ĐKXĐ: x∉{3;-2}

Ta có: \(\frac{4x^2-8x}{-x^2+x+6}<0\)

=>\(\frac{4x\left(x-2\right)}{-\left(x^2-x-6\right)}<0\)

=>\(\frac{x\left(x-2\right)}{x^2-x-6}>0\)

=>\(\frac{x\left(x-2\right)}{\left(x-3\right)\left(x+2\right)}>0\)

TH1: \(\begin{cases}x\left(x-2\right)>0\\ \left(x-3\right)\left(x+2\right)>0\end{cases}\)

=>\(\begin{cases}\left[\begin{array}{l}x>2\\ x<0\end{array}\right.\\ \left[\begin{array}{l}x>3\\ x<-2\end{array}\right.\end{cases}\)

=>x>3 hoặc x<-2

TH2: \(\begin{cases}x\left(x-2\right)<0\\ \left(x-3\right)\left(x+2\right)<0\end{cases}\)

=>\(\begin{cases}0

=>0<x<2

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

Lời giải:

a)

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

\(\Leftrightarrow x^3(x-1)+3x^2(x-1)+8x(x-1)+12(x-1)=0\)

\(\Leftrightarrow (x-1)(x^3+3x^2+8x+12)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x^2+x+6=0\left(1\right)\end{matrix}\right.\)

Đối với (1): \(\Leftrightarrow (x+\frac{1}{2})^2+\frac{23}{4}=0\)

(vô lý vì \((x+\frac{1}{2})^2+\frac{23}{4}\geq \frac{23}{4}>0\) )

Do đó \(x\in\left\{-2;1\right\}\)

b) ĐKXĐ: ......

\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}=\frac{1}{6}\)

\(\Leftrightarrow \frac{1}{(x+1)(x+3)}+\frac{1}{(x+3)(x+5)}=\frac{1}{6}\)

\(\Leftrightarrow \frac{(x+5)+(x+1)}{(x+1)(x+3)(x+5)}=\frac{1}{6}\)

\(\Leftrightarrow \frac{2(x+3)}{(x+1)(x+3)(x+5)}=\frac{1}{6}\Leftrightarrow \frac{2}{(x+1)(x+5)}=\frac{1}{6}\)

\(\Leftrightarrow (x+1)(x+5)=12\)

\(\Leftrightarrow x^2+6x-7=0\)

\(\Leftrightarrow (x-1)(x+7)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\) (thỏa mãn đkxđ)

Vậy \(x\in\left\{-7;1\right\}\)

help meee ɜː

a: \(=\dfrac{x^2-5x+x+4}{x\left(x-2\right)}=\dfrac{x^2-4x+4}{x\left(x-2\right)}=\dfrac{x-2}{x}\)

b: \(=\dfrac{x^2-6x+9+4x^2+8x-4x^2-8x}{\left(x-3\right)\left(x+2\right)}\)

\(=\dfrac{x-3}{x+2}\)

a) \(=\dfrac{x\left(x-5\right)+x+4}{x\left(x-2\right)}=\dfrac{x^2-4x+4}{x\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}=\dfrac{x-2}{x}\)

b) \(=\dfrac{\left(x-3\right)^2+4x\left(x+2\right)-8x-4x^2}{\left(x+2\right)\left(x-3\right)}=\dfrac{x^2-6x+9+4x^2+8x-8x-4x^2}{\left(x+2\right)\left(x-3\right)}\)

\(=\dfrac{x^2-6x+9}{\left(x+2\right)\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{\left(x+2\right)\left(x-3\right)}=\dfrac{x-3}{x+2}\)

Đcm học ngu k biết xài caskov

a) \(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne-3\end{cases}}\)

b) \(P=1+\frac{x+3}{x^2+5x+6}\div\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)

\(\Leftrightarrow P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)

\(\Leftrightarrow P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)\)

\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{6}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow P=1+\frac{\left(x-2\right)\left(x+2\right)}{6\left(x+2\right)}\)

