Question

Bài 1:Giải phương trình và bất phương trình

a) 9/ x²-4 =x-1/x+2 + 3/x-2

b) 1/x-5 - 3/x² -6x+5= 5/x-1

 

Answer 1

a, \(\frac{9}{x^2-4}=\frac{x-1}{x+2}+\frac{3}{x-2}\left(ĐKXĐ:x\ne\pm2\right)\)

\(\frac{9}{\left(x-2\right)\left(x+2\right)}=\frac{x-1}{x+2}+\frac{3}{x-2}\)

\(\frac{9}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

Khử mẫu : \(9=\left(x-1\right)\left(x-2\right)+3\left(x+2\right)\)

Đến đây nhường bn, rất dễ =))

b, \(\frac{1}{x-5}-\frac{3}{x^2-6x+5}=\frac{5}{x-1}\)

\(\frac{1}{x-5}-\frac{3}{\left(x-5\right)\left(x-1\right)}=\frac{5}{\left(x-1\right)}\)

\(\frac{\left(x-1\right)}{x-5}-\frac{3}{\left(x-5\right)\left(x-1\right)}=\frac{5\left(x-5\right)}{\left(x-1\right)\left(x-5\right)}\)

Khử mẫu \(x-1-3=5\left(x-5\right)\)

Tự lm nốt mà cho mk hỏi, đề bài có bpt mà bpt đâu 

Answer 2

\(\frac{9}{x^2-4}=\frac{x-1}{x+2}+\frac{3}{x-2}\left(ĐKXĐ:x\ne2;-2\right)\)

\(< =>\frac{9}{x^2-2^2}=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(< =>\frac{9}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{3x+6}{\left(x+2\right)\left(x-2\right)}\)

\(< =>9=x^2-2x-x+2+3x+6\)

\(< =>x^2-\left(2x+x-3x\right)+\left(2+6-9\right)=0\)

\(< =>x^2-2=0\)\(< =>x^2=2\)

\(< =>x=\pm\sqrt{2}\left(tmđk\right)\)

Vậy tập nghiệm của phương trình trên là \(\pm\sqrt{2}\)

Answer 3

\(\frac{1}{x-5}-\frac{3}{x^2-6x+5}=\frac{5}{x-1}\left(ĐKXĐ:x\ne1;5\right)\)

\(< =>\frac{1}{x-5}-\frac{3}{x^2-x-5x+5}=\frac{5}{x-1}\)

\(< =>\frac{1}{x-5}-\frac{3}{x\left(x-1\right)-5\left(x-1\right)}=\frac{5}{x-1}\)

\(< =>\frac{1}{x-5}-\frac{3}{\left(x-5\right)\left(x-1\right)}=\frac{5}{x-1}\)

\(< =>\frac{x-1}{\left(x-5\right)\left(x-1\right)}-\frac{3}{\left(x-5\right)\left(x-1\right)}=\frac{5x-25}{\left(x-1\right)\left(x-5\right)}\)

\(< =>x-1-3=5x-25\)

\(< =>5x-25-x+4=0\)

\(< =>4x-21=0\)

\(< =>x=\frac{21}{4}=7\left(tmđkxđ\right)\)

Answer 4

\(\frac{9}{x^2-4}=\frac{x-1}{x+2}+\frac{3}{x-2}\)  ( đkxđ : \(x\ne\pm2\))

<=> \(\frac{9}{\left(x+2\right)\left(x-2\right)}=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\)

Khử mẫu

<=> \(9=x^2-3x+2+3x+6\)

<=> \(9=x^2+8\)

<=> \(x^2=1\)

<=> \(x=\pm1\)( tmđk )

\(\frac{1}{x-5}-\frac{3}{x^2-6x+5}=\frac{5}{x-1}\)( đkxđ : \(x\ne5;x\ne1\))

<=> \(\frac{1\left(x-1\right)}{\left(x-5\right)\left(x-1\right)}-\frac{3}{\left(x-5\right)\left(x-1\right)}=\frac{5\left(x-5\right)}{\left(x-5\right)\left(x-1\right)}\)

Khử mẫu

<=> \(x-1-3=5x-25\)

<=> \(x-5x=-25+1+3\)

<=> \(-4x=-21\)

<=> \(x=\frac{21}{4}\)( tmđk )