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

\(ĐKXĐ:x\ne\pm2\)

\(pt\Leftrightarrow\frac{9}{x^2-4}=\frac{x^2-3x+2}{x^2-4}+\frac{3x+6}{x^2-4}\)

\(\Leftrightarrow\frac{9}{x^2-4}=\frac{x^2+8}{x^2-4}\)

\(\Leftrightarrow x^2+8=9\Leftrightarrow x=\pm1\left(tm\right)\)

Vậy pt có 2 nghiệm là 1 và -1

Điều kện :  \(x+2\ne0\) và \(x-2\ne0\Leftrightarrow x=\pm2\)

( Khi đó \(x^2-4=\left(x+2\right)\left(x-2\right)\ne0\) )

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

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

\(\Rightarrow x^2-3x+2+3x+6=9\Leftrightarrow x^2=1\Leftrightarrow x=\pm1\)

Vậy tập nghiệm của PT là: \(S=\left\{-1;1\right\}\)

Chúc bạn học tốt !!!

1) \(\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{4x+15}{9-x^2}\)

ĐKXĐ : \(x\ne\pm3\)

\(\Leftrightarrow\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{-4x-15}{x^2-9}\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\frac{x^2-4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{x^2+3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\frac{x^2-4x+3-x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow-7x+3=-4x-15\)

\(\Leftrightarrow-7x+4x=-15-3\)

\(\Leftrightarrow-3x=-18\)

\(\Leftrightarrow x=6\)( tmđk )

Vậy x = 6 là nghiệm của phương trình

2) 2x + 3 < 6 - ( 3 - 4x )

<=> 2x + 3 < 6 - 3 + 4x

<=> 2x - 4x < 6 - 3 - 3

<=> -2x < 0

<=> x > 0

Vậy nghiệm của bất phương trình là x > 0

\(\dfrac{2x-3}{x-1}< \dfrac{1}{3}\left(đk:x\ne1\right)\)

\(\Leftrightarrow6x-9< x-1\Leftrightarrow5x< 8\Leftrightarrow x< \dfrac{8}{5}\) và ĐK \(x\ne1\)

\(\dfrac{2x-3}{x-1}>\dfrac{1}{3}\left(đk:x\ne1\right)\)

\(\Leftrightarrow x-1< 6x-9\Leftrightarrow5x>8\Leftrightarrow x>\dfrac{8}{5}\) và ĐK \(x\ne1\)

ĐKXĐ: \(x\ge\dfrac{1}{5}\)

\(\Leftrightarrow2x^2+x-3+2x-\sqrt{5x-1}+\sqrt[3]{x-9}+2\le0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+3\right)+\dfrac{4x^2-5x+1}{2x+\sqrt{5x-1}}+\dfrac{x-1}{\sqrt[3]{\left(x-9\right)^2}-2\sqrt[3]{x-9}+4}\le0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+3+\dfrac{4x-1}{2x+\sqrt{5x-1}}+\dfrac{1}{\sqrt[3]{\left(x-9\right)^2}-2\sqrt[3]{x-9}+4}\right)\le0\)

\(\Leftrightarrow x-1\le0\)

\(\Rightarrow\dfrac{1}{5}\le x\le1\)

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Mình không biết sin lỗi vạn

x-1 + x-3 =1 <=> 2x -4=1 tu giai not

a: =>5(2-x)<3(3-2x)

=>10-5x<9-6x

=>x<-1

b: =>2/9x+5/3>=1/5x-1/5+1/3x

=>2/9x+5/3>=8/15x-1/5

=>-14/45x>=-28/15

=>x<=6

\(a,\left(\dfrac{1}{3}\right)^{2x+1}\le9\\ \Leftrightarrow2x+1\ge-2\\ \Leftrightarrow2x\ge-3\\ \Leftrightarrow x\ge-\dfrac{3}{2}\)

\(b,4^x>2^{x-2}\\ \Leftrightarrow2^{2x}>2^{x-2}\\ \Leftrightarrow2x>x-2\\ \Leftrightarrow x>-2\)

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 

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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