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Áp dụng công thức:
Nếu a<b=>a/b<(a+k)/(b+k) (k thuộc N*)
Ta có:\(13^{16}+1x=\frac{13^{16}+1}{13^{17}+1}
Bn nhân cả x và y cho 13 nha
Ta có 10x=1+ 12 / 13^17+1 và 10 y= 1+12 / 13x^16+1
Do 12 / 13^17+1 < 12 / 13^16+1
=>10x<10y
=>x<y
a. \(\frac{7}{15}< \frac{7}{14}=\frac{1}{2};\frac{15}{23}>\frac{15}{30}=\frac{1}{2}\text{ hay }\frac{7}{15}< \frac{1}{2}< \frac{15}{23}\)
Vậy \(\frac{7}{15}< \frac{15}{23}\).
b. \(x=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13x=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)
\(y=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13y=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)
Vì \(13^{17}+1>13^{16}+1\) nên \(\frac{12}{13^{17}+1}< \frac{12}{13^{16}+1}\)
Mà 1 = 1 => \(1+\frac{12}{13^{17}+1}< 1+\frac{12}{13^{16}+1}\text{ hay }13x< 13y\)
=> x < y.
Ta có 13x = \(\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)
13y = \(\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)
Vì 1317 + 1 > 1316 + 1
=> \(\frac{1}{13^{17}+1}< \frac{1}{13^{16}+1}\)
=> \(\frac{12}{13^{17}+1}< \frac{12}{13^{16}+1}\)
=> \(1+\frac{12}{13^{17}+1}< 1+\frac{12}{13^{16}+1}\)
=> 13x < 13y
=> x < y
Vậy x < y
b bạn gõ lại đề giúp mình nhé
a) \(\frac{7}{15}+\frac{9}{10}+\frac{8}{15}-\frac{-1}{10}-\frac{20}{10}+\frac{1}{157}\)
\(=\frac{7}{15}+\frac{9}{10}+\frac{8}{15}+\frac{1}{10}-\frac{20}{10}+\frac{1}{157}\)
\(=\left(\frac{7}{15}+\frac{8}{15}\right)+\left(\frac{9}{10}+\frac{1}{10}\right)-2+\frac{1}{157}\)
\(=1+1-2+\frac{1}{157}\)
\(=2-2+\frac{1}{157}\)
\(=0+\frac{1}{157}=\frac{1}{157}\)
b) \(\frac{1}{13}+\frac{16}{7}+\frac{3}{105}-\frac{9}{7}-\frac{-12}{13}\)
\(=\frac{1}{13}+\frac{16}{7}+\frac{1}{35}-\frac{9}{7}+\frac{12}{13}\)
\(=\left(\frac{1}{13}+\frac{12}{13}\right)+\left(\frac{16}{7}-\frac{9}{7}\right)+\frac{1}{35}\)
\(=1+1+\frac{1}{35}\)
\(=2+\frac{1}{35}\)
\(=\frac{70}{35}+\frac{1}{35}=\frac{71}{35}\)
a. Có: \(\frac{100^{101}+1}{100^{100}+1}>1\Rightarrow\frac{100^{101}+1}{100^{100}+1}>\frac{100^{101}+\left(1+99\right)}{100^{100}+\left(1+99\right)}\)
\(\Rightarrow B>\frac{100^{101}+100}{100^{100}+100}\\ \Rightarrow B>\frac{100\left(100^{100}+1\right)}{100\left(100^{99}+1\right)}\\ \Rightarrow B>\frac{100^{100}+1}{100^{99}+1}=A\\ \Leftrightarrow A< B\)
Vậy A < B
b. Có: \(\frac{13^{16}+1}{13^{17}+1}< 0\Rightarrow\frac{13^{16}+1}{13^{17}+1}< \frac{13^{16}+\left(1+12\right)}{13^{17}+\left(1+12\right)}\)
\(\Rightarrow B< \frac{13^{16}+13}{13^{17}+13}\\ \Rightarrow B< \frac{13\left(13^{15}+1\right)}{13\left(13^{16}+1\right)}\\ \Rightarrow B< \frac{13^{15}+1}{13^{16}+1}=A\\ \Leftrightarrow A>B\)
Vậy A > B
c. Có: \(\frac{1999^{2000}+1}{1999^{1999}+1}>1\Rightarrow\frac{1999^{2000}+1}{1999^{1999}+1}>\frac{1999^{2000}+\left(1+1998\right)}{1999^{1999}+\left(1+1998\right)}\)
\(\Rightarrow B>\frac{1999^{2000}+1999}{1999^{1999}+1999}\\ \Rightarrow B>\frac{1999\left(1999^{1999}+1\right)}{1999\left(1999^{1998}+1\right)}\\ \Rightarrow B>\frac{1999^{1999}+1}{1999^{1998}+1}=A\\ \Leftrightarrow A< B\)
Vậy A < B
Thực hiện phép tính ( bằng cách hợp lí nếu có thể):
a, \(5\frac{4}{13}.15\frac{3}{41}-5\frac{4}{13}.2\frac{3}{41}\)
\(=5\frac{4}{13}\left(15\frac{3}{41}-2\frac{3}{41}\right)\)
\(=15\frac{4}{13}\left(\frac{618}{41}-\frac{85}{41}\right)\)
\(=\frac{69}{13}.13\)
\(=69\)
b, \(6.\left(-\frac{1}{3}\right)^2-\left(\frac{1}{4}:2-\frac{7}{16}.\frac{-4}{21}\right)\)
\(=6.\frac{1}{9}-\left(\frac{1}{8}-\frac{-1}{12}\right)\)
\(=\frac{2}{3}-\left(\frac{3}{24}-\frac{-2}{24}\right)\)
\(=\frac{2}{3}-\frac{5}{24}\)
\(=\frac{16}{24}-\frac{5}{24}\)
\(=\frac{11}{24}\)
Chúc bạn hok tốt!!! lưu khánh huyền
Ta có:
\(A=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13A=\frac{13^{16}+13}{13^{16}+1}=\frac{13^{16}+1+12}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)
\(B=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13B=\frac{13^{17}+13}{13^{17}+1}=\frac{13^{17}+1+12}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)
Ta thấy:
\(13^{16}+1< 13^{17}+1\)
\(\Rightarrow\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)
\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)
hay \(A>B\)
Vậy \(A>B.\)
Ta có: \(\frac{a}{b}< \frac{a+c}{b+c}\)
=> \(B=\frac{13^{16}+1}{13^{17}+1}< \frac{13^{16}+1+12}{13^{17}+1+12}=\frac{13^{16}+13}{13^{17}+13}=\frac{13\left(13^{15}+1\right)}{13\left(13^{16}+1\right)}=\frac{13^{15}+1}{13^{16}+1}=A\)
Vậy: \(A>B\)