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\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow\frac{a}{c}=\frac{b}{d}=\frac{2017a}{2017c}=\frac{2018b}{2018d}=\frac{2018a}{2018c}=\frac{2019b}{2019d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{2017a}{2017c}=\frac{2018b}{2018d}=\frac{2018a}{2018c}=\frac{2019b}{2019d}=\frac{2017a-2018b}{2017c-2018d}=\frac{2018a+2019b}{2018c+2019d}\)
<=>\(\left(2017a-2018b\right)\left(2018c+2019d\right)=\left(2018a+2019b\right)\left(2017c-2018d\right)\)
<=>\(\frac{2017a-2018b}{2018a+2019b}=\frac{2017c-2017d}{2018x+2019d}\)(đpcm)
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 !
Theo tính chất tỉ dãy số bằng nhau thì:
\(\frac{a+b+c-d}{d}=\frac{b+c+d-a}{a}=\frac{c+d+a-b}{b}=\frac{d+a+b-c}{c}=1\)
\(\Leftrightarrow\frac{a+b}{c+d}=\frac{b+c}{d+a}=\frac{c+d}{a+b}=\frac{d+a}{b+c}=1\)
\(\Rightarrow M\Leftrightarrow1+1+1+1=4\)
Ps: Cách mình nhanh hơn nè!
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\)
\(=\frac{a+b+2c+d+a+b+c+2d}{c+d}=\frac{2\left(a+b\right)}{c+d}+3=\)
Tương tự
\(=\frac{2\left(b+c\right)}{d+a}+3=\)
\(=\frac{2\left(c+d\right)}{a+b}+3=\)
\(=\frac{2\left(d+a\right)}{b+c}+3\)
\(\Rightarrow\frac{2\left(a+b\right)}{c+d}+3=\frac{2\left(b+c\right)}{d+a}+3=\frac{2\left(c+d\right)}{a+b}+3=\frac{2\left(d+a\right)}{b+c}+3\)
\(\Rightarrow\frac{2\left(a+b\right)}{c+d}=\frac{2\left(b+c\right)}{d+a}=\frac{2\left(c+d\right)}{a+b}=\frac{2\left(d+a\right)}{b+c}=\)
\(=\frac{2\left(a+b\right)+2\left(b+c\right)+2\left(c+d\right)+2\left(d+a\right)}{c+d+d+a+a+b+b+c}=\frac{4\left(a+b+c+d\right)}{2\left(a+b+c+d\right)}=2\)
\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)
<br class="Apple-interchange-newline"><div id="inner-editor"></div>2a+b+c+da =a+2b+c+db =a+b+2c+dc =a+b+c+2dd =2a+b+c+d+a+2b+c+d+a+b+2c+d+a+b+c+2ca+b+c+d =4
=>2a+b+c+d=4a
=>2a=b+c+d
Tương tự ta có:2b=a+c+d
2c=a+b+d
2d=a+b+c
=>2a+2b=b+c+d+a+c+d=>a+b+2c+2d
=>a+b=2c+2d
=>a+b/c+d=2
Tương tự ta có:b+c/d+a=2
c+d/a+b=2
d+a/b+c=2
=>M=2+2+2+2=8
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