Áp dụng công thức \(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\) ta được:
\(\frac{x+2}{x+6}=\frac{3}{x+1}\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)=3\left(x+6\right)\)
\(\Leftrightarrow x^2+3x+2=3x+18\)
\(\Leftrightarrow x^2=16\)
Vậy \(x\in\left\{4;-4\right\}\)
(x+2)/(x+6)=3/(x+1)
<=> (x+2)(x+1)/(x+6)(x+1)=3(x+6)/(x+6)(x+1)
=>(x+2)(x+1)=3(x+6)
<=> x^2+x+2x+2=3x+18
<=> x^2=16
<=>x^2=4^2 hoặc (-4)^2
<=> x=4 hoặc x=-4
Vậy.........
Ta có:
\(\frac{x+2}{x+6}=\frac{3}{x+1}\)
Áp dụng tỉ lệ thức ta có:
\(\left(x+2\right).\left(x+1\right)=\left(x+6\right).3\)
\(\Rightarrow x.\left(x+2\right)+2.\left(x+1\right)=x.3+6.3\)
\(\Rightarrow x^2+x+2x+2=3x+18\)
\(\Rightarrow x^2+3x+2=3x+18\)
\(\Leftrightarrow x^2+2=18\)
\(\Leftrightarrow x=\sqrt{18-2}=4\)
<=>(x+2)(x+1)=3.(x+6)
<=>\(x^2+x+2x+2=3x+18\)
<=>\(x^2+x+2x-3x=18-2\)
<=>\(x^2=16\)
Vậy \(x\in\left\{4;-4\right\}\)
