Bài 1 tìm x
a, 4 mũ x-5 =64
b, |3.x+7|+5=12
Bài 2: tìm x,y,z biết
a, 6.x=y.5 và y-x=32
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\(3^{m+2}-2^{m+2}+3^m-2^m\left(m\in N\cdot\right)\)
\(=3^m.3^2-2^n.2^2+3^m-2^n=3^m.9-2^n.4+3^m-2^n\)
\(=3^m.9-2^n.4+3^m-2^n=3^m\left(9+1\right)-2^n\left(4+1\right)=3^m.10-2^n.5\)
\(=3^m.10-2^{n-1}.10=10\left(3^m-2^{n-1}\right)⋮10\left(m\inℕ^∗\right)\)
\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\)
\(\Leftrightarrow\frac{a}{b}-\frac{c}{b}+\frac{b}{c}-\frac{a}{c}+\frac{c}{a}-\frac{b}{a}=0\)
\(\Leftrightarrow\frac{a-c}{b}+\frac{b-a}{c}+\frac{c-b}{a}=0\)
Dấu ''='' xảy ra khi : \(a=b=c\)
\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\)
A=(3+3^2+3^3+3^4)+....+(3^2013+3^2014+3^2015+3^2016)
A=3(3+3^2+3^3)+....+3^2013(3+3^2+3^3)
A=(3+3^2+3^3)(3+3^5+3^3^9+....+3^2013)
\(3A=3^2+3^3+3^4+...+3^{2016}+3^{2017}\)
\(3A-A=\left(3^2+3^3+3^4+...+3^{2017}\right)-\left(3+3^2+3^3+...+3^{2016}\right)\)
\(2A=3^{2017}-3\)
\(A=\frac{3^{2017}-3}{2}\)
Bài 1
a) \(4^{x-5}=64\)
\(4^{x-5}=4^3\)
\(=>x-5=3\)
\(x=5+3\)
\(x=8\)
Vậy x=8