Mình làm 2 câu ab thôi nhé!Cách giải các bài tập này đều như nhau!
Giải:
a) \(\frac{x-9}{x}-\frac{x}{x-9}=0\text{⇔}\frac{x-9}{x}=\frac{x}{x-9}\) (ĐKXĐ: x ≠ 0, x ≠ 9)
⇔ (x - 9)2 = x2 ⇔ (x - 9)2 - x2 = 0 ⇔ -9(2x + 9) = 0 ⇔ 2x + 9 = 0 ⇔ x = \(\frac{-9}{2}\)
Vậy phương trình trên có nghiệm là \(\frac{-9}{2}\)
b) \(\frac{x+3}{x-2}=\frac{5}{\left(x-2\right)\left(3-x\right)}\text{⇔}\frac{x+3}{5}=\frac{x-2}{\left(x-2\right)\left(3-x\right)}\text{⇔}\frac{x+3}{5}=\frac{1}{3-x}\) (ĐKXĐ: x ≠ 2, x ≠ 3)
⇔ (x + 3)(x - 3) = -5 ⇔ x2 - 9 = -5 ⇔ x2 = 4 ⇔ x = \(\pm\)2
Vậy phương trình có tập nghiêm S=\(\left\{\pm2\right\}\)
a, \(\frac{x-9}{x}-\frac{x}{x-9}=0\left(đkxđ:x\ne0;9\right)\)
\(< =>\frac{\left(x-9\right)^2}{x\left(x-9\right)}-\frac{x^2}{x\left(x-9\right)}=0\)
\(< =>x^2-18x+81-x^2=0\)
\(< =>18x=81< =>x=\frac{9}{2}\left(tmđk\right)\)
\(\frac{x+3}{x-2}=\frac{5}{\left(x-2\right)\left(3-x\right)}\left(đk:x\ne2;3\right)\)
\(< =>\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\frac{5}{\left(x-2\right)\left(x-3\right)}=0\)
\(< =>x^2-9+5=0\)
\(< =>\left(x+2\right)\left(x-2\right)=0< =>\orbr{\begin{cases}x=2\left(ktm\right)\\x=-2\left(tm\right)\end{cases}}\)
\(\frac{x}{3x-2}-\frac{x}{4x-3}=\frac{x^3}{\left(3x-2\right)\left(4x-3\right)}\left(đkxđ:x\ne\frac{2}{3};\frac{3}{4}\right)\)
\(< =>\frac{x\left(4x-3\right)}{\left(3x-2\right)\left(4x-3\right)}-\frac{x\left(3x-2\right)}{\left(3x-2\right)\left(4x-3\right)}-\frac{x^3}{\left(3x-2\right)\left(4x-3\right)}=0\)
\(< =>4x^2-3x-3x^2+2x-x^3=0\)
\(< =>x^3-x^2+x=0\)
\(< =>x\left(x^2-x+1\right)=0< =>x=0\left(tm\right)\)
Do \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)
\(\frac{x}{x-5}-\frac{x-5}{x}=0\left(đk:x\ne0;5\right)\)
\(< =>\frac{x^2}{\left(x-5\right)x}-\frac{\left(x-5\right)^2}{\left(x-5\right)x}=0\)
\(< =>x^2-x^2+10x-25=0\)
\(< =>10x-25=0< =>x=\frac{25}{10}=\frac{5}{2}\left(tm\right)\)