11000

\(A=3^0+3^1+3^2+\cdots+3^{100}\)

\(3A=3^1+3^2+3^3+\cdots+3^{101}\)

\(3A-A=\left(3^1+3^2+\cdots+3^{101}\right)-\left(3^0+3^1+\cdots+3^{100}\right)\)

\(2A=3^{101}-3^0=3^{101}-1\)

\(A=\frac{3^{101}-1}{2}<3^{101}\)

Vậy \(3^0+3^1+3^2+\cdots+3^{100}<3^{101}\)

\(4\cdot\left(3x-2\right)+\left(2-3x\right)^2=0\)

\(4\cdot\left(3x-2\right)+\left(3x-2\right)^2=0\)

\(\left(3x-2\right)\left(4+3x-2\right)=0\)

\(\left(3x-2\right)\left(3x+2\right)=0\)

\(\rArr\left[\begin{array}{l}3x-2=0\\ 3x+2=0\end{array}\right.\rArr\left[\begin{array}{l}3x=2\\ 3x=-2\end{array}\rArr\left[\begin{array}{l}x=\frac23\\ x=-\frac23\end{array}\right.\right.\)

\(\frac53+\frac{5}{15}+\frac{5}{35}+\ldots+\frac{5}{399}\)

\(=\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+\frac{5}{5\cdot7}+\cdots+\frac{5}{19\cdot21}\)

\(=\frac52\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}_{}+\cdots+\frac{2}{19\cdot21}\right)\)

\(=\frac52\cdot\left(\frac11-\frac13+\frac13-\frac15+\cdots+\frac{1}{19}-\frac{1}{21}\right)\)

\(=\frac52\cdot\left(1-\frac{1}{21}\right)=\frac52\cdot\frac{20}{21}=\frac{50}{21}\)

Noãn

3

củ hành

\(x^2-4x+4-y^2=\left(x-2\right)^2-y^2=\left(x+y-2\right)\left(x-y-2\right)\)

đề là tìm GTLN hay là GTNN bạn

B