Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Áp dụng bất đẳng thức :
\(\dfrac{a}{b}< \dfrac{a+m}{b+m}\)
Ta có :
\(A=\dfrac{10^{101}-1}{10^{102}-1}< \dfrac{10^{101}-1+11}{10^{102}-1+11}=\dfrac{10^{101}+10}{10^{102}+10}=\dfrac{10\left(10^{100}+1\right)}{10\left(10^{101}+1\right)}=\dfrac{10^{100}+1}{10^{101}+1}=B\)
\(\Leftrightarrow A< B\)
Ta có:
\(1-A=1-\dfrac{10^{101}-1}{10^{102}-1}=\dfrac{10^{102}-1\left(10^{101}-1\right)}{10^{102}-1}\) \(=\dfrac{10^{102}-1-10^{101}+1}{10^{102}-2}=\dfrac{10^{102}-10^{101}}{10^{102}-1}\)
\(=\dfrac{10^{101}\left(10-1\right)}{10^{101}\left(10-\dfrac{1}{10^{101}}\right)}=\dfrac{10-1}{10-\dfrac{1}{10^{101}}}=\dfrac{9}{10-\dfrac{1}{10^{101}}}\)\(\left(1\right)\)
\(1-B=1-\dfrac{10^{100}+1}{10^{101}+1}=\dfrac{10^{101}+1-\left(10^{100}+1\right)}{10^{101}+1}\)
\(=\dfrac{10^{101}+1-10^{100}-1}{10^{101}+1}\) \(=\dfrac{10^{101}-10^{100}}{10^{101}+1}=\dfrac{10^{100}\left(10-1\right)}{10^{100}\left(10+\dfrac{1}{10^{100}}\right)}\)
\(=\dfrac{10-1}{10+\dfrac{1}{10^{100}}}=\dfrac{9}{10+\dfrac{1}{100}}\)\(\left(2\right)\)
\(Từ\left(1\right);\left(2\right)\) \(=>A< B\)\(\left(đpcm\right)\)
CHÚC BẠN HỌC TỐT 
Áp dụng tính chất \(\dfrac{a}{b}>1\Leftrightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ta có :
\(B=\dfrac{10^{99}+1}{10^{89}+1}>\dfrac{10^{99}+1+9}{10^{89}+1+9}=\dfrac{10^{99}+10}{10^{98}+10}=\dfrac{10\left(10^{98}+1\right)}{10\left(10^{88}+1\right)}=\dfrac{10^{98}+1}{10^{88}+1}=A\)
\(\Leftrightarrow B>A\)
Ta áp dụng tính chất :
\(\dfrac{a}{b}>1\Leftrightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ta có:
\(B=\dfrac{10^{99}+1}{10^{89}+1}>\dfrac{10^{99}+1+9}{10^{89}+1+9}=\dfrac{10^{99}+10}{10^{89}+10}=\dfrac{10\left(10^{98}+1\right)}{10\left(10^{88}+1\right)}=\dfrac{10^{98}+1}{10^{88}+1}=A\)
\(\Leftrightarrow B>A\)
Chúc bạn học tốt!
T làm biếng lắm; làm C thôi
\(A=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\\ \Rightarrow A< \dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\\ \Rightarrow A^2< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\right)\\ =\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{99}{100}.\dfrac{100}{101}\\ =\dfrac{1}{101}< \dfrac{1}{100}\\ \Rightarrow A< \dfrac{1}{10}\)
Làm tương tự ta được A > 1/15
câu a
\(A=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{30}>\dfrac{20}{30}=\dfrac{2}{3}>\dfrac{1}{3}\)
\(A=\left(\dfrac{1}{11}+..+\dfrac{1}{15}\right)+\left(\dfrac{1}{16}+...+\dfrac{1}{30}\right)< 5.\dfrac{1}{10}+25.\dfrac{1}{15}=\dfrac{1}{2}+\dfrac{5}{3}=\dfrac{8}{6}=\dfrac{4}{3}< \dfrac{5}{2}\)
1)\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2017}{2018}\)
\(B=\dfrac{1}{2018}\)
2)a)\(x^2-2x-15=0\)
\(\Leftrightarrow x^2-2x+1-16=0\)
\(\Leftrightarrow\left(x-1\right)^2-16=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
3)\(\dfrac{a}{b}=\dfrac{d}{c}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a}{b}\cdot\dfrac{d}{c}=\dfrac{ad}{bc}\)
Lại có:\(\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a^2+d^2}{b^2+c^2}\)
\(\Rightarrow\dfrac{a^2+d^2}{b^2+c^2}=\dfrac{ad}{bc}\)
4)Ta có:\(g\left(x\right)=-x^{101}+x^{100}-x^{99}+...+x^2-x+1\)
\(g\left(x\right)=-x^{101}+\left(x^{100}-x^{99}+...+x^2-x+1\right)\)
\(g\left(x\right)=-x^{101}+f\left(x\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=f\left(x\right)+x^{101}-f\left(x\right)=x^{101}\)
Tại x=0 thì f(x)-g(x)=0
Tại x=1 thì f(x)-g(x)=1
\(A=\dfrac{10^{99}+1}{10^{100}+1}\)
\(\Leftrightarrow10A=\dfrac{10\left(10^{99}+1\right)}{10^{100}+1}\)
\(\Leftrightarrow10A=\dfrac{10^{100}+10}{10^{100}+1}=\dfrac{10^{100}+1+9}{10^{100}+1}=1+\dfrac{9}{10^{100}+1}\)
\(B=\dfrac{10^{100}+1}{10^{101}+1}\)
\(\Leftrightarrow10B=\dfrac{10\left(10^{100}+1\right)}{10^{101}+1}\)
\(\Leftrightarrow10B=\dfrac{10^{101}+10}{10^{101}+1}=\dfrac{10^{101}+1+9}{10^{101}+1}=1+\dfrac{9}{10^{101}+1}\)
Do \(\dfrac{9}{10^{100}+1}>\dfrac{9}{10^{101}+1}\) nên \(10A>10B\)
\(\Rightarrow A>B\)
Áp dụng tính chất:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(B=\dfrac{10^{100}+1}{10^{101}+1}< 1\)
\(B< \dfrac{10^{100}+1+9}{10^{101}+1+9}\)
\(B< \dfrac{10^{100}+10}{10^{101}+10}\)
\(B< \dfrac{10\left(10^{99}+1\right)}{10\left(10^{100}+1\right)}\)
\(B< \dfrac{10^{99}+1}{10^{100}+1}=A\)
\(B< A\)