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Ta có:
a/b = c/d => 2018a/2018b = 2018c/2018d = 2018a - 2018c / 2018b- 2018d
a/b = c/d => 2017a/2017b = 2017c/2017d =2017a+ 2017c/ 2017b+ 2017d
=> 2018a-2018c/2018b-2018d = 2017a+2017c/2017b+2017d (=a/b=c/d)
Sửa đề: \(\dfrac{2018a-2019b}{2019a+2020b}=\dfrac{2018c-2019d}{2019c+2020d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2020a}{2020b}=\dfrac{2020c}{2020d}=\dfrac{2019a}{2019c}=\dfrac{2019b}{2019d}=\dfrac{2018a}{2018c}=\dfrac{2018b}{2018d}=\dfrac{2018a-2019b}{2018c-2019d}=\dfrac{2019a+2020b}{2019c+2020d}\\ \Leftrightarrow\dfrac{2018a-2019b}{2019a+2020b}=\dfrac{2018c-2019d}{2019c+2020d}\)
2.
\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2a+2b+2c+2d}=\frac{a+b+c+d}{2\left(a+b+c+d\right)}=\frac{1}{2}\)
\(\Rightarrow a=\frac{2b}{2}=b;b=\frac{2c}{2}=c;c=\frac{2d}{2}=d;d=\frac{2a}{2}=a\)
\(\Rightarrow a=b=c=d\)
Ta có : \(A=\frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}+\frac{2011d-2010a}{b+c}\)
\(=\frac{2011a-2010a}{2a}+\frac{2011a-2010a}{2a}+\frac{2011a-2010a}{2a}+\frac{2011a-2010a}{2a}\)
\(=\frac{4a}{2a}=2\)
3.
\(\left(x-1\right)\left(x-3\right)< 0\)
\(\Rightarrow\hept{\begin{cases}x-1< 0\\x-3>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-1>0\\x-3< 0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x< 1\\x>3\end{cases}}\)( loại ) hoặc \(\hept{\begin{cases}x>1\\x< 3\end{cases}}\)
Vậy \(1< x< 3\)
Đặt \(A=\frac{1}{4\times9}+\frac{1}{9\times14}+\frac{1}{14\times19}+...+\frac{1}{44\times49}\)
Ta có : \(5\times A=\frac{5}{4\times9}+\frac{5}{9\times14}+\frac{5}{14\times19}+...+\frac{5}{44\times49}=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}=\frac{1}{4}-\frac{1}{49}\)
\(=\frac{49}{196}-\frac{4}{196}=\frac{45}{196}\)
\(\Rightarrow A=\frac{9}{196}\)
Đặt \(B=1-3-5-7-...-49=1-\left(3+5+...+49\right)\)
Đặt \(C=3+5+...+49\) ( khoảng cách là 2 )
Số số hạng là : \(\left(49-3\right):2+1=24\)
Tổng C là : \(\left(49+3\right)\times24:2=624\)
\(\Rightarrow B=1-264=-623\)
Vậy \(A=\frac{9}{196}\times\frac{-623}{89}=\frac{-9}{28}\)
Dòng cuối cùng mình không chắc là đúng nhé !
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{a+b+c+d}=1\left(\text{ vì a+b+c+d khác 0}\right)\)
\(\Rightarrow a=b=c=d\)
\(M=\frac{2a-b}{c+b}+\frac{2b-c}{a+d}+\frac{2c-d}{a+b}+\frac{2d-a}{b+c}=\frac{2a-a}{a+a}+\frac{2b-b}{b+b}+\frac{2c-c}{c+c}+\frac{2d-d}{d+d}=\frac{1}{2}.4=2\)
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