1) \(+2x+3y⋮17\)

\(\Rightarrow26x+39y⋮17\)

\(\Rightarrow\left(9x+5y\right)+17x+34y⋮17\)

Mà \(17x+34y⋮17\)

\(\Rightarrow9x+5y⋮17\)

\(+9x+5y⋮17\)

\(\Rightarrow36x+20y⋮17\)

\(\Rightarrow\left(2x+3y\right)+34x+17y⋮17\)

Mà \(34x+17y⋮17\)

\(\Rightarrow2x+3y⋮17\)

Túi ko bt

Ta có :

\(3A=1+\frac{1}{3}+.....+\frac{1}{3^{98}}\)

\(\Rightarrow3A-A=\left(1+\frac{1}{3}+....+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}\right)\)

\(\Rightarrow2A=1-\frac{1}{99}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{198}< \frac{1}{2}\)

\(\Rightarrow A< \frac{1}{2}\)

=> 3A = 1 + 1/3 + 1/3² +...  +1/3⁹⁸

=> 2A = 1 - 1/3⁹⁹

=> A = \(\frac{1-\frac{1}{3^{99}}}{2}\)

Mà \(1-\frac{1}{3^{99}}\) < 1 nên A < \(\frac{1}{2}\)

bài 3 : \(\left\{{}\begin{matrix}ab=2\\bc=3\\ca=54\end{matrix}\right.\)

hiển nhiên a;b;c =0 không phải nghiệm

\(\Leftrightarrow\left(abc\right)^2=2.3.54=18^2\)

\(\Leftrightarrow\left[{}\begin{matrix}abc=-18\\abc=18\end{matrix}\right.\)

abc=-18 => c=-9; a=-6; b=-1/3

abc=18 => c=9; a=6; b=1/3

A=1/3+1/3²+1/3³+...+1/3⁹⁹

3A=1+1/3+1/3²+1/3³+...+1/3⁹⁸

3A-A=1+1/3+1/3²+....+1/3⁹⁹-1/3-1/3²-...-1\3⁹⁸

2A=1-1/3⁹⁸<1

A<1/2(DPCM)

3A=1+1/3+1/3^2+...+1/3^98

3A-A=(1+1/3+1/3^2+...+1/3^98)-(1/3+1/3^2+...+1/3^99)

2A=1-1/3^99<1

Vậy A<1/2 =>ĐPCM

Ta có :

M = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

3M = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

3M - M = ( \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)) - ( \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\))

2M = \(1-\frac{1}{3^{99}}< 1\)

\(\Rightarrow M=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)

3M=1+1/3+1/3^2+....+1/3^98

2M=3M-M=(1+1/3+1/3^2+....+1/3^98)-(1/3+1/3^2+....+1/3^99) = 1-1/3^99 < 1

=> M < 1/2

=> ĐPCM

k mk nha

a, \(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

\(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

\(3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)

\(2C=1-\frac{1}{3^{99}}\)

\(C=\frac{1}{2}-\frac{1}{2.3^{99}}< \frac{1}{2}\)(đpcm)

b, Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)

\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)

\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)

\(2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(6A=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(6A-2A=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)

\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)

\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)

\(4A=3-\frac{397}{3^{100}}\)

\(A=\frac{3}{4}-\frac{397}{4.3^{100}}< \frac{3}{4}\)(đpcm)

nhân C vs 3 sau đó lấy 3C-C sẽ ra đc 2 C = 1 - 1/3⁹⁹ => C= 1/2 - 1/ (2x3⁹⁹ )

1/3<1/2 mà C cộng với 1/9 nữa nên C<1/2