Câu hỏi của Contrim Đẹptrai

Question

bài 1

tìm x trong các tỉ lệ thức sau :

a, $\frac{x-3}{x+5}=\frac{5}{7}$               b,$\frac{7}{x-1}=\frac{x+1}{9}$

c,$\frac{x+4}{20}=\frac{5}{x+4}$            d,\(\frac{x-1}{x+2}=\frac{x-2...

Vũ Minh Tuấn · Accepted Answer

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!Câu trả lời: Bài 1: