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\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+d}+\dfrac{d}{d+a}=2\)
\(1-\dfrac{a}{a+b}-\dfrac{b}{b+c}+1-\dfrac{c}{c+d}-\dfrac{d}{d+a}=0\)
\(\dfrac{b}{a+b}-\dfrac{b}{b+c}+\dfrac{d}{c+d}-\dfrac{d}{d+a}=0\)
\(\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)
<=>b(c+d)(d+a)+d(a+b)(b+c)=0 (vì c≠a)
<=>abc-acd+bd2-b2d=0
<=> (b-d)(ac-bd)=0 <=> ac - bd =0 (vì b≠d) <=> ac = bd
Vậy abcd =(ac)(bd)=(ac)2
\(\hept{\begin{cases}a^2+b^4+c^6+d^8=1\\a^{2016}+b^{2017}+c^{2018}+d^{2019}=1\end{cases}}\)
=> \(0\le a^2;b^4;c^6;d^8\le1\)
=> \(-1\le a;b;c;d\le1\)
=> \(a^{2016}\le a^2\); \(b^{2017}\le b^4\); \(c^{2018}\le c^6\); \(d^8\le d^{2019}\)
=> \(a^{2016}+b^{2017}+c^{2018}+d^{2019}\le a^2+b^4+c^6+d^8\)
Do đó: \(a^{2016}+b^{2017}+c^{2018}+d^{2019}=a^2+b^4+c^6+d^8=1\)
<=> \(a^{2016}=a^2;b^{2017}=b^4;c^{2018}=c^6;d^{2019}=d^8;a^2+b^4+c^6+d^8=1\)
<=> \(\orbr{\begin{cases}a=0\\a=\pm1\end{cases}}\); \(\orbr{\begin{cases}b=0\\b=1\end{cases}}\); \(\orbr{\begin{cases}c=0\\c=\pm1\end{cases}}\); \(\orbr{\begin{cases}d=0\\d=1\end{cases}}\); \(a^2+b^4+c^6+d^8=1\)
<=> \(a=b=c=0;d=1\)hoặc \(a=b=d;c=\pm1\) hoặc \(a=c=d=0;b=1\)hoặc \(b=c=d=0;a=\pm1\).
Tại sao \(0\le a^2;b^4;c^6;d^8\le1\) Lại suy ra \(-1\le a;b;c;d\le1\)????????????????????????
a+b+c+d=0 => a+d= -b-c; (a+b)3=a3+b3+3ab(a+b) => a3+b3=(a+b)3-3ab(a+b)
a3+d3+b3+d3
=(a+d)3- 3ad(a+d)+ (b+c)3-3bc(b+c) (1)
Do a+d=-b-c nên pt (1) trở thành:
-(b+c)3-3ad(-b-c)+ (b+c)3-3bc(b+c)
=3ad(b+c)-3bc(b+c)
=3(b+c)(ad-bc) <đccm>
a) a + b + c + d = 0 \(\Rightarrow a+c=-\left(b+d\right)\)
\(\Rightarrow\)\(\left(a+c\right)^3=-\left(b+d\right)^3\)
\(\Rightarrow\)\(a^3+c^3+3ac\left(a+c\right)=-b^3-d^3-3b\left(b+d\right)\)
\(\Rightarrow\)\(a^3+b^3+c^3+d^3=3ac\left(b+d\right)-3bd\left(b+d\right)\)
\(=3\left(ac-bd\right)\left(b+d\right)\)\(\left(dpcm\right)\)
b) - \(\sqrt{a-b+c}=\sqrt{a}-\sqrt{b}+\sqrt{c}\)
\(\Leftrightarrow\left(\sqrt{a-b+c}+\sqrt{b}\right)^2=\left(\sqrt{a}+\sqrt{c}\right)^2\)
\(\Leftrightarrow b\left(a-b+c\right)=ac\Leftrightarrow\left(b-c\right)\left(a-b\right)=0\Leftrightarrow\orbr{\begin{cases}a=b\\b=c\end{cases}\left(1\right)}\)
- Gia su \(a\le b\le c\), ta có: \(1=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{3}{a}\)
\(\Rightarrow a\le3\Rightarrow a=1,2,3\)
+ Nếu a = 1 thì: \(\frac{1}{b}+\frac{1}{c}=0\left(vl\right)\)
+ Nếu a = 2 thì: \(\frac{1}{b}+\frac{1}{c}=\frac{1}{2}\le\frac{2}{b}\Rightarrow b\le4\)
\(\Rightarrow a=2;b=c=4\)
+ Nếu a = 3 thì: \(\frac{1}{b}+\frac{1}{c}=\frac{2}{3}\le\frac{2}{b}\Rightarrow b\le3\)
\(\Rightarrow a=b=c=3\)
Cac cap (a, b, c) thoa \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)la:
\(\left(2,4,4\right);\left(4,2,4\right);\left(4,4,2\right);\left(3,3,3\right)\)
Kết hợp với \(\left(1\right)\)ta có nghiệm: \(\left(2,4,4\right);\left(4,4,2\right);\left(3,3,3\right)\)