Question

Cho\(\frac{a}{b}\)=\(\frac{c}{d}\).CM\(\frac{4.a-3.b}{4.c-3.d}\)=\(\frac{4.a+3.b}{4.c+3.d}\)
b)\(\frac{a^2+c^2}{b^2+d^2}\)=\(\frac{a^2-c^2}{b^2-d^2}\)

Answer 1

a) Ta có: \(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}=>\frac{4a}{4c}=\frac{3c}{3d}\)

Theo tín chất dãy tỉ số bằng nhau ta có:

\(\frac{4a}{4c}=\frac{3b}{3d}=\frac{4a+3b}{4c+3d}=\frac{4a-3b}{4c-3d}\)(đpcm)

b) Ta có: \(\frac{a}{b}=\frac{c}{d}=>\frac{a^2}{b^2}=\frac{c^2}{d^2}\)

Theo tính chất dãy tỉ số bằng nhau ta có:

\(\frac{a^2}{b^2}=\frac{c^2}{d^2}=>\frac{a^2+c^2}{b^2+d^2}=\frac{a^2-c^2}{b^2-d^2}\)(đpcm)