b: (1/2)^x=1/32
=>(1/2)^x=(1/2)^5
=>x=5
b: (7/5)^x=343/125
=>(7/5)^x=(7/5)^3
=>x=3
b: (-3/2)^x=9/4
=>(-3/2)^x=(-3/2)^2
=>x=2
b: (-2/3)^x=-8/27
=>(-2/3)^x=(-2/3)^3
=>x=3
b: (3/5)^x+98/125=1
=>(3/5)^x=27/125
=>(3/5)^x=(3/5)^3
=>x=3
b1) \(\left(\dfrac{1}{2}\right)^x=\dfrac{1}{32}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^x=\left(\dfrac{1}{2}\right)^5\)
\(\Rightarrow x=5\)
b2) \(\dfrac{343}{125}=\left(\dfrac{7}{5}\right)^x\)
\(\Rightarrow\left(\dfrac{7}{5}\right)^3=\left(\dfrac{7}{5}\right)^x\)
\(\Rightarrow x=3\)
b3) \(\left(-\dfrac{3}{2}\right)^x=\dfrac{9}{4}\)
\(\Rightarrow\left(-\dfrac{3}{2}\right)^x=\left(-\dfrac{3}{2}\right)^2\)
\(\Rightarrow x=2\)
b4) \(\left(-\dfrac{2}{3}\right)^x=-\dfrac{8}{27}\)
\(\Rightarrow\left(-\dfrac{2}{3}\right)^x=\left(-\dfrac{2}{3}\right)^3\)
\(\Rightarrow x=3\)
b5) \(\left(\dfrac{3}{5}\right)^x+\dfrac{98}{125}=1\)
\(\Rightarrow\left(\dfrac{3}{5}\right)^x=1-\dfrac{98}{125}\)
\(\Rightarrow\left(\dfrac{3}{5}\right)^x=\dfrac{27}{125}\)
\(\Rightarrow\left(\dfrac{3}{5}\right)^x=\left(\dfrac{3}{5}\right)^3\)
\(\Rightarrow x=3\)
a) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(x+\dfrac{1}{2}=\dfrac{1}{4}\) hoặc \(x+\dfrac{1}{2}=-\dfrac{1}{4}\)
*) \(x+\dfrac{1}{2}=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{1}{2}\)
\(x=-\dfrac{1}{2}\)
*) \(x+\dfrac{1}{2}=-\dfrac{1}{4}\)
\(x=-\dfrac{1}{4}-\dfrac{1}{2}\)
\(x=-\dfrac{3}{4}\)
Vậy \(x=-\dfrac{3}{4};x=-\dfrac{1}{2}\)
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a) \(\left(\dfrac{3}{4}-7x\right)^2=\dfrac{1}{4}\)
\(\dfrac{3}{4}-7x=\dfrac{1}{2}\) hoặc \(\dfrac{3}{4}-7x=-\dfrac{1}{2}\)
*) \(\dfrac{3}{4}-7x=\dfrac{1}{2}\)
\(7x=\dfrac{3}{4}-\dfrac{1}{2}\)
\(7x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}:7\)
\(x=\dfrac{1}{28}\)
*) \(\dfrac{3}{4}-7x=-\dfrac{1}{2}\)
\(7x=\dfrac{3}{4}+\dfrac{1}{2}\)
\(7x=\dfrac{5}{4}\)
\(x=\dfrac{5}{4}:7\)
\(x=\dfrac{5}{28}\)
Vậy \(x=\dfrac{1}{28};x=\dfrac{5}{28}\)
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a) \(\left(x-\dfrac{5}{6}\right)^2=\dfrac{1}{36}\)
\(x-\dfrac{5}{6}=\dfrac{1}{6}\) hoặc \(x-\dfrac{5}{6}=-\dfrac{1}{6}\)
*) \(x-\dfrac{5}{6}=\dfrac{1}{6}\)
\(x=\dfrac{1}{6}+\dfrac{5}{6}\)
\(x=1\)
*) \(x-\dfrac{5}{6}=-\dfrac{1}{6}\)
\(x=-\dfrac{1}{6}+\dfrac{5}{6}\)
\(x=\dfrac{2}{3}\)
Vậy \(x=\dfrac{2}{3};x=1\)
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a) \(\left(1-\dfrac{2}{3}x\right)^2=\dfrac{4}{9}\)
\(1-\dfrac{2}{3}x=\dfrac{2}{3}\) hoặc \(1-\dfrac{2}{3}x=-\dfrac{2}{3}\)
*) \(1-\dfrac{2}{3}x=\dfrac{2}{3}\)
\(\dfrac{2}{3}x=1-\dfrac{2}{3}\)
\(\dfrac{2}{3}x=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}:\dfrac{2}{3}\)
\(x=\dfrac{1}{2}\)
*) \(1-\dfrac{2}{3}x=-\dfrac{2}{3}\)
\(\dfrac{2}{3}x=1+\dfrac{2}{3}\)
\(\dfrac{2}{3}x=\dfrac{5}{3}\)
\(x=\dfrac{5}{3}:\dfrac{2}{3}\)
\(x=\dfrac{5}{2}\)
Vậy \(x=\dfrac{1}{2};x=\dfrac{5}{2}\)
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*) \(\left(\dfrac{2}{3}-x\right)^2=1:\dfrac{4}{9}\)
\(\left(\dfrac{2}{3}-x\right)^2=\dfrac{9}{4}\)
\(\dfrac{2}{3}-x=\dfrac{3}{2}\) hoặc \(\dfrac{2}{3}-x=-\dfrac{3}{2}\)
*) \(\dfrac{2}{3}-x=\dfrac{3}{2}\)
\(x=\dfrac{2}{3}-\dfrac{3}{2}\)
\(x=-\dfrac{5}{6}\)
*) \(\dfrac{2}{3}-x=-\dfrac{3}{2}\)
\(x=\dfrac{2}{3}+\dfrac{3}{2}\)
\(x=\dfrac{13}{6}\)
Vậy \(x=-\dfrac{5}{6};x=\dfrac{13}{6}\)
c) \(5^{x+1}-5^x=20\)
\(5^x.\left(5-1\right)=20\)
\(5^x.4=20\)
\(5^x=20:4\)
\(5^x=5\)
\(x=1\)
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c) \(2^x+2^{x+4}=544\)
\(2^x.\left(1+2^4\right)=544\)
\(2^x.17=544\)
\(2^x=544:17\)
\(2^x=32\)
\(2^x=2^5\)
\(x=5\)
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c) \(4^{2x+1}+4^{2x}=80\)
\(4^{2x}.\left(4+1\right)=80\)
\(4^{2x}.5=80\)
\(4^{2x}=80:5\)
\(4^{2x}=16\)
\(4^{2x}=4^2\)
\(2x=2\)
\(x=1\)