Bài 2:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=>a=bk; c=dk
\(\frac{a+c}{b+d}=\frac{bk+dk}{b+d}=\frac{k\left(b+d\right)}{b+d}=k\)
\(\frac{a-c}{b-d}=\frac{bk-dk}{b-d}=\frac{k\left(b-d\right)}{b-d}=k\)
Do đó: \(\frac{a+c}{b+d}=\frac{a-c}{b-d}\)
b: \(\frac{a+c}{b+d}=\frac{a-c}{b-d}\)
=>\(\frac{a-c}{a+c}=\frac{b-d}{b+d}\)
c: \(\frac{4a-3b}{a}=\frac{4bk-3b}{bk}=\frac{b\left(4k-3\right)}{bk}=\frac{4k-3}{k}\)
\(\frac{4c-3d}{c}=\frac{4dk-3d}{dk}=\frac{d\left(4k-3\right)}{dk}=\frac{4k-3}{k}\)
Do đó: \(\frac{4a-3b}{a}=\frac{4c-3d}{c}\)
d: \(\frac{7a-11b}{4a+5b}=\frac{7bk-11b}{4bk+5b}=\frac{7k-11}{4k+5}\)
\(\frac{7c-11d}{4c+5d}=\frac{7dk-11d}{4dk+5d}=\frac{d\left(7k-11\right)}{d\left(4k+5\right)}=\frac{7k-11}{4k+5}\)
Do đó: \(\frac{7a-11b}{4a+5b}=\frac{7c-11d}{4c+5d}\)
e: \(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\frac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
\(\frac{ac}{bd}=\frac{bk\cdot dk}{bd}=k^2\)
Do đó: \(\frac{a^2+c^2}{b^2+d^2}=\frac{ac}{bd}\)
Bài 3:
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\frac{x}{y}=\frac{y}{z}=\frac{z}{x}=\frac{x+y+z}{y+z+x}=1\)
=>x=y=z
\(M=\frac{x^{3333}\cdot z^{6666}}{y^{9999}}=\frac{x^{3333}\cdot x^{6666}}{x^{9999}}=1\)
