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a, Ta có : \(10^{15}\cdot11=10^{15}\left(10+1\right)=10^{16}+10^{15}\)
Vì \(10^{16}+10^{15}>10^{16}+10\)
\(\Rightarrow\dfrac{10^{16}+10^{15}}{10^{16}+1}>\dfrac{10^{16}+10}{10^{16}+1}\)
Hay A>B
b, Ta có : \(C=\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}}{10^{10}-1}+\dfrac{1}{10^{10}-1}\)
\(D=\dfrac{10^{10}-1}{10^{13}-3}=\dfrac{10^{10}}{10^{13}-3}+\dfrac{-1}{10^{13}-3}\)
Vì \(\dfrac{10^{10}}{10^{10}-1}>\dfrac{10^{10}}{10^{13}-3};\dfrac{1}{10^{10}-1}>\dfrac{-1}{10^{13}-3}\)
\(\Rightarrow\dfrac{10^{10}+1}{10^{10}-1}>\dfrac{10^{10}-1}{10^{13}-3}\)
Hay C > D
ta có: \(A=\dfrac{10.10^{10}-1}{10.10^{11}-1}=\dfrac{10^{10}-1}{10^{11}-1}\)
so sánh: \(A=\dfrac{10^{10}-1}{10^{11}-1}\)và \(B=\dfrac{10^{10}+1}{10^{11}+1}\)
\(\Rightarrow A< B\)
Ta có :\(a=\dfrac{10^{11}-1}{10^{12}-1}\Rightarrow10a=\dfrac{10^{12}-10}{10^{12}-1}=\dfrac{10^{12}-1-9}{10^{12}-1}=1-\dfrac{9}{10^{12}-1}\)
\(b=\dfrac{10^{10}+1}{10^{11}+1}\Rightarrow10b=\dfrac{10^{11}+10}{10^{11}+1}=\dfrac{10^{11}+1+9}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\)
Ta có : \(1-\dfrac{9}{10^{12}-1}\le1+\dfrac{9}{10^{11}+1}\) hay \(10a< 10b\Rightarrow a< b\)
Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(A=\dfrac{10^{11}-1}{10^{12}-1}< 1\)
\(A< \dfrac{10^{11}-1+11}{10^{12}-1+11}\Rightarrow A< \dfrac{10^{11}+10}{10^{12}+10}\Rightarrow A< \dfrac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}\Rightarrow A< \dfrac{10^{10}+1}{10^{11}+1}=B\)
\(\Rightarrow A< B\)
ta có :
\(A=\dfrac{10^{11}-1}{10^{12}-1}\\ 10A=\dfrac{10^{12}-10}{10^{12}-1}=1-\dfrac{9}{10^{12}-1}\\ =>10A< 1\\ B=\dfrac{10^{10}+1}{10^{11}+1}\\ 10B=\dfrac{10^{11}+10}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\\ =>10B>1\)
=> 10A<10B =>A<B
vậy A bé hơn B
Bạn ơi !
Hàng thứ 2 dưới lên phải viết là : Ta có : 10A < 10B => A < B
Ta có: \(\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{10^{11}-1+11}{10^{12}-1+11}\)
\(\Rightarrow A< \dfrac{10^{11}+10}{10^{12}+10}\)
\(\Rightarrow A< \dfrac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}\)
\(\Rightarrow A< \dfrac{10^{10}+1}{10^{11}+1}\)
\(\Rightarrow A< B\)
Vậy \(A< B\).
