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\(A=\dfrac{a}{abc+ab+a+1}+\dfrac{ba}{abcd+abc+ab+a}+\dfrac{\dfrac{c}{cd}}{\dfrac{acd}{cd}+\dfrac{cd}{cd}+\dfrac{c}{cd}+\dfrac{1}{cd}}+\dfrac{\dfrac{d}{d}}{\dfrac{dab}{d}+\dfrac{ad}{d}+\dfrac{d}{d}+\dfrac{1}{d}}\)
\(A=\dfrac{a}{abc+ab+a+1}+\dfrac{ab}{1+abc+ab+a}+\dfrac{\dfrac{1}{d}}{a+1+\dfrac{1}{d}+\dfrac{1}{cd}}+\dfrac{1}{ab+a+1+\dfrac{1}{d}}\)
Mà \(abcd=1\Rightarrow\dfrac{1}{d}=abc;\dfrac{1}{cd}=ab\)
\(\Rightarrow A=\dfrac{a}{abc+ab+a+a}+\dfrac{ab}{abc+ab+a+1}+\dfrac{abc}{a+1+abc+ab}+\dfrac{1}{ab+a+1+abc}\)
\(\Rightarrow A=\dfrac{a+ab+abc+1}{abc+ab+a+1}=1\)
a ) \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
Nếu : \(a+b+c=0\) thì đẳng thức trên đúng .
\(\Rightarrowđpcm\)
b ) \(a+b+c+d=0\)
\(\Rightarrow a+b=-\left(c+d\right)\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(c+d\right)\left(cb-cd\right)\left(đpcm\right)\)
Chúc bạn học tốt !!!
a ) a^3+b^3+c^3=3abca3+b3+c3=3abc
\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0⇔(a+b)3+c3−3ab(a+b)−3abc=0
\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0⇔(a+b+c)(a2+b2+c2−ab−bc−ac)=0
Nếu : a+b+c=0a+b+c=0 thì đẳng thức trên đúng .(đpcm)
b ) a+b+c+d=0a+b+c+d=0
\Rightarrow a+b=-\left(c+d\right)\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3⇒a+b=−(c+d)⇔(a+b)3=−(c+d)3
\Leftrightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)⇔a3+b3+c3+d3=−3ab(a+b)−3cd(c+d)
\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)⇔a3+b3+c3+d3=3ab(c+d)−3cd(c+d)
\Leftrightarrow a^3+b^3+c^3+d^3=3\left(c+d\right)\left(cb-cd\right)\left(đpcm\right)⇔a3+b3+c3+d3=3(c+d)(cb−cd)(đpcm)
a) \(a^2+b^2+c^2+d^2=ab+bc+ac+cd.\)
<=>\(2a^2+2b^2+2c^2+2d^2=2ab+2ac+2bc+2cd\)
<=>\(2a^2+2b^2+2c^2+2d^2-2ab-2bc-2ac-2cd=0\)
<=>\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)+\left(d^2-2cd+c^2\right)\)=0
<=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2+\left(d-c\right)^2=0\)
=>a=b=c=d
=> ABCD là hình thoi
Giải:
Ta có:
\(\left(a+b+c+d\right)^2=\) \(\left[\left(a+c\right)+\left(b+d\right)\right]^2\)
\(\ge4\left(a+c\right)\left(b+d\right)\) \(=4\left(ab+bc+cd+da\right)\)\(=4\)
\(\Leftrightarrow a+b+c+d\) \(\ge2\left(a,b,c,d>0\right)\)
\(\Rightarrow\dfrac{a^3}{b+c+d}+\dfrac{b+c+d}{8}\) \(+\dfrac{b}{6}+\dfrac{1}{12}\ge\dfrac{2a}{3}\)
Tương tự ta cũng có:
\(\dfrac{b^3}{a+c+d}+\dfrac{a+c+d}{8}+\dfrac{b}{6}+\dfrac{1}{12}\) \(\ge\dfrac{2b}{3}\)
\(\dfrac{c^3}{a+b+d}+\dfrac{a+b+d}{8}+\dfrac{c}{6}+\dfrac{1}{12}\) \(\ge\dfrac{2c}{3}\)
\(\dfrac{d^3}{a+b+c}+\dfrac{a+b+c}{8}+\dfrac{d}{6}+\dfrac{1}{12}\) \(\ge\dfrac{2d}{3}\)
Cộng vế theo vế các BĐT trên ta có:
\(P\ge\dfrac{a+b+c+d}{3}-\dfrac{1}{3}\ge\) \(\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=d=\dfrac{1}{2}\)
