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.

 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?

1) a( b+c) - b(a-c) = ( a+b) c

VT = a( b+c) - b(a-c) 

= ab + ac - ab + bc

= ac + bc

= c(a + b) (=VP)

2)a (b - c)- a (b+d)= - a (c+d)

VT= a (b - c)- a (b+d)

= ab - ac - ab - ad

= -ac - ad

= -a(c + d) (=VP)

a: Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{ab}{cd}=\frac{bk\cdot b}{dk\cdot d}=\frac{b^2}{d^2}\)

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

Do đó: \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

b: \(\frac{7a-4b}{3a+5b}=\frac{7\cdot bk-4b}{3\cdot bk+5b}=\frac{b\left(7k-4\right)}{b\left(3k+5\right)}=\frac{7k-4}{3k+5}\)

\(\frac{7c-4d}{3c+5d}=\frac{7\cdot dk-4d}{3\cdot dk+5d}=\frac{d\left(7k-4\right)}{d\left(3k+5\right)}=\frac{7k-4}{3k+5}\)

Do đó: \(\frac{7a-4b}{3a+5b}=\frac{7c-4d}{3c+5d}\)

c: \(\frac{ac}{bd}=\frac{bk\cdot dk}{bd}=k^2\)

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

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

Do đó; \(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}=\frac{\left(c-a\right)^2}{\left(d-b\right)^2}\)

d: \(\frac{a^3+b^3}{c^3+d^3}=\frac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\frac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}=\frac{b^3}{d^3}\)

\(\frac{\left(a+b\right)^3}{\left(c+d\right)^3}=\frac{\left(bk+b\right)^3}{\left(dk+d\right)^3}=\frac{b^3\left(k+1\right)^3}{d^3\left(k+1\right)^3}=\frac{b^3}{d^3}\)

Do đó: \(\frac{a^3+b^3}{c^3+d^3}=\frac{\left(a+b\right)^3}{\left(c+d\right)^3}\)

Do đó:

Câu hỏi của Adminbird - Toán lớp 7 - Học toán với OnlineMath

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. 

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