ĐKXĐ : \(x\ne\left\{5;-5;0\right\}\)
<=> \(\frac{2}{\left(x-5\right)\left(x+5\right)}-\frac{1}{x\left(x+5\right)}=\frac{4}{x\left(x-5\right)}\)
<=> \(\frac{2x}{x\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x-5\right)\left(x+5\right)}=\frac{4\left(x+5\right)}{x\left(x-5\right)\left(x+5\right)}\)
=> \(2x-x+5=4x+20\)
<=> \(4x-2x+x=5-20\)
<=> \(3x=-15\) <=> \(x=-5\) ( ko t/m )
Vậy pt vô nghiệm.
ĐKXĐ : x khác 0; x khác 5 ; x khác -5
\(\frac{2}{x^2-25}+\frac{1}{x^2+5x}=\frac{4}{x\left(x-5\right)}\Leftrightarrow\frac{2x}{x\left(x-5\right)\left(x+5\right)}+\frac{x-5}{x\left(x-5\right)\left(x+5\right)}=\frac{4\left(x+5\right)}{x\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow\frac{2x+x-5}{x\left(x-5\right)\left(x+5\right)}=\frac{4\left(x+5\right)}{x\left(x-5\right)\left(x+5\right)}\Rightarrow3x-5=4x+20\)
\(\Leftrightarrow3x-4x=20+5\Leftrightarrow-x=25\Leftrightarrow x=-25\)( thỏa mãn ĐKXĐ)
Vậy phương trình có nghiệm x = -25
\(\frac{2}{x^2-25}-\frac{1}{x^2+5x}=\frac{4}{x\left(x-5\right)}\)
<=> \(\frac{2}{\left(x+5\right)\left(x-5\right)}-\frac{1}{x\left(x+5\right)}=\frac{4}{x\left(x-5\right)}\)
<=> \(\frac{2x}{x\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x-5\right)\left(x+5\right)}=\frac{4\left(x+5\right)}{x\left(x-5\right)\left(x+5\right)}\)
<=> \(2x-\left(x-5\right)=4\left(x+5\right)\)
<=> \(2x-x+5=4x+20\)
<=> \(x+5=4x+20\)
<=> \(x-4x=20-5\)
<=> \(-3x=15< =>x=-5\)
Vậy ....
Xin lỗi bn mk thiếu ĐKXĐ :
x khác 0, x khác 5, x khác -5
Do x=-5 ko thoả mãn ĐKXĐ nên phương trình trên vô nghiệm.
Vậy...
\(\frac{2}{x^2-25}-\frac{1}{x^2+5x}=\frac{4}{x\left(x-5\right)}\)
\(\frac{2x}{x\left(x+5\right)\left(x-5\right)}-\frac{x-5}{x\left(x+5\right)\left(x-5\right)}=\frac{5\left(x+5\right)}{x\left(x-5\right)\left(x+5\right)}\)
\(\Rightarrow2x-x+5=4x+20\)
\(\Leftrightarrow2x-x-4x=20-5\)
\(\Leftrightarrow-3x=15\)
\(\Leftrightarrow x=-5\)
Vậy nghiệm pt là x= -5
