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bài 1
Ta có : 2016/2017<1
2017/2018<1
Nên 2016/2017=2017/2018
Bài 1 :
a) Ta có : \(\frac{2016}{2017}=1-\frac{1}{2017}\)
\(\frac{2017}{2018}=1-\frac{1}{2018}\)
Vì \(-\frac{1}{2017}< -\frac{1}{2018}\)nên \(\frac{2016}{2017}< \frac{2017}{2018}\)
b) Ta có : \(\frac{2018}{2017}=1+\frac{1}{2017}\)
\(\frac{2017}{2016}=1+\frac{1}{2016}\)
Vì \(\frac{1}{2017}< \frac{1}{2016}\) nên \(\frac{2018}{2017}< \frac{2017}{2016}\)
Câu 2 :
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{101.103}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{101.103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\frac{102}{103}=\frac{51}{103}\)
a) 2017 x 34 + 2017 x 65 + 2017
= 2017 x ( 34+65+1)
= 2017 x 100
= 201700
b) 3/5 x 7/9 - 3/5 x 2/9
= 3/5 x ( 7/9- 2/9)
= 3/5 x 5/9
= 1/3
a) 2017 x 34 + 2017 x 65 + 2017
= 2017 x (34 + 65 + 1)
= 2017 x 100
= 201700
b) \(\dfrac{3}{5}\) x \(\dfrac{7}{9}\) - \(\dfrac{3}{5}\) x \(\dfrac{2}{9}\)
= \(\dfrac{3}{5}\) x \(\left(\dfrac{7}{9}-\dfrac{2}{9}\right)\)
= \(\dfrac{3}{5}\) x \(\dfrac{5}{9}\)
= \(\dfrac{1}{3}\)
\(\frac{2016\times2018+2}{2016\times2017+2018}=\frac{2016\times\left(2017+1\right)+2}{2016\times2017+2018}=\)\(=\frac{2016\times2017+2016+2}{2016\times2017+2018}=\frac{2016\times2017+2018}{2016\times2017+2018}=1\)
\(898-2017=-1119\)
\(2017--1119=3136\)
Kết quả cuối cùng bằng 3136
Chúc bạn học giỏi
\(\frac{2017}{2018}\times2015+\frac{2017}{2018}\times4-\frac{2017}{2018}.\)
\(=\frac{2017}{2018}\times2015+\frac{2017}{2018}\times4-\frac{2017}{2018}\times1\)
\(=\frac{2017}{2018}\times\left(2015+4-1\right)\)
\(=\frac{2017}{2018}\times2018\)
\(=\frac{2017\times2018}{2018\times1}\)
\(=\frac{2017}{1}=2017\)
tìm x
\(\frac{2017}{2017\cdot\left(x\cdot1\right)}\)=21
dấu chấm là dấu nhân nha ghi đầy đủ cách giải
\(\frac{2017}{2017\cdot\left(x\cdot1\right)}=21\)
\(\Leftrightarrow\frac{1}{x}=21\)
\(\Leftrightarrow x=\frac{1}{21}\)
\(\frac{2016}{2017}< \frac{2017}{2018}\)
Đúng 100%
Đúng 100%
Đúng 100%
\(\frac{2018\cdot2016-1210}{2017\cdot2016+806}=\frac{2017\cdot2016+2016-1210}{2017\cdot2016+806}=\frac{2017\cdot2016+806}{2017\cdot2016+806}=1\)
