(x+1)(x+2)(x+3)(x+4)=24

(x+1)(x+4)(x+2)(x+3)=24

(x\(^2\)+5x+4)(x² +5x+6)=24

Đặt x²+5x+5=t

\(\Rightarrow\)(t+1)(t-1)=24

\(\Rightarrow\) t² -1=24

\(\Rightarrow\) t²-25=0

\(\Rightarrow\) (t-5)(t+5)=0

\(\Rightarrow\) (x²+5x)(x²+5x+10)=0

\(\Rightarrow\) x(x+5)(x+5)²=0

\(\Rightarrow\) x(x+5)³=0

\(\Rightarrow\) x=0 hoặc (x+5)³=0

Vậy x=0 hoặc x= -5

(x² + 3x + 2)(x² + 7x + 12) = 24

=> (x + 1)(x + 2)(x + 3)(x + 4) = 24

=> (x² + 5x + 4)(x² + 5x + 6) = 24

Đặt a = x² + 5x + 4 ta được:

a.(a + 2) = 24 => a² + 2a - 24 = 0 => (a - 4)(a + 6) = 0 => a = 4 hoặc a = -6

+ Với a = 4 => x² + 5x + 4 = 4 => x² + 5x = 0 => x(x + 5) = 0 => x = 0 hoặc x = -5

+ Với a = -6 => x² + 5x + 4 = -6 => x² + 5x + 10 = 0, mà x² + 5x + 10 > 0 => vô nghiệm

                                              Vậy x = 0 , x = -5

         \(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)

\(\Leftrightarrow\frac{\frac{5x+6-2x}{5}}{14}-\frac{x+4}{24}=\frac{\frac{35x+10+9-3x}{5}}{12}+\frac{2}{3}\)

\(\Leftrightarrow\frac{\frac{3x+6}{5}}{14}-\frac{x+4}{24}=\frac{\frac{32x+19}{5}}{12}+\frac{2}{3}\)

\(\Leftrightarrow\left(\frac{3x+6}{5}\cdot\frac{1}{14}\right)-\frac{x+4}{24}=\left(\frac{32x+19}{5}\cdot\frac{1}{12}\right)+\frac{2}{3}\)(CHIA CHO 14 LÀ NHÂN NGHỊCH ĐẢO VỚI 1/14,)                                                                                                                           (CHIA CHO 12 LÀ NHÂN NGHỊCH ĐẢO VỚI 1/12)\(\Leftrightarrow\frac{3x+6}{70}-\frac{x+4}{24}-\frac{32x+19}{60}-\frac{2}{3}=0\)\(\Leftrightarrow\frac{12\left(3x+6\right)-35\left(x+4\right)-14\left(32x+19\right)-2\cdot280}{840}=0\)

 \(\Leftrightarrow12\left(3x+6\right)-35\left(x+4\right)-14\left(32x+19\right)-560=0\)

\(\Leftrightarrow36x+72-35x-140-448x-266-560=0\)

 \(\Leftrightarrow-447x-894=0\Leftrightarrow x=\frac{-894}{447}=-2\)(NHẬN)

 Vậy tập nghiệm của phương trình là : S = { -2 }

tk cho mk nka ! ! ! th@nks ! ! !

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)

\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]=24\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)

\(\Leftrightarrow\left(x^2+5x+5\right)^2-1=24\)

\(\Leftrightarrow\left(x^2+5x+5\right)^2=25\)

Mà \(x^2+5x+5>0\forall x\)

\(\Rightarrow x^2+5x+5=5\Rightarrow x\left(x+5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

Vậy pt có tập nghiệm S={0,-5}

pt <=> (x+1).(x+2).(x+3).(x+4) = 24

<=> [(x+1).(x+4)].[(x+2).(x+3)] = 24

<=> (x^2+5x+4).(x^2+5x+6) = 24

<=> (x^2+5x+5)^2-1 = 24

<=> (x^2+5x+5) = 25

=> x^2+5x+5 = 5 [ vì x^2+5x+5 = (x+2,5)^2-0,25 >= -0,25 > -5 ]

=> x=0 hoặc x=-5 

Vậy pt có tập nghiệm S = {-5;0}

k mk nha

\(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)

\(\Leftrightarrow\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)=24\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)

Đặt \(t=x^2+5x+5\)ta có:

\(\left(t-1\right)\left(t+1\right)=24\)

\(\Leftrightarrow t^2-1=24\)

\(\Leftrightarrow t=\pm-5\)

Với t = 5 thì \(x^2+5x+5=5\Leftrightarrow x.\left(x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

Với t = -5 thì \(x^2+5x+5=-5\Leftrightarrow x^2+5x+10=0\)(vô nghiệm)

\(\Rightarrow PT=\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

1/ \(7x-5=13-5x\)

\(\Leftrightarrow12x=18\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy: \(S=\left\{\dfrac{3}{2}\right\}\)

==========

2/ \(19+3x=5-18x\)

\(\Leftrightarrow21x=-14\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)

Vậy: \(S=\left\{-\dfrac{2}{3}\right\}\)

==========

3/ \(x^2+2x-4=-12+3x+x^2\)

