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\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)
\(\Leftrightarrow\left[x\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]=24\)
\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)=24\)
Dat \(x^2+3x+2=a\left(a>0\right)\)
\(\Leftrightarrow\left(a-2\right)a=24\)
\(\Leftrightarrow a^2-2a-24=0\)
\(\Leftrightarrow a^2-6a+4a-24=0\)
\(\Leftrightarrow\left(a-6\right)\left(a+4\right)=0\\ \left[{}\begin{matrix}a=6\\a=-4\left(Loai\right)\end{matrix}\right.\)
Thay a=6:
\(x^2-3x+4=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
Vayy...
\(x(x+1)(x+2)(x+3)=24\)
\(\Leftrightarrow[x(x+3)][(x+1)(x+2)]-24=0 \)
\(\Leftrightarrow(x^2+3x)(x^2+3x+2)-24=0\)
\(\Leftrightarrow[(x^2+3x+1)-1][(x^2+3x+1)+1]-24=0\)
Đặt \(a=x^2+3x+1\)
\(\Leftrightarrow(a-1)(a+1)-24=0\)
\(\Leftrightarrow (a^2-1)-24=0\)
\(\Leftrightarrow a^2-1-24=0\)
\(\Leftrightarrow a^2-25=0\)
\(\Leftrightarrow(a-5)(a+5)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-5=0\\a+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}a=5\\a=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+3x+1=5\\x^2+3x+1=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+3x-4=0\\x^2+3x+6=0\left(vn\right)\end{matrix}\right.\\ \Leftrightarrow x\left(x+3\right)-4=0\\ \Leftrightarrow x\left(x+3\right)=4\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x+3=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
Vậy tập nghiệm pt \(S=\{-1;4\}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
=( x\(^2\)+x)(x\(^2\)+x -2)=24
đặt x\(^2\)+ x= a\(\Rightarrow\)a(a-2)=24
chuển vế sang rồi tìm a, thay x vào rồi tìm x. tương tự mấy cau trên thui
Ta có : \(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]=24\)\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)(1)
Đặt \(t=x^2+x-1\Rightarrow\hept{\begin{cases}x^2+x=t+1\\x^2+x-2=t-1\end{cases}}\)
Suy ra pt \(\left(1\right)\Leftrightarrow\left(t+1\right)\left(t-1\right)=24\Leftrightarrow t^2-1=24\)
\(\Leftrightarrow t^2=25\Leftrightarrow\left(x^2+x-1\right)=25\)
\(\Leftrightarrow\hept{\begin{cases}x^2+x-1=5\\x^2+x-1=-5\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2+x-6=0\\x^2+x+4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-1\right)x\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left[\left(x-1\right)\left(x+2\right)\right]\left[x\left(x+1\right)\right]-24=0\)
\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x\right)-24=0\)
\(\Leftrightarrow\left[\left(x^2+x-1\right)-1\right]\left[\left(x^2+x-1\right)+1\right]-24=0\)
\(\Leftrightarrow\left(x^2+x-1\right)^2-1^2-24=0\)
\(\Leftrightarrow\left(x^2+x-1\right)^2-5^2=0\)
\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x^2+x+4\right)=0\)
Mà \(x^2+x+4=\left(x+\frac{1}{2}\right)^2+3,75>0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\left(x+3\right)\left(x+1\right)\left(x+2\right)=24\\ =>\left(x^2+3x\right)\left(x^2+3x+3\right)=24\\\)
Đặt \(x^2+3x=a\)ta có
=> \(a\left(a+3\right)=24\\ a^2+3a-24=0\\ \)
cầu phân tích đa thức thành nhân tử di minh tinh dc
X =\(\frac{-3+\sqrt{105}}{2}\)
X = \(\frac{-3-\sqrt{105}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét tích (x+1)(x+2)(x+3)(x+4) là tích của 4 số tự nhiên liên tiếp.
