1)
<=> \(x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
x= 0
x = 3
2) <=> \(x\left(x-3\right)=4\)
=> \(x=\dfrac{4}{x}+3\)
\(2,x^2-3x=4\)
\(\Leftrightarrow x^2-3x-4=0\)
\(\Delta=b^2-4ac=\left(-3\right)^2-4\left(-4\right)=25>0\)
\(\Rightarrow\)Pt có 2 nghiệm pb
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+5}{2}=4\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-3-5}{2}=-1\end{matrix}\right.\)
Vậy \(S=\left\{4;-1\right\}\)
\(3,x^4-5x^2+6=0\)
Đặt \(t=x^2\left(t\ge0\right)\)
Pt trở thành
\(t^2-5t+6=0\)
\(\Delta=b^2-4ac=\left(-5\right)^2-4.6=1>0\)
\(\Rightarrow\)Pt ó 2 nghiệm pb
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+1}{2}=3\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-5-1}{2}-3\end{matrix}\right.\)
\(\Rightarrow t=x^2\Leftrightarrow t=\pm\sqrt{3}\)
Vậy \(S=\left\{\pm\sqrt{3}\right\}\)
\(4,x^3=9x\)
\(\Leftrightarrow x^3-9x=0\)
\(\Leftrightarrow x\left(x^2-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-9=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm3\end{matrix}\right.\)
Vậy \(S=\left\{0;\pm3\right\}\)
\(5,\left(x+2\right)\left(x-3\right)=x^2-4\)
\(\Leftrightarrow x^2-3x+2x-6-x^2+4=0\)
\(\Leftrightarrow-x-2=0\)
\(\Leftrightarrow-x=2\)
\(\Leftrightarrow x=-2\)
Vậy \(S=\left\{-2\right\}\)
\(6,\Leftrightarrow\dfrac{x+11}{x^2-1}-\dfrac{x-1}{x+1}-\dfrac{2\left(x+7\right)}{x+1}=0\)
\(\Leftrightarrow\dfrac{x+11}{\left(x-1\right)\left(x+1\right)}-\dfrac{x-1}{x+1}-\dfrac{2x+14}{x+1}=0\)
\(\Leftrightarrow\dfrac{x+11-\left(x-1\right)^2-\left(2x+14\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x+11-x^2+2x-1-2x^2+2x-14x+14=0\)
\(\Leftrightarrow-3x^2-9x+24=0\)
\(\Delta=b^2-4ac=\left(-9\right)^2-4.\left(-3\right).24=369>0\)
\(\Rightarrow\)Pt có 2 nghiệm pb
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{9+3\sqrt{41}}{-6}=\dfrac{-3-\sqrt{41}}{2}\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{9-3\sqrt{41}}{-6}=\dfrac{-3+\sqrt{41}}{2}\end{matrix}\right.\)
3/<=> \(x^4-2x^2-3x^2+6=0\)
<=> \(x^2\left(x^2-2\right)-3\left(x^2-2\right)=0\)
<=> \(\left(x^3-2\right)\left(x^2-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=\pm\sqrt{3}\end{matrix}\right.\)
4) <=> \(x^3-9x=0\)
<=> \(x\left(x^2-9\right)=0\)
<=> \(x\left(x-3\right)\left(x+3\right)=0\)
=> x = 0 , x = 3 , x =-3
5)
<=> \(\left(x+2\right)\left(x-3\right)=\left(x-2\right)\left(x+2\right)\)
<=> ( x+2 ) ( x- 3 ) - (x-2) (x+2 ) =0
<=> \(\left(x+2\right)\left(x-3-x+2\right)=0\)
<=> \(\left(x+2\right).-1=0\)
=> x = -2
6) đkxđ : x khác 1 , x khác -1
<=> \(\dfrac{x+11}{\left(x-1\right)\left(x+1\right)}-\dfrac{x-1}{x+1}-\dfrac{2\left(x+7\right)}{x+1}=0\)
<=> \(\dfrac{x+11}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{2.\left(x+7\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)
,=> \(x+11-\left(x^2-2x+1\right)-\left(2x+14\right)\left(x-1\right)=0\)
<=> \(x+11-x^2+2x-1-\left(2x^2-2x+14x-14\right)=0\)
<=> \(x+11-x^2+2x-1-2x^2+2x-14x+14=0\)
<=> \(-3x^2-9x+24=0\)
Áp dụng công thức nghiệm bậc 2
\(x=\dfrac{-3\pm\sqrt{41}}{2}\)
