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Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{2a+b}{2a-b}=\dfrac{2bk+b}{2bk-b}=\dfrac{2k+1}{2k-1}\)
\(\dfrac{2c+d}{2c-d}=\dfrac{2dk+d}{2dk-d}=\dfrac{2k+1}{2k-1}\)
=>\(\dfrac{2a+b}{2a-b}=\dfrac{2c+d}{2c-d}\)
b: \(\dfrac{2a+b}{a-2b}=\dfrac{2bk+b}{bk-2b}=\dfrac{2k+1}{k-2}\)
\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{2k+1}{k-2}\)
=>\(\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}\)
Từ \(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\Rightarrow\dfrac{1}{2}.\dfrac{a}{b}=\dfrac{1}{2}.\dfrac{b}{c}=\dfrac{1}{2}.\dfrac{c}{d}=\dfrac{1}{2}.\dfrac{d}{a}\)
⇒ \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=\dfrac{a+b+c+d}{b+c+d+a}=1\)
⇒ \(a=b=c=d\)
Thay b = a ; c = a ; d = a vào biểu thức A ta có:
\(A=\dfrac{2011a-2010a}{2a}+\dfrac{2011a-2010a}{2a}+\dfrac{2011a-2010a}{2a}+\dfrac{2011a-2010a}{2a}\)
\(A=\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}\)
\(A=\dfrac{1}{2}.4=2\)
Vậy A = 2
\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}=\dfrac{a+b+c+d}{2a+2b+2c+2d}=\dfrac{1}{2}\)
=>\(\dfrac{a}{2b}=\dfrac{1}{2}\)=>2a=2b =>a=b
\(\dfrac{b}{2c}=\dfrac{1}{2}\)=>2b=2c =>b=c
\(\dfrac{c}{2d}=\dfrac{1}{2}\)=>2c=2d =>c=d
\(\dfrac{d}{2a}=\dfrac{1}{2}\)=>2d=2a =>d=a
=>a=b=c=d.
*\(\dfrac{2011a-2010b}{c+d}+\dfrac{2011b-2010c}{a+d}+\dfrac{2011c-2010d}{a+b}+\dfrac{2011d-2010a}{b+c}\)
=\(\dfrac{2011a-2010a}{a+a}+\dfrac{2011a-2010a}{a+a}+\dfrac{2011a-2010d}{a+a}+\dfrac{2011a-2010a}{a+a}\)
=\(\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}\)=2
b.\(ĐK:x;y\in Z^+;x;y\ne0\)
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{5}{x}+\dfrac{5}{y}=1\)
\(\Leftrightarrow\dfrac{5}{x}=1-\dfrac{5}{y}\)
\(\Leftrightarrow\dfrac{5}{x}=\dfrac{y-5}{y}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{y-5}\)
\(\Leftrightarrow x=\dfrac{5y}{y-5}\)
\(\Leftrightarrow x=5+\dfrac{25}{y-5}\) ( bạn chia \(5y\) cho \(y-5\) ý )
Để x;y là số nguyên dương thì \(25⋮y-5\) hay \(y-5\in U\left(25\right)=\left\{\pm1;\pm5;\pm25\right\}\)
TH1:
\(y-5=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=30\end{matrix}\right.\) ( tm ) ( bạn thế y=6 vào \(x=5+\dfrac{25}{y+5}\) nhé )
Xét tương tự, ta ra được nghiệm nguyên dương của phương trình:
\(\left\{{}\begin{matrix}x=30\\y=6\end{matrix}\right.\) \(\left\{{}\begin{matrix}x=10\\y=10\end{matrix}\right.\) \(\left\{{}\begin{matrix}x=6\\y=30\end{matrix}\right.\)
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\(TH1:a+b+c+d\ne0\)
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)
\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)
\(\Rightarrow a=b=c=d\)
\(M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\)
\(=1+1+1+1\)
\(=4\)
\(TH2:a+b+c+d=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)
\(M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\)
