Question

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) \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

Answer 1

Bài 1.

a) Nhân 2 vào tỉ số thứ 2 rồi áp dụng tính chất của dãy tỉ số bằng nhau.

Kết quả:

\(\left\{{}\begin{matrix}x=\dfrac{8}{3}\\y=3\\z=\dfrac{8}{3}\end{matrix}\right.\)

b) \(\dfrac{x}{y}=\dfrac{2}{3}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}\)

Theo tính chất dãy tỉ số bằng nhau:

\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{x^2+y^2}{4+9}=\dfrac{52}{13}=4\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=16\\y^2=36\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm4\\y=\pm6\end{matrix}\right.\)

Vậy ...

Bài 2.

a) \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{b}+1=\dfrac{c}{d}+1\Leftrightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)

b) \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{ac}{bd}=\dfrac{c^2}{d^2}\)

\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{ac}{bd}=\dfrac{a^2}{b^2}\)

\(\Leftrightarrow\dfrac{ac}{bd}=\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{a^2+c^2}{b^2+d^2}\)

Vậy ...

Answer 2

2:

b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=i\Rightarrow\left\{{}\begin{matrix}a=bi\\c=di\end{matrix}\right.\)

Ta có:

\(\dfrac{ac}{bd}=\dfrac{c^2i}{d^2i}=\dfrac{c^2}{d^2}=\left(\dfrac{c}{d}\right)^2=i^2\)

\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2i^2+d^2i^2}{b^2+d^2}=\dfrac{i^2\left(b^2+d^2\right)}{b^2+d^2}=i^2\)

Từ đó suy ra \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\) (đpcm)