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\(\frac{1}{15}\)chứ ko phải \(\frac{1}{5}\)nha các bạn
\(\frac{1}{5}+\frac{9}{10}+\frac{14}{15}-\frac{1}{9}-\frac{20}{10}+\frac{1}{157}\)
\(=\frac{3}{15}+\frac{9}{10}+\frac{14}{15}-\frac{1}{9}-\frac{20}{10}+\frac{1}{157}\)
\(=\left(\frac{3}{15}+\frac{14}{15}\right)+\left(\frac{9}{10}-\frac{20}{10}\right)-\frac{1}{9}+\frac{1}{157}\)
\(=\frac{17}{15}+\frac{-11}{10}-\frac{1}{9}+\frac{1}{157}\)
\(=\left(\frac{51}{45}-\frac{5}{45}\right)+\frac{-11}{10}+\frac{1}{157}\)
\(=\frac{46}{45}+\frac{-11}{10}+\frac{1}{157}\)
\(=\left(\frac{92}{90}+\frac{-99}{90}\right)+\frac{1}{157}\)
\(=\frac{-1099}{14130}+\frac{90}{14130}\)
\(=\frac{-1099}{14130}\)
Ta luôn có nếu a>0; b>0 thì \(\frac{a}{b}< \frac{a+m}{b+m}\left(m\in N\right)\)
Áp dụng vào bài toán ta thấy 1011-1 > 0 và 1012-1 > 0 nên
\(A=\frac{10^{11}-1}{10^{12}-1}< \frac{10^{11}-1+11}{10^{12}-1+11}=\frac{10^{11}+10}{10^{12}+10}=\frac{10.\left(10^{10}+1\right)}{10.\left(10^{11}+1\right)}=\frac{10^{10}+1}{10^{11}+1}=B\)
Vậy A < B
Xin lỗi bn nhé bài toán phụ phía trên đang còn 1 đk nữa là a<b
a) 2515 và 810. 330
2515 = (52 ) 15 = 530
810. 330 = (23 )10. 330 = 230. 330 = 630
Vì 530< 630
nên 2515< 810. 330
b) \(\frac{4^{15}}{7^{30}}\)và \(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}\)
\(\frac{4^{15}}{7^{30}}=\frac{\left(2^2\right)^{15}}{7^{30}}=\frac{2^{30}}{7^{30}}\)
\(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}=\frac{\left(2^3\right)^{10}.3^{30}}{7^{30}.\left(2^2\right)^{15}}=\frac{2^{30}.3^{30}}{7^{30}.2^{30}}=\frac{3^{30}}{7^{30}}\)
Vì \(\frac{2^{30}}{7^{30}}< \frac{3^{30}}{7^{30}}\)
nên \(\frac{4^{15}}{7^{30}}< \frac{8^{10}.3^{30}}{7^{30}.4^{15}}\)
a)\(25^{15}=5^{2^{15}}=5^{30}\)
\(8^{10}.3^{30}=2^{3^{10}}.3^{30}=\left(2.3\right)^{30}=6^{30}\)
\(5^{30}< 6^{30}=>25^{15}< 8^{10}.3^{30}\)
b)\(\frac{4^{15}}{7^{30}}=\frac{2^{2^{15}}}{7^{30}}=\frac{2^{30}}{7^{30}}=\left(\frac{2}{7}\right)^{30}\)
\(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}=\frac{2^{30}.3^{30}}{7^{30}.2^{30}}=\frac{6^{30}}{14^{30}}=\left(\frac{6}{14}\right)^{30}=\left(\frac{3}{7}\right)^{30}\)
Vì hai số có mũ bằng 30 nên ta so sánh :\(\frac{2}{7}< \frac{3}{7}\)
=>\(\frac{4^{15}}{7^{30}}< \frac{8^{10}.3^{30}}{7^{30}.4^{15}}\).
a) Ta có: \(25^{15}=\left(5^2\right)^{15}=5^{30}\)
\(8^{10}.3^{30}=\left(2^3\right)^{10}.3^{30}\)\(=2^{30}.3^{30}=6^{30}\)
Vì \(5^{30}< 6^{30}\)nên \(25^{15}< 8^{10}.3^{30}\)
b) Ta có: \(\frac{4^{15}}{7^{30}}=\frac{\left(2^2\right)^{15}}{7^{30}}=\frac{2^{30}}{7^{30}}\)
\(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}=\frac{\left(2^3\right)^{10}.3^{30}}{7^{30}.\left(2^2\right)^{15}}=\frac{2^{30}.3^{30}}{7^{30}.2^{30}}=\frac{3^{30}}{7^{30}}\)
Vì \(2^{30}< 3^{30}\)nên \(\frac{2^{30}}{7^{30}}< \frac{3^{30}}{7^{30}}\)hay \(\frac{4^{15}}{7^{30}}< \frac{8^{10}.3^{30}}{7^{30}.4^{15}}\)
_Học tốt_
Có : 10A = 10^15-10/10^15-11 = (10^15-11)+1/10^15-11 = 1 + 1/10^15-11
10B = 10^15+10/10^15+9 = (10^15+9)+1/10^15+9 = 1 + 1/10^15+9
Vì 10^15-11 < 10^15-9 => 1/10^15-11 > 1/10^15+9 => 10A > 10B
=> A < B
k mk nha