right)^{3x-1}=\left(-\frac{3}{4}\right)^3\)
\(\Leftrightarrow3x-1=3\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\frac{4}{3}\)
b) Đề sai ! Sửa :
\(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{625}\)
\(\Leftrightarrow\left(\frac{4}{5}\right)^{2x+5}=\left(\frac{4}{5}\right)^4\)
\(\Leftrightarrow2x+5=4\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\frac{1}{2}\)
c) \(\frac{\left(x+3\right)^5}{\left(x+5\right)^2}=\frac{64}{27}\)
\(\Leftrightarrow\left(x+3\right)^3=\left(\frac{4}{3}\right)^3\)
\(\Leftrightarrow x+3=\frac{4}{3}\)
\(\Leftrightarrow x=-\frac{5}{3}\)
d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)
\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)
\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)
\(\Leftrightarrow x=\frac{4}{15}\)
17 tháng 8 2019
e)
\(\left(x+3\right)^3=\left(x+3\right)^5\)
\(\Rightarrow\)\(x+3=1;0\)
TH1: TH2
\(x+3=0\) \(x+3=1\)
\(x=-3\) \(x=-2\)
\(x\in\left\{-3;-2\right\}\)
31 tháng 8 2020
\(\frac{x}{\left(-\frac{1}{3}\right)^3}=-\frac{1}{3}\Rightarrow x=\left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right)^3=\left(-\frac{1}{3}\right)^4\)
\(\left(\frac{4}{5}\right)^5\cdot x=\left(\frac{4}{5}\right)^7\)
=> \(x=\frac{\left(\frac{4}{5}\right)^7}{\left(\frac{4}{5}\right)^5}=\left(\frac{4}{5}\right)^2=\frac{16}{25}\)
\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}=\left(\pm\frac{1}{4}\right)^2\)
=> \(\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}}\)
(3x + 1)3 = -27 => (3x + 1)3 = (-3)3 => 3x + 1 = -3 => 3x = -4 => x = -4/3
31 tháng 8 2020
a)\(x:\left(\frac{-1}{3}\right)^3=\frac{-1}{3}\)
\(=>x:\frac{-1}{27}=\frac{-1}{3}\)
\(=>x=\frac{-1}{3}.\frac{-1}{27}=>x=\frac{1}{81}\)
b) \(\left(\frac{4}{5}\right)^5.x=\left(\frac{4}{5}\right)^7\)
\(=>x=\left(\frac{4}{5}\right)^7:\left(\frac{4}{5}\right)^5=>x=\left(\frac{4}{5}\right)^2=\frac{16}{25}\)
c)\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(=>\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\\\left(x+\frac{1}{2}\right)^2=\left(\frac{-1}{4}\right)^2\end{cases}}\)
\(=>\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=\frac{-1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{4}\\x=-1\end{cases}}}\)
d|) \(\left(3x+1\right)^3=-27\)
\(=>\left(3x+1\right)^3=\left(-3\right)^3\)
\(=>3x+1=-3\)
\(=>3x=-4=>x=\frac{-4}{3}\)
cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt:>
S
11 tháng 7 2019
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. \(\left(3x-1\right)^3=\frac{-8}{27}\Leftrightarrow\left(3x-1\right)^3=\left(\frac{-2}{3}\right)^3\)
\(\Leftrightarrow3x-1=\frac{-2}{3}\Leftrightarrow3x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{9}\)
27 tháng 7 2018
\(a,\left(x-3\right)^2=1\)
\(\left(x-3\right)^2=1^2=\left(-1\right)^2\)
\(\hept{\begin{cases}x-3=1\\x-3=-1\end{cases}\Rightarrow\hept{\begin{cases}x=4\\x=2\end{cases}}}\)
\(b,\left(2x-1\right)^3=27\)
\(\left(2x-1\right)^3=3^3\)
\(2x-1=3\)
\(2x=4\)
\(x=2\)
\(c,\left(x+\frac{1}{2}\right)^2=\frac{1}{4}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2=\left(-\frac{1}{2}\right)^2\)
\(\hept{\begin{cases}x+\frac{1}{2}=\frac{1}{2}\\x+\frac{1}{2}=-\frac{1}{2}\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}}\)
\(d,\left(3x-1\right)^3=0\)
\(3x-1=0\)
\(3x=1\)
\(x=\frac{1}{3}\)
27 tháng 7 2018
a,(x-3)2=1
=>x2-9=1
=>x2 = 10
=> x=\(\sqrt{10}\)
b,(2x-1)3=27
=>8x3-1=27
=>8x3=28
=>x3=3,5
=>x=\(\sqrt{3,5}\)
3 tháng 7 2019
a) \(\left|2-\frac{3}{2}x\right|-4=x+2\)
=> \(\left|2-\frac{3}{2}x\right|=x+2+4\)
=> \(\left|2-\frac{3}{2}x\right|=x+6\)
ĐKXĐ : \(x+6\ge0\) => \(x\ge-6\)
Ta có: \(\left|2-\frac{3}{2}x\right|=x+6\)
=> \(\orbr{\begin{cases}2-\frac{3}{2}x=x+6\\2-\frac{3}{2}x=-x-6\end{cases}}\)
=> \(\orbr{\begin{cases}2-6=x+\frac{3}{2}x\\2+6=-x+\frac{3}{2}x\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{2}x=-4\\\frac{1}{2}x=8\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{8}{5}\\x=16\end{cases}}\) (tm)
b) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
=> \(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)
=> \(\left(4x-1\right)^{20}.\left[\left(4x-1\right)^{10}-1\right]=0\)
=> \(\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)
=> \(\orbr{\begin{cases}4x=1\\4x-1=\pm1\end{cases}}\)
=> x = 1/4
hoặc x = 0 hoặc x = 1/2
\(\left(3x-1\right)^3=-\frac{8}{27}\)
\(\Rightarrow3x-1=-\frac{2}{3}\)
\(3x=-\frac{2}{3}+1\)
\(3x=\frac{1}{3}\)
\(x=\frac{1}{3}:3\)
\(x=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}\)
Chúc bạn học tốt
thank