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 \(\dfrac{x}{y}\)
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) \(\Rightarrowđpcm\)
Bài 2:
a, \(\dfrac{3}{x-4}=\dfrac{x+4}{3}\)
\(\Rightarrow\left(x-4\right)\left(x+4\right)=9\)
\(\Rightarrow x^2-16=9\)
\(\Rightarrow x^2=25\)
\(\Rightarrow x=\pm5\)
Vậy \(x=\pm5\)
b, \(\dfrac{x+2}{2}=\dfrac{1}{1-x}\)
\(\Rightarrow\left(x+2\right)\left(1-x\right)=2\)
\(\Rightarrow x-x^2+2-2x=2\)
\(\Rightarrow-x-x^2=0\)
\(\Rightarrow x\left(-1-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\-1-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy x = 0 hoặc x = -1
c, \(\dfrac{x+7}{x+4}=\dfrac{x-1}{x-2}\)
\(\Rightarrow\left(x+7\right)\left(x-2\right)=\left(x+4\right)\left(x-1\right)\)
\(\Rightarrow x^2-2x+7x-14=x^2-x+4x-4\)
\(\Rightarrow x^2+5x-14=x^2+3x-4\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\)
Vậy x = 5
Bài 3:
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)
\(\Rightarrow12x-4y=3x+3y\)
\(\Rightarrow9x=7y\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Vậy \(\dfrac{x}{y}=\dfrac{7}{9}\)
3. 3x-y/x+y=3/4
<=> 4(3x-y)=3(x+y)
<=> 12x-4y-3x-3y=0
<=> 9x=7y
<=> x/y=7/9
2. a) 3/x-4=x+4/3
<=> 9=x^2-16
<=> x^2=25
<=> x=5 hoặc x=-5
b) x+2/2=1/1−x
<=> 2= (x+2)(1-x)
<=> -x-x^2=0
<=> -x(1+x)=0
<=> x=0 hoặc x=-1
c) tương tự =))
