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a, \(A=\frac{10^8+2}{10^8-1}=\frac{10^8-1+3}{10^8-1}=1+\frac{3}{10^8-1}\)
\(B=\frac{10^8}{10^8-3}=\frac{10^8-3+3}{10^8-3}=1+\frac{3}{10^8-3}\)
Vì \(\frac{3}{10^8-1}< \frac{3}{10^8-3}\Rightarrow1+\frac{3}{10^8-1}< 1+\frac{3}{10^8-3}\Rightarrow A< B\)
b, \(A=\frac{10^{15}+1}{10^{16}+1}\Rightarrow10A=\frac{10\left(10^{15}+1\right)}{10^{16}+1}=\frac{10^{16}+10}{10^{16}+1}=\frac{10^{16}+1+9}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)
\(B=\frac{10^{16}+1}{10^{17}+1}\Rightarrow10B=\frac{10\left(10^{16}+1\right)}{10^{17}+1}=\frac{10^{17}+10}{10^{17}+1}=\frac{10^{17}+1+9}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
Vì \(\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\Rightarrow1+\frac{9}{10^{16}+1}>1+\frac{9}{10^{17}+1}\Rightarrow10A>10B0\Rightarrow A>B\)
c, giống câu b
d, giống câu b
e, \(A=\frac{10^{15}+5}{10^{15}-7}=\frac{10^{15}-7+12}{10^{15}-7}=1+\frac{12}{10^{15}-7}\)
\(B=\frac{10^{16}+7}{10^{16}-5}=\frac{10^{16}-5+12}{10^6-5}=1+\frac{12}{10^6-5}\)
Vì \(\frac{12}{10^{15}-7}>\frac{12}{10^{16}-5}\Rightarrow1+\frac{12}{10^{15}-7}>1+\frac{12}{10^{16}-7}\Rightarrow A>B\)
f, \(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
Vì \(\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\Rightarrow1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}\Rightarrow A< B\)
e, Ta có:
\(A-B=\left(\frac{-7}{10^{2013}}+\frac{-15}{10^{2014}}\right)-\left(\frac{-15}{10^{2013}}+\frac{-7}{10^{2014}}\right)\)
\(=\frac{-7}{10^{2013}}+\frac{-15}{10^{2014}}-\frac{-15}{10^{2013}}-\frac{-7}{10^{2014}}\)
\(=\frac{8}{10^{2013}}-\frac{8}{10^{2014}}>0\)
Vậy A > B
Phần a;b;c;d;e;f liên quan tới
\(\frac{a}{b}< \frac{a+c}{b+c}\forall a< b\) \(\frac{a}{b}>\frac{a+c}{b+c}\forall a>b\) phép trừ thì ngược lại
Giải phần g
\(A=\frac{-7}{10^{2013}}+\frac{-7}{10^{2014}}+\frac{-8}{10^{2014}}\)
\(B=\frac{-7}{10^{2013}}+\frac{-8}{10^{2013}}+\frac{-7}{10^{2014}}\)
có đcB>A
k minh nha
\(a.\)
\(A=\)\(\frac{10^{15}+1}{10^{16}+1}\)
\(10A=\) \(\frac{10\left(10^{15}+1\right)}{10^{16}+1}\)
\(10A=\) \(\frac{10^{16}+10}{10^{16}+1}\)
\(10A=\)\(\frac{10^{16}+1+9}{10^{16}+1}\)
\(10A=\frac{10^{16}+1}{10^{16}+1}+\frac{9}{10^{16}+1}\)
\(10A=1+\frac{9}{10^{16}+1}\)
\(B=\frac{10^{16}+1}{10^{17}+1}\)
\(10B=\frac{10\left(10^{16}+1\right)}{10^{17}+1}\)
\(10B=\frac{10^{17}+10}{10^{17}+1}\)
\(10B=\frac{10^{17}+1+9}{10^{17}+1}\)
\(10B=\frac{10^{17}+1}{10^{17}+1}+\frac{9}{10^{17}+1}\)
\(10B=1+\frac{9}{10^{17}+1}\)
\(\Rightarrow10B< 10A\Rightarrow B< A\)\(\text{( vì tự làm ) }\)
xin lỗi hôm qua mk đang làm thì phải đy học zoom học xong quên h mới nhơ ra làm típ :)
b
\(A=\frac{3}{8^3}+\frac{7}{8^4}=\frac{3}{8^3}+\frac{3}{8^4}+\frac{4}{8^4}\)
