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\(\frac{a-3}{a+3}=\frac{b-6}{b+6}\) \(\Rightarrow\)\(\frac{a-3}{b-6}=\frac{a+3}{b+6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a-3}{b-6}=\frac{a+3}{b+6}=\frac{a-3+a+3}{b-6+b-6}=\frac{2a}{2b}=\frac{a}{b}\) (1)
\(\frac{a-3}{b-6}=\frac{a+3}{b+6}=\frac{a-3-a-3}{b-6-b-6}=\frac{-6}{-12}=\frac{1}{2}\) (2)
Từ (1) và (2) suy ra: \(\frac{a}{b}=\frac{1}{2}\)
\(\frac{a-3}{a+3}=\frac{b-6}{b+6}\Rightarrow\left(a-3\right).\left(b+6\right)=\left(b-6\right).\left(a+3\right)\)
\(\Rightarrow ab+6a-3b-18=ab+3b-6a=18\)
\(\Rightarrow b.\left(a-3\right)+6.a-18=a.\left(b-6\right)+3.b-18\)
\(\Rightarrow b.\left(a-3\right)+6a=a.\left(b-6\right)+3b\)
\(\Rightarrow ab-3b=ab-6a+3b-6a\)
\(\Rightarrow ab-3b=ab-3.\left(4a-b\right)\)
\(b=4a-b\Rightarrow2b=4a\Rightarrow b=2a\Rightarrow\frac{a}{b}=\frac{1}{2}\)
Bài 1:
a) \(\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Rightarrow\left(x-3\right).7=\left(x+5\right).5\)
\(\Rightarrow7x-21=5x+25\)
\(\Rightarrow7x-5x=25+21\)
\(\Rightarrow2x=46\)
\(\Rightarrow x=46:2\)
\(\Rightarrow x=23\)
Vậy \(x=23.\)
b) \(\frac{7}{x-1}=\frac{x+1}{9}\)
\(\Rightarrow\left(x+1\right).\left(x-1\right)=7.9\)
\(\Rightarrow x^2-x+x-1=63\)
\(\Rightarrow x^2-1=63\)
\(\Rightarrow x^2=63+1\)
\(\Rightarrow x^2=64\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)
Vậy \(x\in\left\{8;-8\right\}.\)
c) \(\frac{x+4}{20}=\frac{5}{x+4}\)
\(\Rightarrow\left(x+4\right).\left(x+4\right)=5.20\)
\(\Rightarrow\left(x+4\right).\left(x+4\right)=100\)
\(\Rightarrow\left(x+4\right)^2=100\)
\(\Rightarrow x+4=\pm10.\)
\(\Rightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10-4\\x=\left(-10\right)-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)
Vậy \(x\in\left\{6;-14\right\}.\)
Bài 2:
Ta có: \(\frac{a+5}{a-5}=\frac{b+6}{b-6}.\)
\(\Rightarrow\frac{a+5}{b+6}=\frac{a-5}{b-6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{a+5}{b+6}=\frac{a-5}{b-6}=\frac{\left(a+5\right)+\left(a-5\right)}{\left(b+6\right)+\left(b-6\right)}=\frac{\left(a+a\right)+\left(5-5\right)}{\left(b+b\right)+\left(6-6\right)}=\frac{2a}{2b}=\frac{a}{b}\) (1)
\(\frac{a+5}{b+6}=\frac{a-5}{b-6}=\frac{\left(a+5\right)-\left(a-5\right)}{\left(b+6\right)-\left(b-6\right)}=\frac{\left(a-a\right)+\left(5+5\right)}{\left(b-b\right)+\left(6+6\right)}=\frac{10}{12}=\frac{5}{6}\) (2)
Từ (1) và (2) \(\Rightarrow\frac{a}{b}=\frac{5}{6}\left(đpcm\right).\)
Chúc em học tốt!
Bài 1:
$\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt$. Khi đó:
\(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2(bt)^2-3.bt.b+5b^2}{2(bt)^2+3bt.b}=\frac{b^2(2t^2-3t+5)}{b^2(2t^2+3t)}\)
$=\frac{2t^2-3t+5}{2t^2+3t}(1)$
\(\frac{2c^2-3cd+5d^2}{2c^2+3cd}=\frac{2(dt)^2-3.dt.d+5d^2}{2(dt)^2+3dt.d}=\frac{d^2(2t^2-3t+5)}{d^2(2t^2+3t)}=\frac{2t^2-3t+5}{2t^2+3t}(2)\)
Từ $(1);(2)$ suy ra đpcm.
Bài 2:
Từ $\frac{a}{c}=\frac{c}{b}\Rightarrow c^2=ab$. Khi đó:
$\frac{b^2-c^2}{a^2+c^2}=\frac{b^2-ab}{a^2+ab}=\frac{b(b-a)}{a(a+b)}$ (đpcm)
Từ tỉ lệ thức a − 3 a + 3 = b − 6 b + 6 ,ta có:
a − 3 b + 6 = a + 3 b − 6 ⇒ a b + 6 a − 3 b − 18 = a b − 6 a + 3 b − 18 ⇒ 12 a = 6 b ⇒ a b = 1 2 ( d p c m )