BẰng 2 chứ ko phải 22

\(ĐKXĐ:\hept{\begin{cases}x\ne2;x\ne-4\\x\ne1;x\ne-3\end{cases}}\)

\(\frac{24}{x^2+2x-8}-\frac{15}{x^2+2x-3}=2\)

\(\Leftrightarrow\frac{24}{\left(x-2\right)\left(x+4\right)}-\frac{15}{\left(x-1\right)\left(x+3\right)}-2=0\)

\(\Leftrightarrow24\left(x^2+2x-3\right)-15\left(x^2+2x-8\right)-2\left(x^2+2x-8\right)\left(x^2+2x-3\right)=0\)

\(\Leftrightarrow24x^2+48x-72-15x^2-30x+120-2\left(x^4+4x^3-7x^2-22x+24\right)=0\)

\(\Leftrightarrow9x^2+18x+48-2x^4-8x^3+14x^2+44x-48=0\)

\(\Leftrightarrow-2x^4-8x^3+23x^2+62x=0\)

\(\Leftrightarrow-x\left(2x^3+8x^2-23x-62\right)=0\)

\(\Leftrightarrow-x\left(2x^3+4x^2+4x^2+8x-31x-62\right)=0\)

\(\Leftrightarrow x\left[2x^2\left(x+2\right)+4x\left(x+2\right)-31\left(x+2\right)\right]=0\)

\(\Leftrightarrow x\left(x+2\right)\left(2x^2+4x-31\right)=0\)

\(\Leftrightarrow\)\(x=0\)

hoặc \(x=-2\)

hoặc \(2\left(x+1\right)^2-33=0\)

\(\Leftrightarrow\)\(x=0\)(tm)

hoặc \(x=-2\)(tm)

hoặc \(x=-\frac{2\pm\sqrt{66}}{2}\)(tm)

Vậy tập nghiệm của phương trình là \(S=\left\{0;-2;-\frac{2\pm\sqrt{66}}{2}\right\}\)

giúp mình với mình đang cần gấp

ĐKXĐ:

a/ \(\left\{{}\begin{matrix}x^2+2x-8\ne0\\x^2+2x-3\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne-4\\x\ne2\\x\ne1\\x\ne-3\end{matrix}\right.\)

b/ \(x^2-2x+3\ge0\Rightarrow x\in R\)

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

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

\(\Leftrightarrow x\left(x-3\right)+3\left(x-1\right)=x\left(x+1\right)\)

\(\Leftrightarrow x^2-3=x^2+x\)

\(\Leftrightarrow-3=x\)

\(\Leftrightarrow x=-3\)

Vậy: nghiệm phương trình là -3

\(3,\text{ }\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16=0\)

\(\Rightarrow\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)=0-16\)

\(\Rightarrow\text{ Có lẻ thừa số âm }\)

Mà \(\left(x+8\right)>\left(x+6\right)>\left(x+4\right)>\left(x+2\right)\)

Ta có hai trường hợp : 

\(TH\text{ }1\text{ :}\) Có một thừa số âm

\(\Rightarrow\text{ }\left(x+2\right)< 0\)

\(\Rightarrow\text{ }x< -2\)

\(TH\text{ }2\text{ : }\) Có 3 thừa số âm

\(\Rightarrow\text{ }\hept{\begin{cases}\left(x+2\right)< 0\\\left(x+4\right)< 0\\\left(x+6\right)< 0\end{cases}}\)                \(\Rightarrow\text{ }\left(x+2\right)< 0\text{ }\Rightarrow\text{ }x< -2\)

Si thì thôi nha ! Mong bạn thông cảm !

câu a tự quy đồng cùng  mẫu rồi làm thôi :"))

b) \(\left[x.\left(x-1\right)\right].\left[\left(x-2\right).\left(x+1\right)\right]=24\)

\(\Leftrightarrow\left(x^2-x\right).\left(x^2-x-2\right)=24\)

Đặt \(x^2-x=k\), ta có:

\(k.\left(k-2\right)=24\)

\(\Leftrightarrow k^2-2k+1=25\)

\(\Leftrightarrow\left(k-1\right)^2=5^2\Leftrightarrow\orbr{\begin{cases}k-1=5\\k-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}k=6\\k=-4\end{cases}}}\)

\(k=6\Rightarrow x^2-x=6\Rightarrow x^2-x-6=0\)

\(\Rightarrow x^2-3x+2x-6=0\Rightarrow x.\left(x-3\right)+2.\left(x-3\right)=0\)

\(\Rightarrow\left(x+2\right).\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)

\(k=-4\Rightarrow x^2-x+4=0\Rightarrow x^2-x+\frac{1}{4}+\frac{15}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=-\frac{15}{4}\left(\text{loại}\right)\)

c)\(x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow x^4+2x^3+2x^2+4x+3x^2-12=0\)

\(\Leftrightarrow x^3.\left(x+2\right)+2x.\left(x+2\right)+3.\left(x^2-2^2\right)=0\)

\(\Leftrightarrow\left(x+2\right).\left(x^3+5x-6\right)=0\)

\(\Leftrightarrow\left(x+2\right).\left(x^3-x^2+x^2-x+6x-6\right)=0\)

\(\Leftrightarrow\left(x+2\right).\left[x^2.\left(x-1\right)+x.\left(x-1\right)+6.\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right).\left(x-1\right).\left(x^2+x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}\text{vì }x^2+x+6>0\left(\text{tự c/m}\right)}\)

p/s: bn tự kết luận nha :))

1.

\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)

\(\frac{5x-1}{10}+\frac{2x+3}{6}=\frac{x-8}{15}-\frac{x}{30}\)

\(\Rightarrow\frac{3\left(5x-1\right)}{30}+\frac{5\left(2x+3\right)}{30}=\frac{2\left(x-8\right)}{30}-\frac{x}{30}\)

\(\Rightarrow15x-3+10x+15=2x-16-x\)

\(\Rightarrow24x=-28\)

\(\Rightarrow x=-\frac{7}{6}\)

\(a,\frac{2x+4}{10}+\frac{2-x}{15}=\frac{\left(2x+4\right).3}{10.3}+\frac{\left(2-x\right).2}{15.2}\)

\(=\frac{6x+12}{30}+\frac{4-2x}{30}=\frac{6x+12+4-2x}{30}=\frac{4x+16}{30}\)

\(=\frac{4.\left(x+4\right)}{30}=\frac{2\left(x+4\right)}{15}\)

\(b,\frac{3x}{10}+\frac{2x-1}{15}+\frac{2-x}{20}=\frac{3x.6}{10.6}+\frac{\left(2x-1\right).4}{15.4}+\frac{\left(2-x\right).3}{20.3}\)

\(=\frac{18x}{60}+\frac{8x-4}{60}+\frac{6-3x}{60}=\frac{18x+8x-4+6-3x}{60}=\frac{23x+2}{60}\)

\(c,\frac{x+1}{2x-2}+\frac{x^2+3}{2-2x^2}=\frac{x+1}{2\left(x-1\right)}+\frac{x^2+3}{2\left(1-x^2\right)}=\frac{x+1}{2\left(x-1\right)}+\frac{-x^2-3}{2\left(x^2-1\right)}\)

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

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