\(\Leftrightarrow P=1+\frac{x-2}{6}\)

\(\Leftrightarrow P=\frac{x+4}{6}\)

c) Để P = 0

\(\Leftrightarrow\frac{x+4}{6}=0\)

\(\Leftrightarrow x+4=0\)

\(\Leftrightarrow x=-4\)

Để P = 1

\(\Leftrightarrow\frac{x+4}{6}=1\)

\(\Leftrightarrow x+4=6\)

\(\Leftrightarrow x=2\)

d) Để P > 0

\(\Leftrightarrow\frac{x+4}{6}>0\)

\(\Leftrightarrow x+4>0\)(Vì 6>0)

\(\Leftrightarrow x>-4\)

ĐKXĐ: x∅{-2;-8;-4;-6}

Ta có: \(\frac{x^2+4x+6}{x+2}+\frac{x^2+16x+72}{x+8}=\frac{x^2+8x+20}{x+4}+\frac{x^2+12x+42}{x+6}\)

=>\(\frac{x^2+4x+4+2}{x+2}+\frac{x^2+16x+64+8}{x+8}=\frac{x^2+8x+16+4}{x+4}+\frac{x^2+12x+36+6}{x+6}\)

=>\(x+2+\frac{2}{x+2}+x+8+\frac{8}{x+8}=x+4+\frac{4}{x+4}+x+6+\frac{6}{x+6}\)

=>\(\frac{2}{x+2}+\frac{8}{x+8}=\frac{4}{x+4}+\frac{6}{x+6}\)

=>\(1-\frac{2}{x+2}+1-\frac{8}{x+8}=1-\frac{4}{x+4}+1-\frac{6}{x+6}\)

=>\(\frac{x}{x+2}+\frac{x}{x+8}=\frac{x}{x+4}+\frac{x}{x+6}\)

=>\(x\left(\frac{1}{x+2}+\frac{1}{x+8}-\frac{1}{x+4}-\frac{1}{x+6}\right)=0\)

=>\(x\left(\frac{x+8+x+2}{\left(x+8\right)\left(x+2\right)}-\frac{x+4+x+6}{\left(x+4\right)\left(x+6\right)}\right)=0\)

=>x(2x+10)\(\left(\frac{1}{\left(x+8\right)\left(x+2\right)}-\frac{1}{\left(x+4\right)\left(x+6\right)}\right)=0\)

=>\(x\left(x+5\right)\cdot\frac{\left(x+4\right)\left(x+6\right)-\left(x+8\right)\left(x+2\right)}{\left(x+8\right)\left(x+2\right)\left(x+4\right)\left(x+6\right)}=0\)

=>x(x+5)\(\left(x^2+10x+24-x^2-10x-16\right)=0\)

=>8x(x+5)=0

=>x(x+5)=0

=>x=0(nhận) hoặc x=-5(nhận)

a) ĐKXĐ: x khác 0

\(x+\dfrac{5}{x}>0\)

\(\Leftrightarrow x^2+5>0\) ( luôn đúng)

Vậy bất pt vô số nghiệm ( loại x = 0)

d)

\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)

\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2-x-3}{8}\)

\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{-5}{8}\)

\(\Leftrightarrow2x+2-4x+4>-15\)

\(\Leftrightarrow-2x>-21\)

\(\Leftrightarrow x< \dfrac{21}{2}\)

Vậy....................

a)\(x+\dfrac{5}{x}>0\left(ĐKXĐ:x\ne0\right)\)

\(\Leftrightarrow\dfrac{x^2+5}{x}>0\)

Mà \(x^2+5>0\)

\(\Rightarrow x>0\)

d)\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)

\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{2x-2}{12}>\dfrac{-5}{8}\)

\(\Leftrightarrow\dfrac{-x+3}{12}>\dfrac{-5}{8}\)

\(\Leftrightarrow-x+3>-\dfrac{15}{2}\)

\(\Leftrightarrow-x>-\dfrac{21}{2}\)

\(\Leftrightarrow x< \dfrac{21}{2}\)