Những câu hỏi liên quan
Cho
\(\dfrac{2016c-2017b}{2015}\)=\(\dfrac{2017a-2015c}{2016}\)=\(\dfrac{2015b-2016a}{2017}\).
Chứng minh \(\dfrac{a}{2015}\)=\(\dfrac{b}{2016}\)=\(\dfrac{c}{2017}\)
Lời giải:
Ta có \(\frac{2016c-2017b}{2015}=\frac{2017a-2015c}{2016}=\frac{2015b-2016a}{2017}\)
\(\Rightarrow \frac{2015.2016c-2015.2017b}{2015^2}=\frac{2016.2017a-2016.2015c}{2016^2}=\frac{2017.2015b-2017.2016a}{2017^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\( \frac{2015.2016c-2015.2017b}{2015^2}=\frac{2016.2017a-2016.2015c}{2016^2}=\frac{2017.2015b-2017.2016a}{2017^2}\)
\(=\frac{2015.2016c-2015.2017b+2016.2017a-2016.2015c+2017.2015b-2017.2016a}{2015^2+2016^2+2017^2}=0\)
\(\Rightarrow \left\{\begin{matrix} 2015.2016c-2015.2017b=0\\ 2016.2017a-2016.2015c=0\\ 2017.2015b-2016.2016a=0\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} 2016c=2017b\\ 2017a=2015c\\ 2015b=2016a\end{matrix}\right.\Rightarrow \frac{a}{2015}=\frac{b}{2016}=\frac{c}{2017}\)
Ta có đpcm.
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Cho \(\frac{a}{b}\)= \(\frac{c}{d}\). CMR : \(\frac{2015\text{a}-2016b}{2016c+2017\text{d}}\)= \(\frac{2015c-2016\text{d}}{2016\text{d}+2017\text{a}}\)
Cho A= \(\frac{1}{2015}+\frac{2}{2016}+\frac{3}{2017}+...................+\frac{2016}{4030}-2016\) và B= \(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+...................+\frac{1}{4030}\) . Chứng minh rằng \(\frac{A}{B}\) là một số nguyên
Cho A= \(\frac{1}{2015}+\frac{2}{2016}+\frac{3}{2017}+...................+\frac{2016}{4030}-2016\) và B= \(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+...................+\frac{1}{4030}\) . Chứng minh rằng \(\frac{A}{B}\) là một số nguyên
Cho: \(A=\frac{1}{2015}+\frac{2}{2016}+\frac{3}{2017}+..............+\frac{2016}{4030}-2016\)
và \(B=\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+.............+\frac{1}{4030}\)
Chứng minh rằng: \(\frac{A}{B}\) là một số nguyên
\(\left(\frac{1}{2}+\frac{2015}{2016}+\frac{2016}{2017}+1\right)\left(\frac{2105}{2016}+\frac{2016}{2017}+\frac{7}{22}\right)-\left(\frac{1}{2}+\frac{2015}{2016}+\frac{2016}{2017}\right)\left(\frac{2015}{2016}+\frac{2016}{2017}+\frac{7}{22}+1\right)\)
cho A=\(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)+\(\frac{2017}{2018}\)+\(\frac{2018}{2015}\) chứng minh A>4
So sánh hai phân số : A=\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)và B=\(\frac{2015+2016+2017}{2016+2017+2018}\)
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)
\(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Ta có:
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
Cộng vế theo vế, ta có:
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(hay\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
Vậy A > B
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Bạn có nhầm không, tớ thấy cả hai đều giống nhau mà,
Hai cái bằng nhau
Bài 1 cho \(\frac{a}{b}=\frac{c}{d}\)chứng minh
d) \(\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
e) \(\frac{2016a-2017b}{2017c+2018d}=\frac{2016c-2017d}{2017a+2018b}\)