So sánh
A=\(\dfrac{25^{10}+1}{25^{10}-1}\) B=\(\dfrac{25^{10}-1}{25^{10}-3}\)
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NT
1
11 tháng 5 2021
`M=(10^25+1)/(10^26+1)`
`=>10M=(10^26+10)/(10^26+1)=1+9/(10^26+1)``
`CMTT:10N=1+9/(10^27+1)`
Vì `1/(10^26+1)>1/(10^27+1)`
`=>9/(10^26+1)>9/(10^27+1)`
`=>1+9/(10^26+1)>1+9/(10^27+1)`
`=>10M>10N=>M>N`
D
19 tháng 1 2022
2/
a/ \(\dfrac{7}{10}=\dfrac{7.15}{10.15}=\dfrac{105}{150}\)
\(\dfrac{11}{15}=\dfrac{11.10}{15.10}=\dfrac{110}{150}\)
-Vì \(\dfrac{105}{150}< \dfrac{110}{150}\)(105<110)nên \(\dfrac{7}{10}< \dfrac{11}{15}\)
b/ \(\dfrac{-1}{8}=\dfrac{-1.3}{8.3}=\dfrac{-3}{24}\)
-Vì \(\dfrac{-3}{24}>\dfrac{-5}{24}\left(-3>-5\right)\)nên\(\dfrac{-1}{8}>\dfrac{-5}{24}\)
c/\(\dfrac{25}{100}=\dfrac{25:25}{100:25}=\dfrac{1}{4}\)
\(\dfrac{10}{40}=\dfrac{10:10}{40:10}=\dfrac{1}{4}\)
-Vì \(\dfrac{1}{4}=\dfrac{1}{4}\)nên\(\dfrac{25}{100}=\dfrac{10}{40}\)
19 tháng 1 2022
a/ \(\dfrac{7}{10}< \dfrac{11}{15}\)
c/ \(\dfrac{25}{100}=\dfrac{10}{40}\)
27 tháng 9 2021
\(=\dfrac{25}{9}.\dfrac{3}{10}+\dfrac{25}{9}.\dfrac{7}{10}+\dfrac{25}{3}=\dfrac{25}{9}\left(\dfrac{3}{10}+\dfrac{7}{10}\right)+\dfrac{25}{3}=\dfrac{25}{9}+\dfrac{25}{3}=\dfrac{100}{9}\)
C
1
NH
2
YS
27 tháng 10 2018
Xét a, ta có:
2510 = (252)5 = 6255
1010 = (102)5 = 1005
=> (2510 + 1010)25 = (6255 + 1005)25
Xét b, ta có:
2525 = (255)5
1025 = (105)5
=> (2525 + 1025)10 = [(255)5 + (105)5]10
Mình bí mất rồi!
H
20 tháng 7 2021
\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)
\(A=\dfrac{1}{2}-\dfrac{1}{4}\)
\(A=\dfrac{2}{4}-\dfrac{1}{4}\)
\(A=\dfrac{1}{4}\)
7 tháng 3 2021
a, \(\dfrac{10-2x}{2}=\dfrac{25-5x}{5}\)
\(\Leftrightarrow\dfrac{2\left(5-x\right)}{2}=\dfrac{5\left(5-x\right)}{5}\)
\(\Leftrightarrow5-x=5-x\)
\(\Leftrightarrow0x=0\)
⇒ Có vô số giá trị của x thỏa mãn.
Vậy...
b, ĐKXĐ: \(x\ne\pm1\)
\(\dfrac{x-3}{x-1}-\dfrac{2x+1}{x+1}=\dfrac{x-x^2}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+1\right)-\left(2x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-x^2}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow x^2-2x-3-2x^2+x+1=x-x^2\)
\(\Leftrightarrow-2x=2\)
\(\Leftrightarrow x=-1\left(ktm\right)\)
Vậy...
7 tháng 3 2021
a) Ta có: \(\dfrac{10-2x}{2}=\dfrac{25-5x}{5}\)
\(\Leftrightarrow5\left(10-2x\right)=2\left(25-5x\right)\)
\(\Leftrightarrow50-10x=50-10x\)
\(\Leftrightarrow0x=0\)(phương trình có vô số nghiệm)
Vậy: S={x|\(x\in R\)}
\(A=\dfrac{25^{10}+1}{25^{10}-1}=\dfrac{25^{10}-1+2}{25^{10}-1}=\dfrac{25^{10}-1}{25^{10}-1}+\dfrac{2}{25^{10}-1}=1+\dfrac{2}{25^{10}-1}\left(1\right)\)
\(B=\dfrac{25^{10}-1}{25^{10}-3}=\dfrac{25^{10}-3+2}{25^{10}-3}=\dfrac{25^{10}-3}{25^{10}-3}+\dfrac{2}{25^{10}-3}=1+\dfrac{2}{25^{10}-3}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow A< B\)
Ta có : \(\dfrac{25^{10}+1}{25^{10}-1}=\dfrac{25^{10}-1+2}{25^{10}-1}=\dfrac{25^{10}-1}{25^{10}-1}+\dfrac{2}{25^{10}-1}\)\(=1+\dfrac{2}{25^{10}-1}\)
Ta có : \(\dfrac{25^{10}-1}{25^{10}-3}=\dfrac{25^{10}-3+2}{25^{10}-3}=\dfrac{25^{10}-3}{25^{10}-3}+\dfrac{2}{25^{10}-3}=1+\dfrac{2}{25^{10}-3}\)
Vì \(25^{10}-1>25^{10}-3\)
\(\Rightarrow\dfrac{2}{25^{10}-1}< \dfrac{2}{25^{10}-3}\)
\(\Rightarrow1+\dfrac{2}{25^{10}-1}< 1+\dfrac{2}{25^{10}-3}\)
\(\Rightarrow A< B\)
Vậy \(A< B\)