Bài làm:
1) đk: \(x\ne0;x\ne-5\)
Ta có: \(\frac{30}{x}-\frac{30}{x+5}=1\)
\(\Leftrightarrow\frac{30\left(x+5\right)-30x}{x\left(x+5\right)}=1\)
\(\Leftrightarrow x^2+5x=150\)
\(\Leftrightarrow x^2+5x-150=0\)
\(\Leftrightarrow\left(x-10\right)\left(x+15\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-10=0\\x+15=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=10\\x=-15\end{cases}}\)
2) đk: \(x\ne0;x\ne-2\)
Ta có: \(\frac{60}{x}-\frac{60}{x+2}=1\)
\(\Leftrightarrow\frac{60\left(x+2\right)-60x}{x\left(x+2\right)}=1\)
\(\Leftrightarrow x^2+2x=120\)
\(\Leftrightarrow x^2+2x-120=0\)
\(\Leftrightarrow\left(x-10\right)\left(x+12\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-10=0\\x+12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=10\\x=-12\end{cases}}\)
\(\frac{30}{x}-\frac{30}{x+5}=1\)( ĐKXĐ : \(x\ne0;x\ne-5\))
<=> \(30\left(\frac{1}{x}-\frac{1}{x+5}\right)=1\)
<=> \(30\left(\frac{x+5}{x\left(x+5\right)}-\frac{x}{x\left(x+5\right)}\right)=1\)
<=> \(30\left(\frac{5}{x\left(x+5\right)}\right)=1\)
<=> \(\frac{5}{x\left(x+5\right)}=\frac{1}{30}\)
<=> \(5\cdot30=x\left(x+5\right)\)
<=> \(x^2+5x-150=0\)
<=> \(x^2+15x-10x-150=0\)
<=> \(x\left(x+15\right)-10\left(x+15\right)=0\)
<=> \(\left(x-10\right)\left(x+15\right)=0\)
<=> \(\orbr{\begin{cases}x-10=0\\x+15=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-15\end{cases}}\)( tmđk )
Vậy S = { 10 ; -15 }
\(\frac{60}{x}-\frac{60}{x+2}=1\)( ĐKXĐ : \(x\ne0;x\ne-2\))
<=> \(60\left(\frac{1}{x}-\frac{1}{x+2}\right)=1\)
<=> \(60\left(\frac{x+2}{x\left(x+2\right)}-\frac{x}{x\left(x+2\right)}\right)=1\)
<=> \(60\left(\frac{2}{x\left(x+2\right)}\right)=1\)
<=> \(\frac{2}{x\left(x+2\right)}=\frac{1}{60}\)
<=> \(2\cdot60=x\left(x+2\right)\)
<=> \(x^2+2x-120=0\)
<=> \(x^2+12x-10x-120=0\)
<=> \(x\left(x+12\right)-10\left(x+12\right)=0\)
<=> \(\left(x-10\right)\left(x+12\right)=0\)
<=> \(\orbr{\begin{cases}x-10=0\\x+12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-12\end{cases}}\)
Vậy S = { 10 ; -12 }
a, \(\frac{30}{x}-\frac{30}{x+5}=1\) b, \(\frac{60}{x}-\frac{60}{x+2}=1\)
ĐKXĐ: x khác 0 ĐKXĐ: x khác 0
x khác -5 x khác -2
Ta có: Ta có:
\(\frac{30}{x}-\frac{30}{x+5}=1\) \(\frac{60}{x}-\frac{60}{x+2}=1\)
<=>\(\frac{30\left(x+5\right)}{x\left(x+5\right)}-\frac{30x}{x\left(x+5\right)}=\frac{x\left(x+5\right)}{x\left(x+5\right)}\) <=>\(\frac{60\left(x+2\right)}{x\left(x+2\right)}-\frac{60x}{x\left(x+2\right)}=\frac{x\left(x+2\right)}{x\left(x+2\right)}\)
=>30(x+5)-30x=x(x+5) =>60(x+2)-60x=x(x+2)
<=>30x+150-30x=x2+5x <=>60x+120-60x=x2+2x
<=>150=x2+5x <=>120=x2+2x
<=>0=x2+5x-150 <=>0=x2+2x-120
<=>0=x2-10x+15x-150 <=>x2-10x+12x-120
<=>0=(x2-10x)+(15x-150) <=>(x2-10x)+(12x-120)
<=>0=x(x-10)+15(x-10) <=>x(x-10)+12(x-10)
<=>0=(x-10)(x+15) <=>(x-10)(x+12)
<=>x-10=0 hoặc x+15=0 <=>x-10=0 hoặc x+12=0
1, x-10=0 2,x+15=0 1, x-10=0 2, x+12=0
<=>x=10 <=>x=-15 <=>x=10 <=>x=-12
( thỏa mãn ĐKXĐ) (thỏa mãn ĐKXĐ)
Vậy tâp nghiệm của PT là S={10;-15} Vậy tập ngiệm của PT là S={10;-12}