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\(=\frac{\left(2^{31}.3^{16}\right).2^{19}.3^{45}+\left(2^{31}.3^{16}\right).2^{59}}{\left(2^{31}.3^{16}\right).2^{20}.3^{45}+\left(2^{31}.3^{16}\right)}\\ =\frac{\left(2^{31}.3^{16}\right).\left(2^{19}.3^{45}+2^{59}\right)}{\left(2^{31}.3^{16}\right).\left(2^{20}.3^{45}+1\right)}\\ =\frac{2^{19}.3^{45}+2^{59}}{2^{20}.3^{45}+1}\)
\(2^{19}.\frac{3^{45}+2^{40}}{2^{20}.3^{45}+1}\)
so sánh
ta có
333444=3334.111=(3334)111
444333=4443.111=(4443)111
So sánh 3334và 4443 ta có
3334=(3.111)4=34 . 1114=81.1114
4443=(4.111)3=43. 1113=64. 1113
Vì 81.1114>64.1113
Suy ra 333444>444333
\(\frac{2^{50}.3^{61}+2^{90}.3^{16}}{2^{51}.3^{61}+2^{31}.3^{61}}\)
\(=\frac{3^{61}\left(2^{50}+2^{90}\right)}{3^{61}\left(2^{51}+2^{31}\right)}\)
\(=\frac{2^{50}+2^{90}}{2^{51}+2^{31}}\)
\(=\frac{2^{31}\left(2^{19}+2^{59}\right)}{2^{31}\left(1+2^{20}\right)}\)
\(=\frac{2^{19}+2^{59}}{1+2^{20}}\)
5x + 1 - 5x = 500
=> 5x - (-1) - 5x = 500
=> 5x - 5x - (-1) = 500
=> 0 - (-1) = 500
=> 1 = 500
=> Sai đề
làm bừa thui,ai tích mình mình tích lại
Số số hạng là :
Có số cặp là :
50 : 2 = 25 ( cặp )
Mỗi cặp có giá trị là :
99 - 97 = 2
Tổng dãy trên là :
25 x 2 = 50
Đáp số : 50
\(\frac{2^{50}.3^{61}+2^{90}.3^{16}}{2^{51}.3^{61}+2^{91}.3^{16}}\)
\(\Leftrightarrow\frac{2^{50}+2^{90}}{2^{50}.2+2^{90}.2}\)
\(\Leftrightarrow\frac{1+1}{2+2}\)
\(\Leftrightarrow\frac{1}{2}\)
P/s: Không chắc nhé! '-'
a) \(\frac{7^3.5^8}{49.25^4}=\frac{7^3.5^8}{7^2.5^8}=7\)
b) \(\frac{3^9.25.5^3}{15.625.3^8}=\frac{3^9.5^2.5^3}{3.5.5^4.3^8}=\frac{3^9.5^5}{3^9.5^5}=1\)
c) \(\frac{2^{50}.3^{61}+2^{90}.3^{16}}{2^{51}.3^{61}+2^{91}.3^{16}}=\frac{2^{50}.3^{16}\left(3^{45}+2^{40}\right)}{2^{51}.3^{16}\left(3^{45}+2^{40}\right)}=\frac{1}{2}\)
d) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2\)
\(=\left(\frac{-1}{10}\right)^2+\left(\frac{11}{10}\right)^2\)
\(=\frac{1}{100}+\frac{121}{100}=\frac{122}{100}=\frac{61}{50}\)
Ta thấy: \(\left(x-5\right)^{88}\ge0\)
\(\left(x+y+3\right)^{496}\ge0\)
\(\Rightarrow\left(x-5\right)+\left(x+y+3\right)^{496}\ge\) ( Đó là điều đương nhiên )
Vậy: \(x;y\in R\)
\(\left(x-5\right)^{88}+\left(x+y+z\right)^{496}\ge0\)0
Dấu "=" xảy ra kih và chỉ khi \(\hept{\begin{cases}\left(x-5\right)^{88}\\\left(x+y+3\right)^{496}\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\5+y+3=0\end{cases}}}\)\(\Leftrightarrow\hept{\begin{cases}x=5\\y=-8\end{cases}}\)
\(\frac{2^{50}.3^{61}+2^{90}.3^{16}}{2^{51}.3^{61}+2^{31}.3^{16}}=\frac{2^{50}.3^{16}+3^{45}+2^{50}+2^{40}.3^{16}}{2^{31}+3^{20}+2^{31}.3^{16}}\)
\(=556758,4881\)
=\(\frac{2^{50}.+2^{90}}{2^{51}+2^{31}}=\frac{2^{19}}{2^{39}}=\frac{1}{1048576}\)
K nha