\(B=\frac{3}{8^4}+\frac{7}{8^3}=\frac{3}{8^4}+\frac{3}{8^3}+\frac{4}{8^3}\)
Vì \(\frac{4}{8^4}< \frac{4}{8^3}\)=.> A < B
Bài 1:
Ta có:
\(\left(\frac{1}{10}\right)^{15}=\left(\frac{1}{5}\right)^{3.5}=\left(\frac{1}{125}\right)^5\)
\(\left(\frac{3}{10}\right)^{20}=\left(\frac{3}{10}\right)^{4.5}=\left(\frac{81}{10000}\right)^5\)
Lại có:
\(\frac{1}{125}=\frac{80}{10000}< \frac{81}{10000}\Rightarrow\left(\frac{1}{125}\right)^5< \left(\frac{81}{10000}\right)^5\)
\(\Rightarrow\left(\frac{1}{10}\right)^{15}< \left(\frac{3}{10}\right)^{20}\)
Bài 2:
Ta có:
\(A=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13A=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)
\(B=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13B=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)
Mà \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)
\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)
\(\Rightarrow13A>13B\Rightarrow A>B\)
Minh chi biet lam cau b thoi ak
b) Giai:
B=10^16+1 tren 10^17 +1 <10^16+1+9 tren 10^17+1+9
ma 10^16+1+9 tren 10^17+1+9 = 10^16+10 tren 10^17+10
=10(10^15+1) tren 10(10^16+1)
=10^15+1 tren 10^16+1 =A
=>A>B
Cho y kien voi!
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
a) Đặt \(A=\frac{7^{15}}{1+7+7^2+...+7^{14}}\)
Đặt \(B=1+7+7^2+...+7^{14}\)
\(\Rightarrow7B=7+7^2+...+7^{15}\)
\(\Rightarrow7B-B=6B=7^{15}-1\)
\(\Rightarrow B=\frac{7^{15}-1}{6}\)
\(\Rightarrow A=\frac{7^{15}-1+1}{\frac{7^{15}-1}{6}}=\left(7^{15}-1\right).\frac{6}{7^{15}-1}+\frac{6}{7^{15}-1}=6+\frac{6}{7^{15}-1}\)
Tự làm tiếp nha
Đặt S=1/12+1/13+1/14+1/15+...+1/23
ta có 1/12+1/13+1/14+1/15+...+1/22+1/23 = (1/12+1/13+1/14+...+1/17)+(1/18+1/19+...+1/23)
đặt A=1/12+1/13+1/14+...+1/17
ta có
1/13<1/12
1/14<1/12
..........................
.........................
1/17<1/12
=>A<1/12+1/12+1/12+....+1/12 (có 6 phân số)
=>A<1x6/12
=>A<1/2 (1)
Đặt B=1/18+1/19+...+11/23
ta có
1/19<1/18
1/20<1/18
...........................
..........................
1/23<1/18
=> B<1/18+1/18+1/18+...+1/18 (có 6 phân số)
=>B<1x 6/18
=>B<1/3 (2)
từ 1 và 2 =>S=A+B<1/2+1/3
=>S<5/6 (dpcm)
k cho mình nhé
Đặt S=1/12+1/13+1/14+1/15+...+1/23
ta có 1/12+1/13+1/14+1/15+...+1/22+1/23 = (1/12+1/13+1/14+...+1/17)+(1/18+1/19+...+1/23)
đặt A=1/12+1/13+1/14+...+1/17
ta có
1/13<1/12
1/14<1/12
..........................
.........................
1/17<1/12
=>A<1/12+1/12+1/12+....+1/12 (có 6 phân số)
=>A<1x6/12
=>A<1/2 (1)
Đặt B=1/18+1/19+...+11/23
ta có
1/19<1/18
1/20<1/18
...........................
..........................
1/23<1/18
=> B<1/18+1/18+1/18+...+1/18 (có 6 phân số)
=>B<1x 6/18
=>B<1/3 (2)
từ 1 và 2 =>S=A+B<1/2+1/3
=>S<5/6 (dpcm)
k cho mình nhé
nói chuyện với mk đi
nói làm j