Mà ta thấy 24 = 1 . 2 . 3 . 4
Vậy x + 1 = 1 ; x + 2 = 2 ; x + 3 = 3 ; x + 4 = 4
Do đó x = 0
(x+1)(x+2)(x+3)(x+4)= 24
<=> (x+1)(x+2)(x+3)(x+4)-24=0
<=>(x+1)(x+4)(x+2)(x+3)-24=0
<=>(x2+5x+4)(x2+5x+6)-24=0
Đặt t=x2+5x+4 ta được:
t.(t+2)-24=0
<=>t2+2t-24=0
<=>t2-4t+6t-24=0
<=>t.(t-4)+6.(t-4)=0
<=>(t-4)(t+6)=0
<=>t-4=0 hoặc t+6=0
thay t=x2+5x+4 ta được:
x2+5x=0 hoặc x2+5x+10=0
Vì x2+5x+10=x2+2.x.5/2+25/4+15/4
=(x+5/2)2+15/4>0
nên
x2+5x=0
<=>x.(x+5)=0
<=>x=0 hoặc x=-5
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có :
\(\left|x+1\right|+\left|x-1\right|=1+\left|\left(x-1\right)\left(x+1\right)\right|\)
\(\Leftrightarrow\left|x-1\right|\left|x+1\right|-\left|x-1\right|-\left|x+1\right|+1=0\)
\(\Leftrightarrow\left(\left|x-1\right|-1\right)\left(\left|x+1\right|-1\right)=0\Leftrightarrow\orbr{\begin{cases}\left|x-1\right|=1\\\left|x+1\right|=1\end{cases}}\)
\(\Leftrightarrow x\in\left\{-2,0,2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x+1).(x+2).(x+3).(x+4) - 24 = 0
(x2 + 5x + 4).(x2 + 5x + 6) - 24 = 0
(x2 + 5x + 5-1).(x2 + 5x + 5 + 1) - 24 = 0
(x2 + 5x + 5)2 - 1 - 24 = 0
(x2 + 5x + 5 - 5).(x2 + 5x + 5 + 5) = 0
x.(x+5) .(x2 + 5x + 10) = 0
=> x = 0
x+ 5 = 0 => x = -5
\(x^2+5x+10>0\)
KL:..
(x+1)(x+2)(x+3)(x+4) - 24 = 0
<=> [(x+1)(x+4)][(x+2)(x+3)] - 24 =0
<=> (x^2+4x+x+4)(x^2+3x+2x+6) - 24 = 0
<=> (x^2+5x+4)(x^2+5x+6) - 24 = 0
Đặt x^2+5x+5 = a, ta có
(a-1)(a+1) - 24 = 0
<=> a^2 - 1 - 24 = 0
<=> a^2 - 25 =0
<=> a = 5
hay x^2 + 5x + 5 = 5
<=> x(x+5) = 5 - 5 = 0
<=> x=0 hoặc x+5 = 0 <=> x= -5
Vậy tập ngh của p.tr là S = { 0; -5 }
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{x^2+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}+\dfrac{1}{x^2+14x+48}=\dfrac{4}{105}\)
\(\Leftrightarrow\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}=\dfrac{8}{105}\)
\(\Leftrightarrow\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)+\left(\dfrac{1}{x+2}-\dfrac{1}{x+4}\right)+\left(\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)+\left(\dfrac{1}{x+6}-\dfrac{1}{x+8}\right)=\dfrac{8}{105}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{105}\)
\(\Leftrightarrow\dfrac{8}{x\left(x+8\right)}=\dfrac{8}{105}\)
\(\Leftrightarrow x\left(x+8\right)=105\)
\(\Leftrightarrow x^2+8x-105=0\)
\(\Leftrightarrow x^2-7x+15x-105=0\)
\(\Leftrightarrow x\left(x-7\right)+15\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-15\end{matrix}\right.\)
Thử lại ta có nghiệm của phương trình trên là \(x=7\text{v}à\text{x}=15\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 $x=-3$
2
x(x - 1) ( x + 1) (x + 2 ) = 24
<=> [x(x + 1)][(x - 1)(x + 2)] = 24
<=> (x^2 + x)(x^2 + x - 2) = 24
đặt x^2 + x = a
<=> a(a - 2) = 24
<=> a^2 - 2a = 24
<=> a^2 - 2a - 24 = 0
<=> a^2 + 4a - 6a - 24 = 0
<=> a(a + 4) - 6(a + 4) = 0
<=> (a - 6)(a + 4) = 0
<=> a = 6 hoặc a = -4
a = 6 => x^2 + x = 6
<=> x^2 + x - 6 = 0
<=> (x + 3)(x - 2) = 0
<=> x = - 3 hoặc x = 2
a = -4 => x^2 + x + 4 = 0
mà x^2 + x + 4 > 0
=> vô lí
vậy x = -3 hoặc x = 2