\(\Leftrightarrow-x=-8\)

\(\Leftrightarrow x=8\)

Vậy: \(S=\left\{8\right\}\)

===========

4/ \(-\left(x+5\right)=3\left(x-5\right)\)

\(\Leftrightarrow-x-5=3x-15\)

\(\Leftrightarrow-4x=-10\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

Vậy: \(S=\left\{\dfrac{5}{2}\right\}\)

==========

5/ \(3\left(x+4\right)=\left(-x+4\right)\)

\(\Leftrightarrow3x+12=-x+4\)

\(\Leftrightarrow4x=-8\)

\(\Leftrightarrow x=-2\)

Vậy: \(S=\left\{-2\right\}\)

[----------]

1. \(7x-5=13-5x\) \(\Leftrightarrow12x=18\Leftrightarrow x=\dfrac{3}{2}\)

2. \(19+3x=5-18x\Leftrightarrow21x=-14\Leftrightarrow x=-\dfrac{2}{3}\)

3. \(x^2+2x-4=-12+3x+x^2\Leftrightarrow-x=-8\Leftrightarrow x=8\)

4. \(-\left(x+5\right)=3\left(x-5\right)\Leftrightarrow-x-5=3x-15\Leftrightarrow4x=10\Leftrightarrow x=\dfrac{5}{2}\)

5. \(3\left(x+4\right)=-x+4\Leftrightarrow3x+12=-x+4\Leftrightarrow4x=-8\Leftrightarrow x=-2\)

1) Ta có: \(7x-5=13-5x\)

\(\Leftrightarrow12x=18\)

hay \(x=\dfrac{3}{2}\)

2) Ta có: \(19+3x=5-18x\)

\(\Leftrightarrow21x=-14\)

hay \(x=-\dfrac{2}{3}\)

3) Ta có: \(x^2+2x-4=x^2+3x-12\)

\(\Leftrightarrow3x-12=2x-4\)

hay x=8

4) Ta có: \(-\left(x+5\right)=3\left(x-5\right)\)

\(\Leftrightarrow-x-5-3x+15=0\)

\(\Leftrightarrow-4x=-10\)

hay \(x=\dfrac{5}{2}\)

Câu 1: 

a) Ta có: 7x+21=0

\(\Leftrightarrow7x=-21\)

hay x=-3

Vậy: S={-3}

b) Ta có: 3x-2=2x-3

\(\Leftrightarrow3x-2-2x+3=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

Vậy: S={-1}

c) Ta có: 5x-2x-24=0

\(\Leftrightarrow3x=24\)

hay x=8

Vậy: S={8}

Câu 2: 

a) Ta có: \(\left(2x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-1\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{2};1\right\}\)

b) Ta có: \(\left(2x-3\right)\left(-x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\-x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\-x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=7\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};7\right\}\)

c) Ta có: \(\left(x+3\right)^3-9\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left[\left(x+3\right)^2-9\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+3-3\right)\left(x+3+3\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-6\end{matrix}\right.\)

Vậy: S={0;-3;-6}

\(\frac{3x^2-7x+5}{x^2-x-x}-x+\frac{1}{x+1}< 0\Leftrightarrow\frac{x^2-6x+11}{\left(x-2\right)\left(x+1\right)}< 0\Leftrightarrow\frac{\left(x-3\right)^2+2}{\left(x-2\right)\left(x+1\right)}< 0\)

=> (x-2)(x+1)<0 ( vì (x-3)^2+2>0 lđ)

lại có x+1>x-2 => x-2<0 và x+1>0

=> -1<x<2

học tốt

Cho mình làm lại nha:

\(\frac{3x^2-7x+5}{\left(x+1\right)\left(x-2\right)}< \frac{2x+2-1}{x+1}.\)

\(\Leftrightarrow\frac{3x^2-7x+5}{\left(x+1\right)\left(x-2\right)}-\frac{2x+1}{x+1}< 0.\)

\(\Leftrightarrow\frac{3x^2-7x+5-\left(2x+1\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}< 0.\)

\(\Leftrightarrow\frac{3x^2-7x+5-2x^2+4x-x+2}{\left(x+1\right)\left(x-2\right)}< 0.\)

\(\Leftrightarrow\frac{x^2-4x+4+3}{\left(x+1\right)\left(x-2\right)}< 0.\)

\(\Leftrightarrow\frac{\left(x-2\right)^2+3}{\left(x+1\right)\left(x-2\right)}< 0\Leftrightarrow\left(x+1\right)\left(x-2\right)< 0.\)

ta có x+1>x-2 => x+1>0;x-2<0 => -1<x<2

đọc lộn xíu xin lỗi nha

học tốt

a³-b³ = (a-b)(a²-ab+b²) , áp dung hằng đẳng thức rồi phân tích nha bạn 

1:

a: =>3x=6

=>x=2

b: =>4x=16

=>x=4

c: =>4x-6=9-x

=>5x=15

=>x=3

d: =>7x-12=x+6

=>6x=18

=>x=3

2:

a: =>2x<=-8

=>x<=-4

b: =>x+5<0

=>x<-5

c: =>2x>8

=>x>4