So sánh 2^300 và 3^200

Bài 1:

\(2^{300}=\left(2^3\right)^{100}=8^{100}< 9^{100}=\left(3^2\right)^{100}=3^{200}\)

Vậy \(2^{300}< 3^{200}\)

Bài 2:

a.

\(2^{3x+1}=32^x\)

\(2^{3x+1}=\left(2^5\right)^x\)

\(2^{3x+1}=2^{5x}\)

\(3x+1=5x\)

\(3x-5x=-1\)

\(-2x=-1\)

\(x=\frac{-1}{-2}\)

\(x=\frac{1}{2}\)

b.

\(3^{2+x}=27^{3x-1}\)

\(3^{x+2}=\left(3^3\right)^{3x-1}\)

\(3^{x+2}=3^{3\times\left(3x-1\right)}\)

\(x+2=3\times\left(3x-1\right)\)

\(x+2=9x-3\)

\(9x-x=3+2\)

\(8x=5\)

\(x=\frac{5}{8}\)

Bài 4:

\(2^2+4^2+6^2+...+18^2+20^2\)

\(=\left(1\times2\right)^2+\left(2\times2\right)^2+\left(3\times2\right)^2+...+\left(9\times2\right)^2+\left(10\times2\right)^2\)

\(=1^2\times2^2+2^2\times2^2+3^2\times2^2+...+9^2\times2^2+10^2\times2^2\)

\(=2^2\times\left(1^2+2^2+3^3+...+9^2+10^2\right)\)

\(=4\times385\)

\(=1540\)

Bài 5:

a.

\(\left(\frac{4}{5}\right)^{2x+7}=\frac{256}{625}\)

\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)

\(2x+7=4\)

\(2x=4-7\)

\(2x=-3\)

\(x=-\frac{3}{2}\)

b.

\(\left(4x-3\right)^4=\left(4x-3\right)^2\)

TH1:

\(4x-3=0\)

\(4x=3\)

\(x=\frac{3}{4}\)

TH2:

\(4x-3=1\)

\(4x=1+3\)

\(4x=4\)

\(x=\frac{4}{4}\)

\(x=1\)

TH3:

\(4x-3=-1\)

\(4x=-1+3\)

\(4x=2\)

\(x=\frac{2}{4}\)

\(x=\frac{1}{2}\)

Vậy \(x\in\left\{\frac{1}{2};\frac{3}{4};1\right\}\)