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 ?

a) \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)

<=> \(a^2-2a+1+b^2-2b+1+c^2-2c+1=0\)

<=> \(\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)

Tổng 3 số không âm bằng 0 <=> a=b=c=1

b) \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2ac+2bc=3ab+3ac+3bc\)

<=> \(a^2-ab+b^2-bc+c^2-ac=0\)

<=> \(2a^2-2ab+2b^2-2bc+2c^2-2ac=0\)

<=> \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Tổng 3 số không âm bằng 0 <=> a=b=c

#NguyễnHoàngTiến ơi cảm ơn bạn đã giúp mình nhưng cho mình hỏi left với right trong bài của bạn có nghĩa là gì vậy hả, mình không hiểu lắm.

Do a/b=c/d  ⇔ ad=bc

1) Ta có: (a+c)b=ab+bc

               (b+d)a=ab+ad

Do bc=ad nên ab+ad=ab+bc

Suy ra (a+c)b=(b+d)a   (đpcm)

2) Ta có: (b+d)c=bc+dc

               (a+c)d=ad+cd

Do bc=ad nên bc+dc=ad+cd

Suy ra (b+d)c=(b+d)c   (đpcm)

3)Ta có:(a+b)(c-d)=ac-ad+bc-bd=(ac-bd)-(ad-bc)

             (a-b)(c+d)=ac+ad-bc-bd=(ac-bd)+(ad-bc)

Do ad=bc  ⇔ ad-bc=0 nên (ac-bd)-(ad-bc)=(ac-bd)+(ad-bc)

⇔(a+b)(c-d)= (a-b)(c+d) (đpcm)

A) \(\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt,c=dt\)

\(\frac{a}{a+b}=\frac{bt}{bt+b}=\frac{t}{t+1},\frac{c}{c+d}=\frac{dt}{dt+d}=\frac{t}{t+1}\)

suy ra đpcm. 

\(\frac{a-b}{c-d}=\frac{bt-b}{dt-d}=\frac{b}{d},\frac{a+b}{c+d}=\frac{bt+b}{dt+d}=\frac{b}{d}\)

suy ra đpcm. 

B) \(\frac{a+3c}{b+3d}=\frac{a+c}{b+d}=\frac{\left(a+3c\right)-\left(a+c\right)}{\left(b+3d\right)-\left(b+d\right)}=\frac{2c}{2d}=\frac{c}{d}\)

\(\frac{a+3c}{b+3d}=\frac{a+c}{b+d}=\frac{\left(a+3c\right)-3\left(a+c\right)}{\left(b+3d\right)-3\left(b+d\right)}=\frac{-2a}{-2b}=\frac{a}{b}\)

suy ra đpcm. 

\(\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\) ; \(\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{b}.\dfrac{a}{b}.\dfrac{a}{b}=\dfrac{a^3}{b^3}\)

 \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)

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

CÁC CẬU ƠI GIÚP MIK VS!!!!!!

Lê Minh Tuấn bn tham khảo nha:

 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) (dpcm)

cảm ơn OoO Ledegill2 OoO

Ta có :

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)

Áp dụng t/c dãy tỉ số bawg nhau ta có :

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)

\(\Leftrightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{b}{c}\right)^3=\left(\dfrac{c}{d}\right)^3=\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{b}.\dfrac{a}{b}.\dfrac{a}{b}=\dfrac{b}{c}.\dfrac{b}{c}.\dfrac{b}{c}=\dfrac{c}{d}.\dfrac{c}{d}.\dfrac{c}{d}=\dfrac{a}{d}\)

\(\Leftrightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\left(đpcm\right)\)

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