Question

[Trần Thanh Tâm]

Bài 1: Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh

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

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

c) \(\dfrac{a-c}{a}=\dfrac{b-d}{b}\)

d) \(\dfrac{3a+5b}{2a-7b}=\dfrac{3c+5d}{2c-7d}\)

e) \(\dfrac{\left(a+b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}\)

f) \(\left(\dfrac{a-b}{c-d}\right)^{2012}=\dfrac{a^{2012}+b^{2012}}{c^{2012}+d^{2012}}\)

Bài 2: Tìm x, biết

a) \(\dfrac{3}{x-4}=\dfrac{x+4}{3}\)

b) \(\dfrac{x+2}{2}=\dfrac{1}{1-x}\)

c) \(\dfrac{x+7}{x+4}=\dfrac{x-1}{x-2}\)

Bài 3: Cho tỉ lệ thức \(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)

Tìm giá trị của tỉ số \(\dfrac{x}{y}\)

Answer 1

[Nguyễn Huy Tú]

Bài 1:

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

a, Ta có: \(\dfrac{a+c}{c}=\dfrac{bk+dk}{dk}=\dfrac{\left(b+d\right)k}{dk}=\dfrac{b+d}{d}\)

\(\Rightarrowđpcm\)

b, Ta có: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\) (1)

\(\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=\dfrac{k\left(b-d\right)}{b-d}=k\) (2)

Từ (1), (2) \(\Rightarrowđpcm\)

c, Ta có: \(\dfrac{a-c}{a}=\dfrac{bk-dk}{bk}=\dfrac{k\left(b-d\right)}{bk}=\dfrac{b-d}{b}\)

\(\Rightarrowđpcm\)

d, Ta có: \(\dfrac{3a+5b}{2a-7b}=\dfrac{3bk+5b}{2bk-7b}=\dfrac{b\left(3k+5\right)}{b\left(2k-7\right)}=\dfrac{3k+5}{2k-7}\)(1)

\(\dfrac{3c+5d}{2c-7d}=\dfrac{3dk+5d}{2dk-7d}=\dfrac{d\left(3k+5\right)}{d\left(2k-7\right)}=\dfrac{3k+5}{2k-7}\) (2)

Từ (1), (2) \(\Rightarrowđpcm\)

e, Sai đề

f, \(\left(\dfrac{a-b}{c-d}\right)^{2012}=\left(\dfrac{bk-b}{dk-d}\right)^{2012}=\left[\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right]^{2012}=\dfrac{b^{2012}}{d^{2012}}\)(1)

\(\dfrac{a^{2012}+b^{2012}}{c^{2012}+d^{2012}}=\dfrac{b^{2012}k^{2012}+b^{2012}}{d^{2012}k^{2012}+d^{2012}}=\dfrac{b^{2012}\left(k^{2012}+1\right)}{d^{2012}\left(k^{2012}+1\right)}=\dfrac{b^{2012}}{d^{2012}}\) (2)

Từ (1), (2) \(\Rightarrowđpcm\)

Answer 2

[Hà Linh]

Hâm mộ :)))))