Cách 2:
Ta có: \(10A=\dfrac{10^{12}-10}{10^{12}-1}=1-\dfrac{9}{10^{12}-1}\)
\(10B=\dfrac{10^{11}+10}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\)
Vì \(\dfrac{9}{10^{12}-1}< \dfrac{9}{10^{11}+1}\Rightarrow1-\dfrac{9}{10^{12}-1}< 1+\dfrac{9}{10^{11}+1}\)
\(\Rightarrow10A< 10B\Rightarrow A< B\)
Vậy A < B
c) E = \(\dfrac{4116-14}{10290-35}\) và K = \(\dfrac{2929-101}{2.1919+404}\)
E = \(\dfrac{4116-14}{10290-35}\)
E = \(\dfrac{14.\left(294-1\right)}{35.\left(294-1\right)}\)
E = \(\dfrac{14}{35}\)
K = \(\dfrac{2929-101}{2.1919+404}\)
K = \(\dfrac{101.\left(29-1\right)}{101.\left(38+4\right)}\)
K = \(\dfrac{29-1}{34+8}\)
K = \(\dfrac{28}{42}\) = \(\dfrac{2}{3}\)
Ta có : E = \(\dfrac{14}{35}\) và K = \(\dfrac{2}{3}\)
\(\dfrac{14}{35}\) = \(\dfrac{42}{105}\)
\(\dfrac{2}{3}\) = \(\dfrac{70}{105}\)
Vậy E < K
Các câu còn lại tương tự
\(A=\frac{10^{11}-1}{10^{12}-1}< \frac{10^{11}-1+11}{10^{12}-1+11}\) theo công thức \(\frac{a}{b}< \frac{a+m}{b+m}\)
\(A< \frac{10^{11}+10}{10^{12}+10}=\frac{10^{10}\left(10+1\right)}{10^{11}\left(10+1\right)}=\frac{10^{10}}{10^{11}}\)
\(\Rightarrow\frac{10^{10}}{10^{11}}=\frac{10^{10}\cdot10^{12}}{10^{11}\cdot10^{12}}=\frac{10^{22}}{10^{23}}\)
\(\Leftrightarrow A< \frac{10^{10}}{10^{11}}=\frac{10^{11}}{10^{12}}\)
Lại áp dụng công thức \(\frac{a}{b}< \frac{a+m}{b+m}\)
\(A< \frac{10^{10}}{10^{11}}=\frac{10^{11}}{10^{12}}< \frac{10^{11}+1}{10^{12}+1}=B\)
\(\Leftrightarrow A< B\)
Hoặc \(A< \frac{10^{11}-1+2}{10^{12}-1+2}=\frac{10^{12}+1}{10^{12}+1}\)
..... (EZ)
\(10A=\frac{10\left(10^{11}-1\right)}{10^{12}-1}=\frac{10^{12}-10}{10^{12}-1}=1-\frac{9}{10^{12}-1}\)
\(10B=\frac{10\left(10^{10}+1\right)}{10^{11}+1}=\frac{10^{11}+10}{10^{11}+1}=1+\frac{9}{10^{11}+1}\)
Vì \(1-\frac{9}{10^{12}-1}< 1+\frac{9}{10^{11}+1}\Rightarrow10A< 10B\)
\(\Rightarrow A< B\)
Ta có: \(10C=\dfrac{10^{12}-10}{10^{12}-1}=1-\dfrac{9}{10^{12}-1}\)
\(10D=\dfrac{10^{11}+10}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\)
\(\dfrac{9}{10^{12}-1}< \dfrac{9}{10^{12}+1}\Rightarrow1-\dfrac{9}{10^{12}-1}< 1+\dfrac{9}{10^{11}+1}\Rightarrow10A< 10B\)
\(\Rightarrow A< B\)
Vậy A < B
10C=\(\dfrac{10.\left(10^{11}-1\right)}{10^{12}-1}\)=\(\dfrac{10^{12}-10}{10^{12}-1}\)=\(\dfrac{10^{12}-1-9}{1012-1}\)=1-\(\dfrac{9}{10^{12}-1}\)<1 (1)
10D=\(\dfrac{10.\left(10^{10}+1\right)}{10^{11}+1}\)=\(\dfrac{10^{11}+10}{10^{11}+1}\)=\(\dfrac{10^{11}+1+9}{10^{11}+1}\)=1+\(\dfrac{9}{10^{11}+1}\)>1 (2)
(1),(2)=> 1-\(\dfrac{9}{10^{12}-1}\)<1+\(\dfrac{9}{10^{11}+1}\)
hay 10C<10D
=>C<D