giải pt
1.(x^2-x+1)(X^2-x+2)=2
2.X(x+2)(x+3)(x+5)=280
3.(x+3)(x+4)(X+5)=x
4.\(\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}=\dfrac{1}{9}\) 5. 12/x^2-12/x^2+2=1 6.(x^2-6x)^2+14(x-3)^2=81 7.(x^2+5x)^2-2(x^2+5x)-24=0
8. x^2+2x+3=(x^2+x+1)(X^4+x^2+4)
2. \(x\left(x+2\right)\left(x+3\right)\left(x+5\right)=280\)
\(\Leftrightarrow x\left(x+5\right)\left(x+2\right)\left(x+3\right)=280\)
\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+6\right)=280\)
Đặt \(x^2+5x+3=t\)
\(\Rightarrow\left(t-3\right)\left(t+3\right)=280\)
\(\Leftrightarrow t^2-9=280\)
\(\Leftrightarrow t^2=289\Leftrightarrow\left[{}\begin{matrix}t=17\\t=-17\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+3=17\\x^2+5x+3=-17\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x-14=0\\x^2+5x+20=0\end{matrix}\right.\)
\(\Leftrightarrow x^2+5x-14=0\text{(vì }x^2+5x+20=\left(x+\dfrac{5}{2}\right)^2+\dfrac{55}{4}>0\forall x\text{)}\)
\(\Leftrightarrow x^2-2x+7x-14=0\)
\(\Leftrightarrow x\left(x-2\right)+7\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\)
\(\Leftrightarrow\) x - 2 = 0 hoặc x + 7 = 0
\(\Leftrightarrow\) x = 2 hoặc x = - 7
Vậy x = 2 hoặc x = -7.
3. \(\left(x+3\right)\left(x+4\right)\left(x+5\right)=x\)
\(\Leftrightarrow\left(x+3\right)\left(x+4\right)\left(x+5\right)-x=0\)
\(\Leftrightarrow x^3+12x^2+47x+60-x=0\)
\(\Leftrightarrow x^3+12x^2+46x+60=0\)
\(\Leftrightarrow x^3+6x^2+6x^2+36x+10x+60=0\)
\(\Leftrightarrow x^2\left(x+6\right)+6x\left(x+6\right)+10\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x^2+6x+10\right)=0\)
\(\Leftrightarrow x+6=0\text{(vì }x^2+6x+10=\left(x+3\right)^2+1>0\forall x\text{)}\)
\(\Leftrightarrow x=-6\)
Vậy x = -6.
4.\(\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}=\dfrac{1}{9}\)
\(\Leftrightarrow2\left[\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}\right]=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+6}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{6}{x\left(x+6\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow2x\left(x+6\right)=54\)
\(\Leftrightarrow2x^2+12x-54=0\)
\(\Leftrightarrow2x^2-6x+18x-54=0\)
\(\Leftrightarrow2x\left(x-3\right)+18\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+18\right)=0\)
\(\Leftrightarrow2\left(x-3\right)\left(x+9\right)=0\)
\(\Leftrightarrow\) x - 3 = 0 hoặc x + 9 = 0
\(\Leftrightarrow\) x = 3 hoặc x = -9
Vậy x = 3 hoặc x = -9.
1. \(\left(x^2-x+1\right)\left(x^2-x+2\right)=2\)
Đặt \(x^2-x+1=t\)
\(\Rightarrow t\left(t+1\right)=2\)
\(\Leftrightarrow t^2+1=2\)
\(\Leftrightarrow t^2=1\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=1\\x^2-x+1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x=0\\x^2-x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x^2-x=0\text{ (vì }x^2-x+2=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\forall x\text{)}\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\) x = 0 hoặc x - 1 = 0
\(\Leftrightarrow\) x = 0 hoặc x = 1
Vậy x = 0 hoặc x = 1.
6.\(\left(x^2-6x\right)^2+14\left(x-3\right)^2=81\)
\(\Leftrightarrow\left(x^2-6x\right)^2+14\left(x^2-6x+9\right)=81\)
Đặt \(x^2-6x=t\)
\(\Rightarrow t^2+14\left(t+9\right)=81\)
\(\Leftrightarrow t^2+14t+126-81=0\)
\(\Leftrightarrow t^2+14t+45=0\)
\(\Leftrightarrow t^2+5t+9t+45=0\)
\(\Leftrightarrow t\left(t+5\right)+9\left(t+5\right)=0\)
\(\Leftrightarrow\left(t+5\right)\left(t+9\right)=0\)
\(\Leftrightarrow\) t + 5 = 0 hoặc t + 9 = 0
\(\Leftrightarrow\) t = -5 hoặc t = -9
Vậy t = -5 hoặc t = -9.
