vào câu hỏi tương tự yk 

quách anh thư mk vào r mà ko có

Ta có : \(\left(100^{99}+99^{99}\right)^{100}>\left(100^{99}+99^{99}\right)^{99}\times100^{99}=\left(100^{100}+100\times99^{99}\right)^{99}>\left(100^{100}+99^{100}\right)^{99}\)

\(\Rightarrow A>B\)

Vậy \(A>B\)

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 \(34x+17y⋮17\)

\(\Rightarrow2x+3y⋮17\)

Bài 1: \(\left(\frac{-1}{16}\right)^{100}=\frac{1}{\left(2^4\right)^{100}}=\frac{1}{2^{400}}>\frac{1}{2^{500}}=\left(\frac{-1}{2}\right)^{500}.\)

Bài 2: \(100^{99}+1>100^{68}+1\Rightarrow\frac{1}{100^{99}+1}< \frac{1}{100^{68}+1}\Rightarrow\frac{-99}{100^{99}+1}>\frac{-99}{100^{68}+1}\)

\(\Rightarrow100+\frac{-99}{100^{99}+1}>100+\frac{-99}{100^{68}+1}\Rightarrow\frac{100^{100}+1}{100^{99}+1}>\frac{100^{69}+1}{100^{68}+1}\)

                                              Bài giải

Ta có : \(9^{99}=\left(9^{11}\right)^9\)

Vì \(\left(9^{11}\right)^9>99^9\text{ }\left[\left(81\cdot9^9\right)^9>99^9\right]\text{ }\Rightarrow\text{ }9^{99}>99^9\)

                                              Bài giải

Ta có : \(9^{99}=\left(9^{11}\right)^9\)

Vì \(\left(9^{11}\right)^9>99^9\text{ }\left[\left(81\cdot9^9\right)^9>99^9\right]\text{ }\Rightarrow\text{ }9^{99}>99^9\)

đề ý 63.1,2-21.3,6+1=0=>B=0
 

Để ý ta thấy 63.1,2-21.3,6+1=0

=>A=0