1. Tìm x, y biết:
a) \(\dfrac{x-1}{5}=\dfrac{y-2}{3}=\dfrac{z-2}{2}\) và x + 2y - z = 6
b) \(\dfrac{x}{y}=\dfrac{2}{3}\) và x2 + y2 = 52
2. Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) . Chứng minh rằng:
a) \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
b) \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
Bài 1:
a) ta có: \(\frac{x-1}{5}=\frac{y-2}{3}=\frac{z-2}{2}=\frac{2y-4}{6}\)
ADTCDTSBN
có: \(\frac{x-1}{5}=\frac{2y-4}{6}=\frac{z-2}{2}=\frac{x-1+2y-4-z+2}{5+6-2}\)\(=\frac{\left(x+2y-z\right)-\left(1+4-2\right)}{9}=\frac{6-3}{9}=\frac{3}{9}=\frac{1}{3}\)
=>...
bn tự tính típ nhé!
b) ta có: \(\frac{x}{y}=\frac{2}{3}\Rightarrow\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x^2}{4}=\frac{y^2}{9}\)
ADTCDTSBN
có: \(\frac{x^2}{4}=\frac{y^2}{9}=\frac{x^2+y^2}{4+9}=\frac{52}{13}=4\)
=>...
Bài 2:
a) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)
\(\Rightarrow\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a+b}{b}=\frac{c+d}{b}\left(đpcm\right)\)
b) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{ac}{bd}\) (*)
mà \(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)
Từ (*) \(\Rightarrow\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\left(đpcm\right)\)