a) Đặt x^2+x+1=y. Ta có: y(y+1)=12=>y^2+y-12=0
<=> y^2+4y-3y-12=0 <=> (y+4)(y-3)=0
<=> y+4=0 hoặc y-3=0
<=> y=-4 hoặc y=3
a.
Đặt t = x2 + x + 1 \(\left(t>0\right)\). Pt trở thành:
\(t\left(t+1\right)=12\)
\(\Leftrightarrow t^2+t+\dfrac{1}{4}=\dfrac{49}{4}\)
\(\Leftrightarrow\left(t+\dfrac{1}{2}\right)^2=\dfrac{49}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}t+\dfrac{1}{2}=\dfrac{7}{2}\\t+\dfrac{1}{2}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=3\\t=-4\left(loai\right)\end{matrix}\right.\)
\(\Rightarrow x^2+x+1=3\Rightarrow x^2+x+\dfrac{1}{4}=\dfrac{9}{4}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{9}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}t+\dfrac{1}{2}=\dfrac{3}{2}\\t+\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-2\end{matrix}\right.\)
Vậy S = ...........
Theo mình giải đenta nhanh hơn :)
Với đa thức bậc 2 ax2 + bx + c = 0 thì kiểm tra \(\Delta=\) b2 - 4ac
+ \(\Delta>0\Rightarrow2n_o\left[{}\begin{matrix}x=\dfrac{-b+\sqrt{\Delta}}{2a}\\x=\dfrac{-b-\sqrt{\Delta}}{2}\end{matrix}\right.\)
+ \(\Delta=0\Rightarrow2n_o-kep:\dfrac{-b}{2a}\)
+ \(\Delta< 0\Rightarrow0n_o\)
b.
Đặt t = x2 + x. Pt trở thành:
\(x\left(x+1\right)\left(t+1\right)=42\)
\(\Leftrightarrow t\left(t+1\right)=42\)
\(\Leftrightarrow t^2+t-42=0\)
\(\Delta=1^2-4\cdot1\cdot\left(-42\right)=169\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{-1+\sqrt{169}}{2\cdot1}=6\\t=\dfrac{-1-\sqrt{169}}{2\cdot1}=-7\end{matrix}\right.\)
* \(t=6\Leftrightarrow x^2+x=6\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Delta'=1^2-4\cdot1\cdot\left(-6\right)=25\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1+\sqrt{25}}{2\cdot1}=2\\x=\dfrac{-1-\sqrt{25}}{2\cdot1}=-3\end{matrix}\right.\)
* \(t=-7\Leftrightarrow x^2+x=-7\)
\(\Leftrightarrow x^2+x+7=0\)
\(\Delta''=1^2-4\cdot1\cdot7=-27< 0\)
\(\Rightarrow\) Pt vô nghiệm
Vậy S = .........................
b) x(x+1)(x^2+x+1)=42 =>(x^2+x)(x^2+x+1)=42
=> y(y+1)-42=0 =>y^2+y-42=0 =>y^2+7y-6y-42=0
=> (y+7)(y-6)=0 => y+7=0 hoặc y-6=0
=> y=-7 hoặc y=6
c: =>(x^2+x)(x^2+x-2)=24
=>(x^2+x)^2-2(x^2+x)-24=0
=>(x^2+x-6)(x^2+x+4)=0
=>x^2+x-6=0
=>(x+3)(x-2)=0
=>x=2 hoặc x=-3
d: =>(x^2+1)^2+x(x^2+1)+2x(x^2+1)+2x^2=0
=>(x^2+1)(x^2+x+1)+2x(x^2+x+1)=0
=>x^2+2x+1=0
=>(x+1)^2=0
=>x+1=0
=>x=-1
