cho A=\(\frac{2^{18}-3}{2^{20}-3}\)
B=\(\frac{2^{20}-3}{2^{22}-3}\)
so sanh a va b nha
ai giai nhanh tui cho 5 k
cho A=\(\frac{2^{18}-3}{2^{20}-3}\)
B=\(\frac{2^{20}-3}{2^{22}-3}\)
so sanh A và B nha
ai giải nhanh tui cho 5 k nha
So Sanh : a) (1+2+3+4+5)^2 va 1^3+2^3+3^3+4^3+5^3
b) 19^4 va 16*18*20*22
so sánh
A=\(\frac{2^{18-3}}{2^{20}-3}\)và B= \(\frac{2^{20}-3}{2^{22}-3}\)
So sánh:
a)A=\(\frac{19^{30}+5}{19^{31}+5}\):B=\(\frac{19^{31}+5}{19^{32}+5}\)
b)A=\(\frac{2^{18}-3}{2^{20}-3}\):B=\(\frac{2^{20}-3}{2^{22}-3}\)
Lưu ý :Ko tính kết quả chỉ so sánh
Thankyouverymuch
So sánh A và B biét
A=\(\frac{19^{30}+5}{10^{31}+5}\)và B=\(\frac{19^{31}+5}{19^{32}+5}\)
A= \(\frac{2^{18}-3}{2^{20}-3}\)và B = \(\frac{2^{20}-3}{2^{22}-3}\)
A = \(\frac{1+5+5^2+.......+5^9}{1+5+5^2+.....+5^8}\) B = \(\frac{1+3+3^2+.....+3^9}{1+3+3^2+.......+3^8}\)
a, \(B=\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+90}{19^{32}+5+90}=\frac{19^{31}+95}{19^{32}+95}=\frac{19\left(19^{30}+5\right)}{19\left(19^{31}+5\right)}=\frac{19^{30}+5}{19^{31}+5}=A\)
b, Ta có: \(\frac{1}{A}=\frac{2^{20}-3}{2^{18}-3}=\frac{2^2.\left(2^{18}-3\right)+9}{2^{18}-3}=4+\frac{9}{2^{18}-3}\)
\(\frac{1}{B}=\frac{2^{22}-3}{2^{20}-3}=\frac{2^2\left(2^{20}-3\right)+9}{2^{20}-3}=4+\frac{9}{2^{20}-3}\)
Vì \(\frac{9}{2^{18}-3}>\frac{9}{2^{20}-3}\)\(\Rightarrow\frac{1}{A}>\frac{1}{B}\Rightarrow A< B\)
c, Câu hỏi của truong nguyen kim
so sanhs \(a=\frac{2^{18}-3}{2^{20}-3}vàB=\frac{2^{20}-3}{2^{22}-3}\)
\(A=\frac{2^{18}-3}{2^{20}-3}=1-\frac{2^2}{2^{20}-3}\)
\(B=\frac{2^{20}-3}{2^{22}-3}=1-\frac{2^2}{2^{22}-3}\)
Vì \(\frac{2^2}{2^{20}-3}>\frac{2^2}{2^{22}-3}\) nên A < B
Sa sánh A và B:
a) A=\(\frac{19^{30}+5}{19^{31}+5}\)và B=\(\frac{19^{31}+5}{19^{32}+5}\)
b)A=\(\frac{2^{18}-3}{2^{20}-3}\)và B=\(\frac{2^{20}-3}{2^{22}-3}\)
Lưu ý: ko tính kết quả chỉ so sánh thôi
Càng rõ càng tốt
Thankyouverymuch
so sank
\(A=\frac{2^{18}-3}{2^{20}-2}vaB=\frac{2^{20}-3}{2^{22}-3}\)
Ta có :
\(\dfrac{1}{A}=\dfrac{2^{20}-3}{2^{18}-3}=\dfrac{2^2.\left(2^{18}-3\right)+9}{2^{18}-3}=4+\dfrac{9}{2^{18}-3}\left(1\right)\)
\(\dfrac{1}{B}=\dfrac{2^{22}-3}{2^{20}-3}=\dfrac{2^2.\left(2^{20}-3\right)+9}{2^{20}-3}=4+\dfrac{9}{2^{20}-3}\left(2\right)\)
Từ (1) và (2) ta có \(\dfrac{1}{A}>\dfrac{1}{B}\Leftrightarrow A< B\)
So sanh A va B