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\(A=\frac{10^{29}+10^{10}}{10^{30}+10^{10}}=\frac{10^{10}.\left(10^{19}+1\right)}{10^{10}.\left(10^{20}+1\right)}=\)\(\frac{10^{19}+1}{10^{20}+1}\)
\(\Leftrightarrow10A=1+\frac{9}{10^{20}+1}\)
\(B=\frac{10^{30}+10^{10}}{10^{31}+10^{10}}=\frac{10^{10}.\left(10^{20}+1\right)}{10^{10}.\left(10^{21}+1\right)}=\frac{10^{20}+1}{10^{21}+1}\)
\(\Leftrightarrow10B=1+\frac{9}{10^{21}+1}\)
Vì \(1+\frac{9}{10^{20}+1}>1+\frac{9}{10^{21}+1}\Rightarrow10A>10B\Leftrightarrow A>B\)
Ta có :
\(A=\frac{10^{11}-1}{10^{12}-1}\) \(B=\frac{10^{11}+1}{10^{11}+1}\)
\(10A=\frac{10^{12}-10}{10^{12}-1}\) \(10B=\frac{10^{11}+10}{10^{11}+1}\)
\(10A=\frac{10^{12}-1-9}{10^{12}-1}\) \(10B=\frac{10^{11}+1+9}{10^{11}+1}\)
\(10A=1-\frac{9}{10^{12}-1}\) \(10B=1+\frac{9}{10^{11}+1}\)
Ta thấy : \(1-\frac{9}{10^{12}-1}< 1\) mà \(1+\frac{9}{10^{11}+1}>1\)
\(\Rightarrow A< B\)
Vậy \(A< B\)
Ủng hộ mk nha !!! ^_^
\(A=\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)
\(B=\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)
\(\frac{13}{10^7-7}>\frac{13}{10^8-7}\Rightarrow\frac{10^7+5}{10^7-8}>\frac{10^8+6}{10^8-7}\)
linh ới mi còn kém lắm tui lm đc rùi nha mà ko cần nhìn cái j nha ^.^ hề hề
Ta có \(A=\frac{10^{11}-1}{10^{12}-1}\)
=> \(10A=\frac{10^{12}-10}{10^{12}-1}\)
=>\(10A=\frac{\left(10^{12-1}\right)-9}{10^{12}-1}\)
=>\(10A=1-\frac{9}{10^{12}-1}\) ( 1 )
Ta có \(B=\frac{10^{10}+1}{10^{11}+1}\)
=>\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{\left(10^{11}+1\right)+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}\) ( 2 )
Từ 1 và 2 => 10A < 10B => A < B
a. 1-2+3-4+5-6+7-8+9-10 1+2+3+4+5+6+7+8+9+10
= (-1)+ (-1)+(-1)+(-1) (-1) =(10+1)10/2
=-5 =55
1-2+3-4+...+9-10/1+2+3+...+9+10=-5/55=-1/11
bài này là bài mấy vậy
\(10A=\frac{10\left(10^{29}+10^{10}\right)}{10^{30}+10^{10}}=\frac{10^{30}+10^{11}}{10^{30}+10^{10}}=1+\frac{10^{11}-10^{10}}{10^{30}+10^{10}}\)
\(10B=\frac{10\left(10^{30}+10^{10}\right)}{10^{31}+10^{10}}=\frac{10^{31}+10^{11}}{10^{31}+10^{10}}=1+\frac{10^{11}-10^{10}}{10^{31}+10^{10}}\)
\(10^{30}+10^{10}< 10^{31}+10^{10}\Rightarrow\frac{10^{11}-10^{10}}{10^{30}+10^{10}}>\frac{10^{11}-10^{10}}{10^{31}+10^{10}}\)
\(\Rightarrow10A=1+\frac{10^{11}-10^{10}}{10^{30}+10^{10}}>10B=1+\frac{10^{11}-10^{10}}{10^{31}+10^{10}}\)
\(\Rightarrow A>B\)