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Sửa đề : \(\left(-\frac{4}{3}\right)^n.\left(\frac{16}{9}\right)^2=\left(-\frac{64}{27}\right)^2\)
=> \(\left(-\frac{4}{3}\right)^n.\left[\left(-\frac{4}{3}\right)^2\right]^2=\left[\left(-\frac{4}{3}\right)^3\right]^2\)
=> \(\left(-\frac{4}{3}\right)^n.\left(-\frac{4}{3}\right)^4=\left(-\frac{4}{3}\right)^6\)
=> \(\left(-\frac{4}{3}\right)^n=\left(-\frac{4}{3}\right)^2\)
=> n = 2
\(a.\dfrac{3^{27}}{9^6.3^{16}}=\dfrac{3^{27}}{3^{12}.3^{16}}=\dfrac{3^{27}}{3^{28}}=\dfrac{1}{3}\)
\(\left(x-\dfrac{5}{2}\right)^2=\dfrac{9}{4}\\ \Rightarrow x-\dfrac{5}{2}=\pm\dfrac{3}{2}\)
\(TH1:x-\dfrac{5}{2}=\dfrac{3}{2}\Rightarrow x=\dfrac{3}{2}+\dfrac{5}{2}=\dfrac{8}{2}=4\)
\(TH2:x-\dfrac{5}{2}=-\dfrac{3}{2}\Rightarrow x=-\dfrac{3}{2}+\dfrac{5}{2}=\dfrac{2}{2}=1\)
a: \(=\dfrac{3^{27}}{3^{12}\cdot3^{16}}=\dfrac{1}{3}\)
a)
\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)
Vậy \(x = \frac{{ - 3}}{2}\).
b)
\(\begin{array}{l}\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right):\frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right).\frac{3}{2}\\x = \frac{{ - 5}}{6}\end{array}\)
Vậy \(x = \frac{{ - 5}}{6}\).
c)
\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)
Vậy \(x = \frac{4}{5}\)
d)
\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)
Vậy \(x = \frac{{ - 2}}{5}\).
Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.
\(\Leftrightarrow\left(\dfrac{4}{3}\right)^{150}:x=\left(-\dfrac{4}{3}\right)^{135}\)
\(\Leftrightarrow x=\left(\dfrac{4}{3}\right)^{150}:\left(-\dfrac{4}{3}\right)^{135}=-\left(\dfrac{4}{3}\right)^{15}\)
a) 32.x+2=1342176728
32.x=134217728-2
32.x=134217726
x=134217726:32
x=4194303,938
a, ĐKXĐ:\(x\ge1\)
\(\sqrt{x-1}=3\\ \Rightarrow x-1=9\\ \Rightarrow x=10\)
\(b,x^2-64=0\\ \Rightarrow\left(x-8\right)\left(x+8\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\\ c,x^2+16=25\\ \Rightarrow x^2=9\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ d,ĐKXĐ:x\ge0\\ \left|\sqrt{x}-3\right|+3=9\\ \Rightarrow\left|\sqrt{x}-3\right|=6\\ \Rightarrow\left[{}\begin{matrix}\sqrt{x}-3=-6\\x-3=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\sqrt{x}=-3\left(vô.lí\right)\\x=9\left(tm\right)\end{matrix}\right.\)
(-3/4)63x-1=(3/4)^3
3x-1=3+1
3x=3=1
x=4;3
x=4/3
Vậy x=4/3
\(\left(x+1\right)^2=81\)
\(\Rightarrow\left(x+1\right)^2=9^2\)
\(\Rightarrow x+1=9\)
\(\Rightarrow x=9-1=8\)
Vậy x = 8
b, \(\left(x+5\right)^3=-64\)
\(\Rightarrow\left(x+5\right)^3=\left(-4\right)^3\)
\(\Rightarrow x+5=-4\)
\(\Rightarrow x=\left(-4\right)-5\)
\(\Rightarrow x=-9\)
Vậy x = -9
c, \(\left(2x-3\right)^2=9\)
\(\Rightarrow\left(2x-3\right)^2=3^2\)
\(\Rightarrow2x-3=3\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
Vậy x = 3
d, \(\left(4x+1\right)^3=27\)
\(\Rightarrow\left(4x+1\right)^3=3^3\)
\(\Rightarrow4x+1=3\)
\(\Rightarrow4x=2\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
\(\left(\dfrac{9}{16}\right)^5\cdot x=\left(\dfrac{27}{64}\right)^3\)
\(\Leftrightarrow\left(\dfrac{3}{4}\right)^{10}\cdot x=\left(\dfrac{3}{4}\right)^9\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^9:\left(\dfrac{3}{4}\right)^{10}\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^{-1}\)
\(\Rightarrow x=\dfrac{4}{3}\)
Vậy x=4/3
(\(\dfrac{9}{16}\))5\(\times\) \(x\) = (\(\dfrac{27}{64}\))3
\(x\) = (\(\dfrac{27}{64}\))3 : (\(\dfrac{9}{16}\))5
\(x\) = (\(\dfrac{3^3}{2^6}\))3: (\(\dfrac{3^2}{2^4}\))5
\(x\) = \(\dfrac{3^9}{2^{18}}\) : \(\dfrac{3^{10}}{2^{20}}\)
\(x\) = \(\dfrac{3^9}{2^{18}}\) \(\times\) \(\dfrac{2^{20}}{3^{10}}\)
\(x\) = \(\dfrac{2^2}{3}\)
\(x\) = \(\dfrac{4}{3}\)