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a/b+c+d =b/c+d+a=c/d+a+b=d/a+b+c
=>a+b+c+d/3(a+b+c+d)=1/3
có thể P=4
Lời giải:
Nếu $a+b+c+d=0$ thì $a+b+c=-d$
Khi đó: $P=\frac{-d}{d}=-1$
Nếu $a+b+c+d\neq 0$ thì áp dụng tính chất dãy tỉ số bằng nhau thì:
$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1$
$\Rightarrow a=b=c=d$
$\Rightarrow P=\frac{d+d+d}{d}=\frac{3d}{d}=3$
Áp dụng t/c dttsbn:
\(\dfrac{a+b+c-2020d}{d}=\dfrac{b+c+d-2020a}{a}=\dfrac{c+d+a-2020b}{b}=\dfrac{d+a+b-2020c}{c}=\dfrac{3\left(a+b+c+d\right)-2020\left(a+b+c+d\right)}{a+b+c+d}=-2017\)
\(\Rightarrow\left\{{}\begin{matrix}a+b+c-2020d=-2017d\\b+c+d-2020a=-2017a\\c+d+a-2020b=-2017b\\d+a+b-2020c=-2017c\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a+b+c=3d\\b+c+d=3a\\c+d+a=3b\\d+a+b=3c\end{matrix}\right.\Rightarrow a=b=c=d\)
\(F=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\\ F=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=4\)
⇒ab+c+d+1=bc+d+a+1=cd+a+b+1=da+b+c+1⇒ab+c+d+1=bc+d+a+1=cd+a+b+1=da+b+c+1
⇒a+b+c+db+c+d=b+c+d+ac+d+a=c+d+a+bd+a+b=d+a+b+ca+b+c⇒a+b+c+db+c+d=b+c+d+ac+d+a=c+d+a+bd+a+b=d+a+b+ca+b+c
Xét a+b+c+d≠0a+b+c+d≠0 suy ra a=b=c=da=b=c=d. Khi đó:
P=a+bc+d+b+cd+a+c+db+a+d+ab+c=1+1+1+1=4P=a+bc+d+b+cd+a+c+db+a+d+ab+c=1+1+1+1=4
Xét a+b+c+d=0a+b+c+d=0 suy ra ⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩a+b=−(c+d)b+c=−(d+a)c+d=−(b+a)d+a=−(b+c){a+b=−(c+d)b+c=−(d+a)c+d=−(b+a)d+a=−(b+c). Khi đó:
P=a+bc+d+b+cd+a+c+db+a+d+ab+c=(−1)+(−1)+(−1)+(−1)=−4