\(=-\dfrac{c+d}{c+d}-\dfrac{d+a}{d+a}-\dfrac{a+b}{a+b}-\dfrac{b+c}{b+c}\)
\(=-1-1-1-1\)
\(=-4\)
TH1: \(a+b+c+d\ne0\)
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)
\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\Rightarrow a=b=c=d\)
\(P=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
\(\Rightarrow P=1+1+1+1\)
\(\Rightarrow P=4\)
TH2: \(a+b+c+d=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)
\(P=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
\(\Rightarrow P=\dfrac{-\left(c+d\right)}{c+d}+\dfrac{-\left(d+a\right)}{d+a}+\dfrac{-\left(a+b\right)}{a+b}+\dfrac{-\left(b+c\right)}{b+c}\)
\(\Rightarrow P=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(\Rightarrow P=-4\)
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Từ \(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\Rightarrow\dfrac{1}{2}\cdot\dfrac{a}{b}=\dfrac{1}{2}\cdot\dfrac{b}{c}=\dfrac{1}{2}\cdot\dfrac{c}{d}=\dfrac{1}{2}\cdot\dfrac{d}{a}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=\dfrac{a+b+c+d}{b+c+d+a}=1\)
\(\Rightarrow a=b=c=d\)
Thay \(b=a;c=a;d=a\) vào biểu thức A ta có;
\(A=\dfrac{2011a-2010a}{2a}+\)\(\dfrac{2011a-2010a}{2a}+\)\(\dfrac{2011a-2010a}{2a}+\)\(\dfrac{2011a-2010a}{2a}\)
\(A=\)\(\dfrac{a}{2a}+\)\(\dfrac{a}{2a}+\)\(\dfrac{a}{2a}+\)\(\dfrac{a}{2a}\)
\(A=\dfrac{1}{2}\cdot4=2\)
Vậy \(A=2\)
Ta có:
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
⇔ \(\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1\)
\(=\dfrac{a+b+c+2d}{d}-1\)
⇔ \(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)
Nếu a+b+c+d=0
⇒a+b=−(c+d);c+b=−(a+d);c+d=−(a+b);a+d=−(c+b)
Thay vào M, ta có:
\(M=\dfrac{a+b}{-\left(a+b\right)}=\dfrac{b+c}{-\left(b+c\right)}=\dfrac{c+d}{-\left(c+d\right)}=\dfrac{a+d}{-\left(a+d\right)}=-1\)
Nếu a+b+c+d ≠0
⇒ \(a=b=c=d\)
Thay vào M, ta có
\(M=\dfrac{a+b}{a+b}=\dfrac{b+c}{b+c}=\dfrac{c+d}{c+d}=\dfrac{d+a}{d+a}=1\)
Vì \(a+b+c+d\ne0\) nên áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=\dfrac{a+b+c+d}{b+c+d+a}=1\)
\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=d\\d=a\end{matrix}\right.\) \(\Rightarrow a=b=c=d\) (1)
Thay (1) vào P, ta có:
\(P=\dfrac{2a-a}{a+a}+\dfrac{2a-a}{a+a}+\dfrac{2a-a}{a+a}=\dfrac{2a-a}{a+a}\)
\(\Rightarrow P=\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=2\)
Vậy P = 2
Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=k\)
\(\Rightarrow\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}.\dfrac{d}{a}=k^4\)
\(\Rightarrow k=\pm1\)
- Với \(k=1\) :
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}\)
\(\Rightarrow a=b=c=d\)
- Với \(k=-1\) :
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=-1\)
\(\Rightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-d\\d=-a\end{matrix}\right.\)
\(\Rightarrow a=-b=c=-d\)
\(\Rightarrow P=\dfrac{2a+a}{2a+a}+\dfrac{-2a-a}{-2a-a}+\dfrac{2a+a}{2a+a}+\dfrac{-2a-a}{-2a-a}\)
\(\Rightarrow P=4\)