Bài 1:

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

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

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

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

Bài 2:

a: Đặt \(\frac{a}{b}=\frac{b}{c}=k\)

=>b=ck; a=bk=ck^2

\(\frac{a^2+b^2}{b^2+c^2}=\frac{\left(ck^2\right)^2+\left(ck\right)^2}{\left(ck\right)^2+c^2}=\frac{c^2k^2\left(k^2+1\right)}{c^2\left(k^2+1\right)}=k^2\)

\(\frac{a}{c}=\frac{ck^2}{c}=k^2\)

Do đó: \(\frac{a}{c}=\frac{a^2+b^2}{b^2+c^2}\)

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

=>\(\begin{cases}c=dk\\ b=ck=dk\cdot k=dk^2\\ a=bk=dk^2\cdot k=dk^3\end{cases}\)

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

\(=\left\lbrack\frac{dk\left(k^2+k-1\right)}{d\left(k^2+k-1\right)}\right\rbrack^3=k^3\)

\(\frac{a}{d}=\frac{dk^3}{d}=k^3\)

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

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

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=6\left(a^2+b^2+c^2\right)-6\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)

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

\(\Leftrightarrow a=b=c\).

ta có (a+b+c)^2= a^2+b^2+c^2+2ab+2bc+2ac

nếu (a+b+c)^2=3(a^2+b^2+c^2)

=> 3a^2+3b^2+3c^2=a^2+b^2+c^2 +2ab+2bc+2ac

=> 2a^2+2b^2+2c^2=2ab+2bc+2ac

=>a^2+b^2+c^2=ab+bc+ac

=>a.a+b.b+c.c=ab+bc+ac

=>a=b=c

=> đpcm

bạn lớp 7 mà học kém quá nhỉ

dễ ot

b,c=1