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: \(10^{2017}< 10^{2018}\Rightarrow10^{2017}+1< 10^{2018}+1\)

\(\Rightarrow\dfrac{1}{10^{2017}+1}>\dfrac{1}{10^{2018}+1}\Rightarrow\dfrac{9}{10^{2017}+1}>\dfrac{9}{10^{2018}+1}\)

\(\Rightarrow1+\dfrac{9}{10^{2017}+1}>1+\dfrac{9}{10^{2018}+1}\Rightarrow10A>10B\Rightarrow A>B\)

Đặt :

\(A=\dfrac{10^{2017}+1}{10^{2016}+1}\)

\(B=\dfrac{10^{2018}+1}{10^{2017}+1}\)

Ta thấy :

\(\left\{{}\begin{matrix}A=\dfrac{10^{2017}+1}{10^{2016}+1}>1\\B=\dfrac{10^{2018}+1}{10^{2017}+1}>1\end{matrix}\right.\)

Áp dung tính chất \(\dfrac{a}{b}>1\Leftrightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\)

\(B=\dfrac{10^{2018}+1}{10^{2017}+1}>\dfrac{10^{2018}+1+9}{10^{2017}+1+9}=\dfrac{10^{2018}+10}{10^{2017}+10}=\dfrac{10\left(10^{2017}+1\right)}{10\left(10^{2016}+1\right)}=\dfrac{10^{2017}+1}{10^{2016}+1}=A\) \(\Leftrightarrow B>A\)

Vậy ......................

Ta có:A=\(\dfrac{-21}{10^{2016}}\)+\(\dfrac{-12}{10^{2017}}\)

= \(\dfrac{-12}{10^{2016}}\)+\(\dfrac{-9}{10^{2016}}\)+\(\dfrac{-12}{10^{2017}}\).

B=\(\dfrac{-12}{10^{2016}}\)+\(\dfrac{-21}{10^{2017}}\)

=\(\dfrac{-12}{10^{2016}}\)+\(\dfrac{-9}{10^{2017}}\)+ \(\dfrac{-12}{10^{2017}}\)

Khi đó để so sánh A và B ta chỉ cần so sánh:\(\dfrac{-9}{10^{2016}}\)và \(\dfrac{-9}{10^{2017}}\)vì A và B cùng có:

\(\dfrac{-12}{10^{2016}}\)+\(\dfrac{-12}{10^{2017}}\).

Do:\(\dfrac{9}{10^{2016}}\)>\(\dfrac{9}{10^{2017}}\).

Suy ra:\(\dfrac{-9}{10^{2016}}\)<\(\dfrac{-9}{10^{2017}}\).

Từ đó ta suy ra được: A< B

TA CÓ:

A = \(-\dfrac{21}{10^{2016}}+-\dfrac{12}{10^{2017}}=\dfrac{-210+-12}{10^{2017}}=-\dfrac{222}{10^{2017}}\)

B = \(\dfrac{-12}{10^{2016}}+\dfrac{-21}{10^{2017}}=\dfrac{-120+-21}{10^{2017}}=\dfrac{-141}{10^{2017}}>\dfrac{-222}{10^{2017}}\)

=> B>A

\(A>B\)

Đúng 100%

Đúng 100%

Đúng 100%

Có \(A=\frac{10^{2017}+1-3}{10^{2017}+1}=1-\frac{3}{10^{2017}+1}\)

\(B=\frac{10^{2017}+3-3}{10^{2017}+3}=1-\frac{3}{10^{2017}+3}\)

Có 10²⁰¹⁷+1<10²⁰¹⁷+3

=> \(\frac{3}{10^{2017}+1}>\frac{3}{10^{2017}+3}\)

=>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

Sửa đề: \(C=1+3^1+3^2+...+3^{100}\)

b) Ta có: \(C=1+3^1+3^2+...+3^{100}\)

\(\Leftrightarrow3\cdot C=3+3^2+...+3^{101}\)

\(\Leftrightarrow C-3\cdot C=1+3+3^2+...+3^{100}-3-3^2-...-3^{100}-3^{101}\)

\(\Leftrightarrow-2\cdot C=1-3^{101}\)

hay \(C=\dfrac{3^{101}-1}{2}\)

b) Ta có: C=1+31+32+...+3100C=1+31+32+...+3100

⇔3⋅C=3+32+...+3101⇔3⋅C=3+32+...+3101

⇔C−3⋅C=1+3+32+...+3100−3−32−...−3100−3101⇔C−3⋅C=1+3+32+...+3100−3−32−...−3100−3101

⇔−2⋅C=1−3101

Bài 1 :

a, Ta có :

\(\dfrac{a}{b}< \dfrac{c}{d}\Leftrightarrow ad< bc\)

\(\Leftrightarrow ad+ab< bc+ab\)

\(\Leftrightarrow a\left(b+d\right)< b\left(a+c\right)\)

\(\Leftrightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}\) \(\left(1\right)\)

Mà \(ad< bc\)

\(\Leftrightarrow ad+cd< bc+cd\)

\(\Leftrightarrow d\left(a+c\right)< c\left(b+d\right)\)

\(\Leftrightarrow\dfrac{a+c}{b+d}< \dfrac{c}{d}\) \(\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\rightarrowđpcm\)

b) \(\dfrac{-1}{3}=\dfrac{-16}{48}< \dfrac{-15}{48};\dfrac{-14}{48};\dfrac{-13}{48}< \dfrac{-12}{48}=\dfrac{-1}{4}\)

Ta thấy :

\(\left\{{}\begin{matrix}A=\dfrac{10^{2017}+1}{10^{2016}+1}>1\\B=\dfrac{10^{2018}+1}{10^{2017}+1}>1\end{matrix}\right.\)

Áp dụng tính chất \(\dfrac{a}{b}>1\Leftrightarrow\dfrac{a+m}{b+m}\) ta có :

\(B=\dfrac{10^{2018}+1}{10^{2017}+1}>\dfrac{10^{2018}+1+9}{10^{2017}+1+9}=\dfrac{10^{2018}+10}{10^{2017}+10}=\dfrac{10\left(10^{2017}+1\right)}{10\left(10^{2016}+1\right)}=\dfrac{10^{2017}+1}{10^{2016}+1}=A\)

\(\Leftrightarrow B>A\)

Bài b :

\(-\dfrac{1}{3}=-\dfrac{24}{72}\) và \(-\dfrac{1}{4}=-\dfrac{18}{72}\)

Ta có :

\(-\dfrac{24}{12}< -\dfrac{23}{12}< -\dfrac{22}{12}< -\dfrac{21}{12}< -\dfrac{20}{12}< -\dfrac{19}{12}< -\dfrac{18}{12}\)

Vậy 5 số cần tìm là ...................