cho mn hỏi
so sánh \(\frac{19^{19}-5}{19^{20}+4}vs\frac{19^{20}-5}{19^{21}+4}\)
so sánh
A=\(\frac{19^{20}+5}{19^{20}-8}\)và B=\(\frac{19^{21}+6}{19^{21}-7}\)
Ta có:
\(A=\frac{19^{20}+5}{19^{20}-8}=\frac{19^{20}-8+13}{19^{20}-8}=1+\frac{13}{19^{20}-8}\)
\(B=\frac{19^{21}+6}{19^{21}-7}=\frac{19^{21}-7+13}{19^{21}-7}=1+\frac{13}{19^{21}-7}\)
Vì \(19^{20}-8< 19^{21}-7\Rightarrow\frac{13}{19^{20}-8}>\frac{13}{19^{21}-7}\)
\(\Rightarrow A>B\)
A =B NHA ! maivananh
ĐỀU = 1 CẢ .
Ta có : \(A=\frac{19^{20}+5}{19^{20}-8}=\frac{19^{20}-8+13}{19^{20}-8}=\frac{19^{20}-8}{19^{20}-8}+\frac{13}{19^{20}-8}=1+\frac{13}{19^{20}-8}\)
\(B=\frac{19^{21}+6}{19^{21}-7}=\frac{19^{21}-7+13}{19^{21}-7}=\frac{19^{21}-7}{19^{21}-1}+\frac{13}{19^{21}-7}=1+\frac{13}{19^{21}-7}\)
Mà : \(\frac{13}{19^{20}-8}< \frac{13}{19^{21}-7}\)
Nên : \(A< B\)
so sánh
A=\(\frac{19^{20}+5}{19^{20}-8}\)
B=\(\frac{19^{21}+6}{19^{21}-7}\)
Cho \(A=\frac{19^{20}+5}{19^{20}-8}\) và \(B=\frac{19^{21}+6}{19^{21}-7}\)
So sánh 2 phân số A và B
thông điệp nhỏ:
hay khi ko muốn tích
ai tích mình tích lại nha nha
SO SÁNH A VÀ B BIẾT:
\(A=\frac{19^{20}+5}{19^{20}-8}\)
\(B=\frac{19^{21}+6}{19^{21}-7}\)
So sánh hai phân số sau:
A=\(\frac{19^{20}+5}{19^{20}-8}\)
B=\(\frac{19^{21}+6}{19^{21}-7}\)
giải chi tiết nhé để có tick
A= \(\frac{19^{20}+5}{19^{20}-8}=\frac{19^{20}-8+13}{19^{20}-8}=1+\frac{13}{19^{20}-8}\)
B= \(\frac{19^{21}+6}{19^{21}-7}=\frac{19^{21}-7+13}{19^{21}-7}=1+\frac{13}{19^{21}-7}\)
Mà \(\frac{13}{19^{20}-8}>\frac{13}{19^{21}-7}\) nên A > B
k nha
A=19^20+5/19^20-8 >1
=> 19^20+5/19^20-8> 19^20+5+1+19/19^20-8+1+19 B=19^20+5+1+19/19^20-8+1+19 =19^21+6/19^21-7
=> A>B
so sánh
A=\(\frac{19^{20}+5}{19^{20}-8}\)
B=\(\frac{19^{20}+6}{19^{20}-7}\)
Ta có: \(A=\frac {19^{20}+5}{19^{20}-8}=\frac {19^{20}-8+13}{19^{20}-8}=1+\frac {13}{19^{20}-8}\)
\(B=\frac {19^{20}+6}{19^{20}-7}=\frac {19^{20}-7+13}{19^{20}-7}=1+\frac {13}{19^{20}-7}\)
Vì \(19^{20}-8<19^{20}-7\) nên \(\frac {13}{19^{20}-8}>\frac {13}{19^{20}-7}\)
\(\Rightarrow\)\(1+\frac{13}{19^{20}-8}>1+\frac{13}{19^{20}-7}\) Hay \(A>B\)
Vậy A>B
ta có A = \(\frac{19^{20}+5}{19^{20}-8}=\frac{19^{20}-8+13}{19^{20}-8}=1+\frac{13}{19^{20}-8}\)
và B = \(\frac{19^{20}+6}{19^{20}-7}=\frac{19^{20}-7+13}{19^{20}-7}=1+\frac{13}{19^{20}-7}\)
vì \(\frac{13}{19^{20}-8}>\frac{13}{19^{20}-7}\)\(\Rightarrow1+\frac{13}{19^{20}-8}>1+\frac{13}{19^{20}-7}\)\(\Rightarrow A>B\)
Cho \(S=\frac{17}{18}+\frac{18}{19}+\frac{19}{20}+\frac{20}{17}\)
SO sánh S với 4
\(ChoM=\frac{18}{18+19+20}+\frac{19}{18+19+21}+\frac{20}{18+19+21}+\frac{21}{18+19+21}.\)
Chứng tỏ rằng 1<M<2
Có \(\frac{18}{18+19+20}>\frac{18}{18+19+20+21}\)
\(\frac{19}{18+19+21}>\frac{19}{18+19+20+21}\)
\(\frac{20}{18+19+21}>\frac{20}{18+19+20+21}\)
\(\frac{21}{18+19+21}>\frac{21}{18+19+20+21}\)
=> \(\frac{18}{18+19+20}+\frac{19}{18+19+21}+\frac{20}{18+19+21}+\frac{21}{18+19+21}>\frac{18}{18+19+20+21}+\frac{19}{18+19+20+21}+\frac{20}{18+19+20+21}+\frac{21}{18+19+20+21}\)
=> \(\frac{18}{18+19+20}+\frac{19}{18+19+21}+\frac{20}{18+19+21}+\frac{21}{18+19+21}>\frac{18+19+20+21}{18+19+20+21}\)
=>\(\frac{18}{18+19+20}+\frac{19}{18+19+21}+\frac{20}{18+19+21}+\frac{21}{18+19+21}>1\)
=>M>1
Còn lại mình không biết, đúng thì tick nha
201819 - \(\frac{5}{2018^{19}+1}\)
So sánh với
201920 - \(\frac{4}{2019^{20}+2}\)