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1. Đặt $x^2+x=a$ thì pt trở thành:
$a^2+4a=12$
$\Leftrightarrow a^2+4a-12=0$
$\Leftrightarrow (a-2)(a+6)=0$
$\Leftrightarrow a-2=0$ hoặc $x+6=0$
$\Leftrightarrow x^2+x-2=0$ hoặc $x^2+x+6=0$
Dễ thấy $x^2+x+6=0$ vô nghiệm.
$\Rightarrow x^2+x-2=0$
$\Leftrightarrow (x-1)(x+2)=0$
$\Leftrightarrow x=1$ hoặc $x=-2$
2.
$x(x-1)(x+1)(x+2)=24$
$\Leftrightarrow [x(x+1)][(x-1)(x+2)]=24$
$\Leftrightarrow (x^2+x)(x^2+x-2)=24$
$\Leftrightarrow a(a-2)=24$ (đặt $x^2+x=a$)
$\Leftrightarrow a^2-2a-24=0$
$\Leftrightarrow (a+4)(a-6)=0$
$\Leftrightarrow a+4=0$ hoặc $a-6=0$
$\Leftrightarrow x^2+x+4=0$ hoặc $x^2+x-6=0$
Nếu $x^2+x+4=0$
$\Leftrightarrow (x+\frac{1}{2})^2=\frac{1}{4}-4<0$ (vô lý - loại)
Nếu $x^2+x-6=0$
$\Leftrightarrow (x-2)(x+3)=0$
$\Leftrightarrow x-2=0$ hoặc $x+3=0$
$\Leftrightarrow x=2$ hoặc $x=-3$
\(a)\left(x^2+x\right)^2+4\left(x^2+x\right)=12\\ \Leftrightarrow\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)
Đặt \(t=x^2+x\left(t\ge0\right)\)
\(\Leftrightarrow t^2+4t-12=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-6\end{matrix}\right.\)
Với \(t=2\Rightarrow x^2+x=2\Rightarrow x^2-x-2=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Với \(t=-6\Rightarrow x^2+x=-6\Rightarrow x^2+x+6=0\Rightarrow x\notin\)
Vậy...
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b; x(x-1)(x+1)(x+2)-24
=(x2+x)(x2+x-2)-24
Đặt x2+x=k khi đó k(k-2)-24=k2-2k-24
=(k2-2k+1)-25=(k-1)2-52
=(k-1-5)(k-1+5)=(k-6)(k+4)
c; (x+2)(x-2)(x2-10)-72
=(x2-4)(x2-10)-72
Đặt x2-7=k khi đó (k-3)(k+3)-72=k2-9-72
=k2-81=(k-9)(k+9)=(x2-7-9)(x2-7+9)
=(x2-16)(x2+2)=(x-4)(x+4)(x2+2)
d; (x-7)(x-5)(x-4)(x-2)-72
=(x2-9x+14)(x2-9x+8)-72
Đặt x2-9x+11=k khi đó (k+3)(k-3)-72=k2-9-72
=k2-81=(k-9)(k+9)=(x2-9x+11-9)(x2-9x+11+9)
=(x2-9x+2)(x2-9x+20)
=(x2-9x+2)(x2-4x-5x+20)
=(x2-9x+2)(x-4)(x-5)
a, ( x2 + x )2 - 14 ( x2 + x ) + 24
= (x2 + x)2 - 2(x2 + x) -12(x2 + x) + 24
= (x2 + x).(x2 + x -2) - 12(x2 + x -2)
= (x2 + x -2).(x2 + x -12)
= (x2 + 2x - x - 2).(x2 + 4x - 3x - 12)
=[x.(x+2)-(x+2)].[x.(x+4)-3(x+4)]
= (x+2).(x-1).(x+4).(x-3)
= x4 + 2x3 - 13x2 - 14x + 24
b, ( x2 + x )2 + 4x2 + 4x - 12
= x4 + 2x3 + x2 + 4x2 + 4x -12
= x4 + 2x3 + 5x2 + 4x -12
c, x4 + 2x3 + 5x2 + 4x - 12
= x4 - x3 + 3x3 - 3x2 + 8x2 - 8x +12x -12
= x3(x-1) + 3x2(x-1) + 8x(x-1) + 12(x-1)
= (x-1) . (x3 + 3x2 + 8x +12)
= (x-1) . ( x3 +2x2 + x2 + 2x + 6x +12)
= (x-1). [x2(x+2) + x(x+2) + 6(x+2)]
= (x-1).(x+2).(x2 + x+ 6)
a/\(\left(x^2-x\right)^2+4\left(x^2-x\right)-12.\)
cho \(\left(x^2-x\right)=a\)
\(\Rightarrow a^2+4a-12\)
\(=a^2+6a-2a-12\)
\(=\left(a^2+6a\right)-\left(2a+12\right)\)
\(=a\left(a+6\right)-2\left(a+6\right)\)
\(=\left(a+6\right)\left(a-2\right)\)
\(=\left(x^2-x+6\right)\left(x^2-x-2\right)\)
b/ \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24\)
\(=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
Gọi \(x^2+5x+5=a\)
\(\Rightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=\left(a-1\right)\left(a+1\right)-24\)
\(=a^2-1-24\)
\(=a^2-25\)
\(=\left(a-5\right)\left(a+5\right)\)
\(\Rightarrow\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
Câu a:
\((x^2+x)^2+4(x^2+x)=12\)
\(\Leftrightarrow (x^2+x)^2+4(x^2+x)+4=16\)
\(\Leftrightarrow (x^2+x+2)^2=16\)
\(\Rightarrow \left[\begin{matrix} x^2+x+2=4\\ x^2+x+2=-4\end{matrix}\right.\Rightarrow \left[\begin{matrix} x^2+x-2=0\\ x^2+x+6=0\end{matrix}\right.\)
Với \(x^2+x-2=0\Leftrightarrow (x-1)(x+2)=0\Rightarrow \left[\begin{matrix} x=1\\ x=-2\end{matrix}\right.\)
Với \(x^2+x+6=0\Leftrightarrow (x^2+x+\frac{1}{4})+\frac{23}{4}=0\)
\(\Leftrightarrow (x+\frac{1}{2})^2=\frac{-23}{4}<0\) (vô lý- loại)
Vậy \(x\in \left\{-2;1\right\}\)
Câu b:
\(x(x-1)(x+1)(x+2)=24\)
\(\Leftrightarrow [x(x+1)][(x-1)(x+2)]=24\)
\(\Leftrightarrow (x^2+x)(x^2+x-2)=24\)
\(\Leftrightarrow a(a-2)=24\) (đặt \(x^2+x=a\) )
\(\Leftrightarrow a^2-2a-24=0\)
\(\Leftrightarrow (a-6)(a+4)=0\Rightarrow \left[\begin{matrix} a-6=0\\ a+4=0\end{matrix}\right.\)
Nếu \(a-6=0\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow (x-2)(x+3)=0\Rightarrow \left[\begin{matrix} x=2\\ x=-3\end{matrix}\right.\)
Nếu \(a+4=0\Leftrightarrow x^2+x+4=0\Leftrightarrow (x+\frac{1}{2})^2=\frac{-15}{4}<0\) (vô lý)
Vậy............