Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{10^{99}+1}{10^{100}+1}\)
\(\Leftrightarrow10A=\dfrac{10\left(10^{99}+1\right)}{10^{100}+1}\)
\(\Leftrightarrow10A=\dfrac{10^{100}+10}{10^{100}+1}=\dfrac{10^{100}+1+9}{10^{100}+1}=1+\dfrac{9}{10^{100}+1}\)
\(B=\dfrac{10^{100}+1}{10^{101}+1}\)
\(\Leftrightarrow10B=\dfrac{10\left(10^{100}+1\right)}{10^{101}+1}\)
\(\Leftrightarrow10B=\dfrac{10^{101}+10}{10^{101}+1}=\dfrac{10^{101}+1+9}{10^{101}+1}=1+\dfrac{9}{10^{101}+1}\)
Do \(\dfrac{9}{10^{100}+1}>\dfrac{9}{10^{101}+1}\) nên \(10A>10B\)
\(\Rightarrow A>B\)
Áp dụng tính chất:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(B=\dfrac{10^{100}+1}{10^{101}+1}< 1\)
\(B< \dfrac{10^{100}+1+9}{10^{101}+1+9}\)
\(B< \dfrac{10^{100}+10}{10^{101}+10}\)
\(B< \dfrac{10\left(10^{99}+1\right)}{10\left(10^{100}+1\right)}\)
\(B< \dfrac{10^{99}+1}{10^{100}+1}=A\)
\(B< A\)
\(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow2B=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{97}+\left(\dfrac{1}{2}\right)^{98}\)
\(\Rightarrow2B-B=1-\left(\dfrac{1}{2}\right)^{99}\)
\(B=1-\left(\dfrac{1}{2}\right)^{99}\)
\(2,\)
\(a,\dfrac{45^{10}.2^{10}}{75^{15}}\)
\(=\dfrac{5^{10}.9^{10}.2^{10}}{25^{15}.3^{15}}\)
\(=\dfrac{5^{10}.3^{20}.2^{10}}{5^{30}.3^{15}}\)
\(=\dfrac{5^{10}.3^{15}.\left(3^5.2^{10}\right)}{5^{10}.3^{15}.\left(5^{20}\right)}\)
\(=\dfrac{3^5.2^{10}}{5^{20}}\)
\(b,\dfrac{2^{15}.9^4}{6^3.8^3}\)
\(=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5\)
\(c,\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{4^{10}.2^{10}+4^{10}}{4^4.2^4+4^4.4^7}=\dfrac{4^4.\left(4^6.2^{10}+4^6\right)}{4^4.\left(2^4+4^7\right)}\)
\(=\dfrac{4^{11}+4^6}{4^8.4^7}=\dfrac{4^6.\left(4^5+1\right)}{4^6.\left(4^2-4\right)}=\dfrac{1024+1}{16-4}=\dfrac{1025}{12}\)
\(d,\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}=\dfrac{3^{44}.3^{17}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
\(3,\)
\(a,\left(2x+4\right)^2=\dfrac{1}{4}\)
\(\left(2x+4\right)^2=\left(\dfrac{1}{2}\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x+4=\dfrac{1}{2}\\2x+4=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{2}-4=\dfrac{-7}{2}\\2x=\dfrac{-1}{2}-4=\dfrac{-9}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-7}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-7}{4};\dfrac{-9}{4}\right\}\)
\(b,\left(2x-3\right)^2=36\)
\(\left(2x-3\right)^2=6^2=\left(-6\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=6+3=9\\2x=-6+3=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{9}{2};\dfrac{-3}{2}\right\}\)
\(c,5^{x+2}=628\)
\(5^{x+2}=5^4\)
\(\Rightarrow x+2=4\)
\(\Rightarrow x=4-2=2\)
Vậy \(x=2\)
\(d,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Rightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)
\(\Rightarrow\left(x-1\right)^{x+2}.\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{0;1;2\right\}\)
Bài 1:
B= \(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{99}\)
2B= \(2.[\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}]\)
2B= \(1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{98}\)
⇒2B-B= \(1-\left(\dfrac{1}{2}\right)^{99}\)
B= 1
Vậy B=1
Bài 2:
a, \(\dfrac{45^{10}.2^{10}}{75^{15}}\)= \(\dfrac{\left(3^2.5\right)^{10}.2^{10}}{\left(3.5^2\right)^{15}}=\dfrac{3^{20}.5^{10}.2^{10}}{3^{15}.5^{30}}=\dfrac{3^5.2^{10}}{5^{20}}\)
b, \(\dfrac{2^{15}.9^4}{6^3.8^3}=\dfrac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^3.\left(2^3\right)^3}=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5\)
