6.3^x-1 + 5 = 167
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\(6.3^2+8^7:8^5-1^{2020}\)
\(=6.9+8^2-1\)
\(=54+64-1\)
\(=118-1\)
\(=117\)
\(6\cdot3^2+8^7:8^5-1^{2020}\)
\(=6\cdot9+8^{7-5}-1\)
\(=54+8^2-1\)
\(=53+64\)
\(=117\)
(52 - 24 ).x= 81- 54
(25-16).x= 27
9.x=27
x=27:9
x=3
Vậy x=3
Nhớ nha.....
\(=\dfrac{3^5\cdot2^5\cdot3^6}{2^6\cdot3^{10}}-\dfrac{1}{2}=\dfrac{3}{2}-\dfrac{1}{2}=1\)
\(\left(2x+1\right)^3=125\\ \Rightarrow\left(2x+1\right)^3=5^3\\ \Rightarrow2x+1=5\\ \Rightarrow2x=4\\ \Rightarrow x=2.\\ b,\left(2x-1\right)^4=16\\ \Rightarrow\left(2x-1\right)^4=2^4\\ \Rightarrow2x-1=2\\ \Rightarrow2x=3\\ \Rightarrow x=\dfrac{3}{2}.\\ c,6.3^x-2.3^x=36\\ \Rightarrow3^x.\left(6-2\right)=36\\ \Rightarrow3^x.4=36\\ \Rightarrow3^x=9\\ \Rightarrow3^x=3^2\\ \Rightarrow x=2.\\ d,2^{x+1}-2^x=32\\ \Rightarrow2^x.\left(2-1\right)=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5.\)
\(a.\dfrac{3^{27}}{9^6.3^{16}}=\dfrac{3^{27}}{3^{12}.3^{16}}=\dfrac{3^{27}}{3^{28}}=\dfrac{1}{3}\)
\(\left(x-\dfrac{5}{2}\right)^2=\dfrac{9}{4}\\ \Rightarrow x-\dfrac{5}{2}=\pm\dfrac{3}{2}\)
\(TH1:x-\dfrac{5}{2}=\dfrac{3}{2}\Rightarrow x=\dfrac{3}{2}+\dfrac{5}{2}=\dfrac{8}{2}=4\)
\(TH2:x-\dfrac{5}{2}=-\dfrac{3}{2}\Rightarrow x=-\dfrac{3}{2}+\dfrac{5}{2}=\dfrac{2}{2}=1\)
a: \(=\dfrac{3^{27}}{3^{12}\cdot3^{16}}=\dfrac{1}{3}\)
\(6.3^{x-1}+5=167\)
\(6.3^{x-1}=167-5\)
\(6.3^{x-1}=162\)
\(3^{x-1}=162:6\)
\(3^{x-1}=27\)
\(3^{x-1}=3^3\)
\(x-1=3\)
\(x=3+1\)
\(x=4\)
\(6.3^{x-1}+5=167\)
\(\Leftrightarrow6.3^{x-1}=162\)
\(\Leftrightarrow3^{x-1}=27=3^3\)
\(\Rightarrow x-1=3\)
\(\Leftrightarrow x=2\)