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Thấy \(a+b+c+d=0\Rightarrow\left\{{}\begin{matrix}a=-b-c-d\\b=-a-c-d\\c=-a-b-d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ab-cd=-b^2-bc-bd-cd=\text{-(b + c) (b + d)=(a+d)(b+d)}\\bc-ad=-ca-c^2-cd-ad=\text{-(a + c) (c + d)=(b+d)(c+d)}\\ca-bd=-a^2-ab-ad-bd=\text{-(a + b) (a + d)}=\left(c+d\right)\left(a+d\right)\end{matrix}\right.\)\(\Rightarrow\)x=(a+d)(b+d)(c+d)
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Theo đề, ta có:
\(\left\{{}\begin{matrix}a\ge c+d\\b\ge c+d\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a-c\ge d\ge0\\b-d\ge c\ge0\end{matrix}\right.\)
\(\Rightarrow\left(a-c\right)\left(b-d\right)\ge cd\)
\(\Leftrightarrow ab-bc-ad+cd\ge cd\)
\(\Leftrightarrow\) \(ab\ge ad+bc\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(ad=bc;a,b,c,d>0\)
\(\Rightarrow2\sqrt{ad}=2\sqrt{bc}\)
Khi đó : \(\frac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}}\) \(=\frac{1}{\left(\sqrt{a}+\sqrt{d}\right)+\left(\sqrt{b}+\sqrt{c}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{d}\right)-\left(\sqrt{b}+\sqrt{c}\right)}{\left[\left(\sqrt{a}+\sqrt{d}\right)+\left(\sqrt{b}+\sqrt{c}\right)\right].\left[\left(\sqrt{a}+\sqrt{d}\right)-\left(\sqrt{b}+\sqrt{c}\right)\right]}\)
\(=\frac{\sqrt{a}+\sqrt{d}-\sqrt{b}-\sqrt{c}}{\left(\sqrt{a}+\sqrt{d}\right)^2-\left(\sqrt{b}+\sqrt{c}\right)^2}\) \(=\frac{\sqrt{a}+\sqrt{d}-\sqrt{b}-\sqrt{c}}{a+d+2\sqrt{ad}-b-c-2\sqrt{bc}}\)
\(=\frac{\sqrt{a}+\sqrt{d}-\sqrt{b}-\sqrt{c}}{a+d-b-c}\) ( Do \(2\sqrt{ad}=2\sqrt{bc}\) )
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(a+b+c+d=0\)
\(\Rightarrow b+c=-\left(a+d\right)\)
\(\Rightarrow\left(b+c\right)^2=\left(a+d\right)^2\)
\(\Rightarrow\left(b+c\right)^2-\left(a+d\right)^2=0\)
\(\Rightarrow b^2+c^2+2bc-a^2-d^2-2ad=0\)
Lại có :
\(a^3+b^3+c^3+d^3\)
\(=\left(a+b\right)\left(a^2+d^2-ad\right)+\left(b+c\right)\left(b^2+c^2-bc\right)\)
\(=\left(b+c\right)\left(b^2+c^2-bc\right)-\left(b+c\right)\left(a^2+d^2-ad\right)\)
\(=\left(b+c\right)\left[\left(b^2+c^2-bc\right)\left(a^2+d^2-ad\right)\right]\)
\(=\left(b+c\right)\left[\left(b^2+c^2-bc-a^2-d^2-2ad\right)+3ad-3bc\right]\)
\(=\left(b+c\right)\left[0+3\left(ad-bc\right)\right]\)
\(=3\left(b+c\right)\left(ad-bc\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đề bài cho a,b,c,d khác 1 phải không?
Vì ac –a-c =b2-2b nên ac–a-c +1=b2-2b+1 hay (a-1).(c-1) =(b-1)2
suy ra: (a-1)/(b-1) =(b-1)/(c-1). (1)
Tương tự ta có (b-1).(d-1) =(c-1)2 suy ra: (b-1)/(c-1) =(c-1)/(d-1) (2)
Từ (1) và (2) suy ra: (a-1)/(b-1) = (c-1)/(d-1) = (a+c-2)/(b+d-2)=(a-c)/(b-d)
Suy ra : (a+c-2). (b-d) = (b+d-2).(a-c)
Khai triển, chuyển vế và rút gọn được: 2bc+2a+2d= 2ad +2b+2c
Suy ra: ad +b+c= bc+a+d