c,\(\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2.4\right)^{10}+4^{10}}{\left(2.4\right)^4+4^{11}}=\dfrac{2^{10}.4^{10}+4^{10}}{2^4.4^4+4^{11}}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6+4^6.4^5}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(4^5+1\right)}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(2^{10}+1\right)}=4^4=256\)
d, \(\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}=\dfrac{\left(3^4\right)^{11}.3^{17}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\dfrac{3^{44}.3^{17}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
Bài 3:
a, \(\left(2x+4\right)^2=\dfrac{1}{4}\)
\(\left(2x+4\right)^2=\left(\dfrac{1}{2}\right)^2\)
\(2x+4=\dfrac{1}{2}\)
\(2x=\dfrac{1}{2}-4\)
\(2x=-\dfrac{7}{2}\)
\(x=-\dfrac{7}{2}:2\)
\(x=-\dfrac{7}{2}.\dfrac{1}{2}\)
\(x=-\dfrac{7}{4}\)
b, \(\left(2x-3\right)^2=36\)
\(\left(2x-3\right)^2=6^2\)
\(2x-3=6\)
\(2x=9\)
\(x=\dfrac{9}{2}\)
c, \(5^{x+2}=625\)
\(5^{x+2}=5^4\)
\(x+2=4\)
\(x=2\)
Áp dụng bất đẳng thức :
\(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\left(a;b;m\in N;b\ne0\right)\)
Ta có : \(B=\dfrac{10^{2007}+1}{10^{2008}+1}< 1\)
\(\Leftrightarrow B=\dfrac{10^{2007}+1}{10^{2008}+1}< \dfrac{10^{2007}+1+9}{10^{2008}+1+9}=\dfrac{10^{2007}+10}{10^{2008}+10}=\dfrac{10\left(10^{2006}+1\right)}{10\left(10^{2007}+1\right)}=\dfrac{10^{2006}+1}{10^{2007}+1}=A\)
\(\Leftrightarrow B< A\)
10a=10^2017+10/10^2017+1
10b=10^2018+10/10^2018+1
cậu tự so sánh nhé vậy là dễ rồi
Ta có: \(A=\dfrac{10^{2016}+1}{10^{2017}+1}\Rightarrow10A=\dfrac{10\left(10^{2016}+1\right)}{10^{2017}+1}=\dfrac{10^{2017}+10}{10^{2017}+1}\)
\(=\dfrac{10^{2017}+1+9}{10^{2017}+1}=\dfrac{10^{2017}+1}{10^{2017}+1}+\dfrac{9}{10^{2017}+1}=1+\dfrac{9}{10^{2017}+1}\)
Tương tự ta cũng có: \(10B=1+\dfrac{9}{10^{2018}+1}\)
Lại có: \(10^{2017}< 10^{2018}\Rightarrow10^{2017}+1< 10^{2018}+1\)
\(\Rightarrow\dfrac{1}{10^{2017}+1}>\dfrac{1}{10^{2018}+1}\Rightarrow\dfrac{9}{10^{2017}+1}>\dfrac{9}{10^{2018}+1}\)
\(\Rightarrow1+\dfrac{9}{10^{2017}+1}>1+\dfrac{9}{10^{2018}+1}\Rightarrow10A>10B\Rightarrow A>B\)
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 
a. Xét phân số trung gian là \(\dfrac{72}{78}\) , ta thấy:
\(\dfrac{72}{73}>\dfrac{72}{78}\)
\(\dfrac{58}{78}< \dfrac{72}{78}\)
\(\Rightarrow\dfrac{72}{73}>\dfrac{58}{78}\)
b. Xét phân số trung gian là \(\dfrac{n}{n+2}\) , ta thấy:
\(\dfrac{n}{n+3}< \dfrac{n}{n+2}\)
\(\dfrac{n}{n+2}< \dfrac{n+1}{n+2}\)
\(\Rightarrow\dfrac{n}{n+3}< \dfrac{n+1}{n+2}\)
c. Ta có: \(\dfrac{10^{11}-1}{10^{12}-1}< 1\) (vì tử < mẫu)
\(\Rightarrow\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{\left(10^{11}-1\right)+11}{\left(10^{12}-1\right)+11}=\dfrac{10^{11}+10}{10^{12}+10}=\dfrac{10^{10}+1}{10^{11}+1}\)
Vậy \(\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{10^{10}+1}{10^{11}+1}\)
d. Xét phân số trung gian là \(\dfrac{1}{4}\) , ta thấy:
\(\dfrac{12}{47}>\dfrac{12}{48}=\dfrac{1}{4}\)
\(\dfrac{19}{77}< \dfrac{19}{76}=\dfrac{1}{4}\)
\(\Rightarrow\dfrac{12}{47}>\dfrac{19}{77}\)
Có:\(10A=\dfrac{10^{16}+10}{10^{16}+1}=\dfrac{10^{16}+1+9}{10^{16}+1}=\dfrac{10^{16}+1}{10^{16}+1}+\dfrac{9}{10^{16}+1}=1+\dfrac{9}{10^{16}+1}\)
\(10B=\dfrac{10^{17}+10}{10^{17}+1}=\dfrac{10^{17}+1+9}{10^{17}+1}=\dfrac{10^{17}+1}{10^{17}+1}+\dfrac{9}{10^{17}+1}=1+\dfrac{9}{10^{17}+1}\)
\(1+\dfrac{9}{10^{16}+1}>1+\dfrac{9}{10^{17}+1}\Rightarrow A>B\)
Vậy \(A>B\)
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!