Ta có : \(a+b+c+d=0\)

\(\Leftrightarrow a+b=-c-d\)

\(\Leftrightarrow\left(a+b\right)^3=\left(-c-d\right)^3\)

\(\Leftrightarrow a^3+b^3+3ab.\left(a+b\right)=-c^3-d^3+3cd.\left(c+d\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3cd.\left(c+d\right)-3ab.\left(a+b\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3.cd.\left(a+b\right)+3ab.\left(c+d\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3.\left(c+d\right)\left(cd+ab\right)\)

Ta có : 

⇔a3+b3+c3+d3=3.cd.(a+b)+3ab.(c+d)

 a+b+c+d=0 
=>a+b=-(c+d) 
=> (a+b)^3=-(c+d)^3 
=> a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d) 
=> a^3+b^3+c^3+d^3=-3ab(a+b)-3cd(c+d) 
=> a^3+b^3+c^3+d^3=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d)) 
==> a^3 +b^^3+c^3+d^3==3(c+d)(ab-cd) (đpcm)

hey you, còn câu b,c?

ở đây có ai thích sơn tùng không ?

\(\dfrac{a^3}{b}+ab+\dfrac{b^3}{c}+bc+\dfrac{c^3}{a}+ca\ge2\sqrt{\dfrac{a^4b}{b}}+2\sqrt{\dfrac{b^4c}{c}}+2\sqrt{\dfrac{c^4a}{a}}=2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ca\)

áp dụng AM GM ta có a^3/b+ab>=2a^2

chứng minh tương tự => a^3/b+b^3/c+c^3/a>=2(a^2+b^2+c^2)-(ab+bc+ca)

mà ta có a^2+b^2+c^2>=(ab+bc+ca)

=>a^3/b+b^3/c+c^3/a>= ab+bc+ca

"=" xảy ra khi a=b=c

Câu hỏi của ✰✰ βєsէ ℱƐƝƝIƘ ✰✰ - Toán lớp 8 - Học toán với OnlineMath

Ta co: \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)

=> \(\frac{a}{c}.\frac{b}{d}=\frac{a-b}{c-d}.\frac{b}{d}=\frac{a-b}{c-d}.\frac{a-b}{c-d}\)

=>. \(\frac{ab}{cd}=\left(\frac{a-b}{c-d}\right)^2\)

Ta co: \(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)

          \(\Rightarrow\frac{\left(a+c\right)^3}{\left(b+d\right)^3}=\frac{a^3}{b^3}=\frac{c^3}{d^3}=\frac{a^3-c^3}{b^3-d^3}\)

\(a+b+c+d=0\Rightarrow a+b=-\left(c+d\right)\)

\(\Rightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)

\(\Rightarrow\left(a+b\right)^3+\left(c+d\right)^3=0\)

\(\Rightarrow a^3+b^3+3ab\left(a+b\right)+c^3+d^3+3cd\left(c+d\right)=0\)

\(\Rightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)\)

\(\Rightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\) (do \(a+b=-\left(c+d\right)\)

\(\Rightarrow a^3+b^3+c^3+d^3=3\left(ab-cd\right)\left(c+d\right)\)

\(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=bk;c=dk\)

\(\Rightarrow\frac{a-b}{c-d}=\frac{bk-b}{dk-d}=\frac{b\left(k-1\right)}{d\left(k-1\right)}=\frac{b}{d}\Rightarrow\left(\frac{a-b}{c-d}\right)^2=\frac{b^2}{d^2}\)

=> Sai đề.