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 ?

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

Lời giải:

Đặt $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=t$

$t^3=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{a}{d}(1)$

Áp dụng tính chất dãy tỉ số bằng nhau:

$t^3=\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}(2)$

Từ $(1);(2)$ ta có đpcm.

1: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=b\cdot k;c=d\cdot k\)

\(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\)

\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\)

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

2: \(\dfrac{2a+b}{a-2b}=\dfrac{2\cdot bk+b}{bk-2b}=\dfrac{b\left(2k+1\right)}{b\left(k-2\right)}=\dfrac{2k+1}{k-2}\)

\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{d\left(2k+1\right)}{d\left(k-2\right)}=\dfrac{2k+1}{k-2}\)

Do đó: \(\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}\)

3: \(\dfrac{a+b}{a-b}=\dfrac{bk+b}{bk-b}=\dfrac{b\left(k+1\right)}{b\cdot\left(k-1\right)}=\dfrac{k+1}{k-1}\)

\(\dfrac{c+d}{c-d}=\dfrac{dk+d}{dk-d}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\)

Do đó: \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)

4: \(\dfrac{5a+3b}{5c+3d}=\dfrac{5\cdot bk+3b}{5dk+3d}=\dfrac{b\left(5k+3\right)}{d\left(5k+3\right)}=\dfrac{b}{d}\)

\(\dfrac{5a-3b}{5c-3d}=\dfrac{5\cdot bk-3b}{5\cdot dk-3d}=\dfrac{b\left(5k-3\right)}{d\left(5k-3\right)}=\dfrac{b}{d}\)

Do đó: \(\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)

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

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

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

=>\(\frac{a}{d}=\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{a}{d}=\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)

=>\(\frac{a}{d}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)

=>ĐPCM

Trl nhanh cho mik vs ạ!! Mik đag cần gấp!!:33

a) Đề phải là \(\frac{c}{a-c}=\frac{d}{b-d}\) chứ.

Ta có: \(\frac{a}{b}=\frac{c}{d}.\)

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

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

\(\Rightarrow\frac{b}{d}-1=\frac{a}{c}-1\)

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

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

\(\Rightarrow\frac{d}{b-d}=\frac{c}{a-c}\left(đpcm1\right).\)

c) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)

\(\Rightarrow\frac{2a}{2b}=\frac{3c}{3d}.\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

\(\frac{2a}{2b}=\frac{3c}{3d}=\frac{2a+3c}{2b+3d}\) (1)

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

Từ (1) và (2) \(\Rightarrow\frac{2a+3c}{2b+3d}=\frac{2a-3c}{2b-3d}\left(đpcm\right).\)

Chúc bạn học tốt!