(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

(x^2+3x+2)(x^2+7X+12)=24

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

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

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

đặt \(x^2+5x+5=a\)=> ta có phương trình \(\Leftrightarrow\left(a-1\right)\left(a+1\right)=24\)

\(\Leftrightarrow a^2-1=24\)\(\Leftrightarrow a^2=25\Leftrightarrow a=\orbr{\begin{cases}5\\-5\end{cases}}\)

+)\(x^2+5x+5=5\)\(\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\)

\(\Leftrightarrow x=\orbr{\begin{cases}5\\0\end{cases}}\)

+) \(x^2+5x+5=-5\)\(\Leftrightarrow x^2+5x+10=0\)\(\Rightarrowđenta=5^2-4.10=-15< 0\Rightarrow ptvonghiem\)

vậy \(x=\orbr{\begin{cases}0\\5\end{cases}}\)

( x^2 + 3x + 2 )( x^2 + 7x + 12 ) = 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 x² + 5x + 5 = a = ta có : \(\Leftrightarrow\left(a-1\right)\left(a+1\right)=24\)

\(\Leftrightarrow a^2-1=24\Leftrightarrow a^2=25\Leftrightarrow a=\orbr{\begin{cases}5\\-5\end{cases}}\)

+)\(x^2+5x+5=5\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\)

\(\Leftrightarrow x=\orbr{\begin{cases}5\\0\end{cases}}\)

+)\(x^2+5x+5=-5\Leftrightarrow x^2+5x+10=0\)

\(\Rightarrowđenta=5^2-4.10=-15< 0\Rightarrow ptvonghiem\)

\(Vay.x=\orbr{\begin{cases}5\\0\end{cases}}\)

(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

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

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

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

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

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

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)

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

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

\(\Rightarrow\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)-24=0\)

\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)

\(\Rightarrow\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-24=0\)

\(\Rightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

Đặt \(x^2+5x+4=t\Rightarrow x^2+5x+6=t+2\) ta được:

\(t\left(t+2\right)-24=0\Rightarrow t^2+2t-24=0\)

\(\Rightarrow t^2-4t+6t-24=0\Rightarrow\left(t-4\right)\left(t-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}t-4=0\\t-6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}t=4\\t=6\end{matrix}\right.\)

Vì \(t=x^2+5x+4\) nên

\(\left[{}\begin{matrix}x^2+5x+4=4\\x^2+5x+6=6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\left(x+5\right)=0\\x\left(x+5\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy................

Chúc bạn học tốt!!!

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

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

\(\Leftrightarrow\left(x^2+5x+4\right)^2+2\left(x^2+5x+4\right)-24=0\)

\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+10\right)=0\)

b/ ĐKXĐ; ...

\(\Leftrightarrow\frac{x^2}{x^2+4x+4}+12x+5=3x^2+6x+2\)

\(\Leftrightarrow\frac{x^2+\left(12x+5\right)\left(x^2+4x+4\right)}{x^2+4x+4}=3x^2+6x+2\)

\(\Leftrightarrow\frac{\left(4x+10\right)\left(3x^2+6x+2\right)}{x^2+4x+4}=3x^2+6x+2\)

\(\Leftrightarrow\left[{}\begin{matrix}3x^2+6x+2=0\\\frac{4x+10}{x^2+4x+4}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3x^2+6x+2=0\\x^2=6\end{matrix}\right.\)

a) Ta có :

(x² + 3x + 2)(x² + 7x + 12) = 24

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

⇔ ( x + 1 ) ( x + 4 ) ( x + 2 ) ( x + 3 ) - 24 = 0

⇔ ( x² + 5x + 4 ) ( x² + 5x + 6 ) - 24 = 0

Đặt t = x² + 5x + 4, ta có :

t ( t + 2 ) - 24 = 0

⇔ t² + 2t +1 - 25 = 0

⇔ ( t + 1 )² - 5² = 0

⇔ ( t - 4 ) ( t + 6 ) = 0

\(\Leftrightarrow\left[{}\begin{matrix}t-4=0\\t+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=4\\t=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+4=4\\x^2+5x+4=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+5x=0\\x^2+5x+10=0\end{matrix}\right.\)

Sau đó tìm x bạn tự làm nha

Ý b) là - 3 à !?

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

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

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

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

Đặt: \(x^2+5x+5=y\) ta được: \(\left(y-1\right)\left(y+1\right)=24\)

\(\Leftrightarrow y^2=25\Leftrightarrow\left[{}\begin{matrix}y=-5\\y=5\end{matrix}\right.\)

Với: \(y=-5\Rightarrow x^2+5x+5=-5\Leftrightarrow x^2+5x+10=0\)

Có: \(\Delta=-15< 0\) vô nghiệm.

Với: \(y=5\Rightarrow x^2+5x+5=5\Leftrightarrow\left(x+5\right)x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy pt có tập \(n_0S=\left\{0;-5\right\}\)

\(b,\frac{x^2}{\left(x+2\right)^2}=3x^2-6x-3\left(Đkxđ:x\ne-2\right)\)

\(\Leftrightarrow x^2=\left(x+2\right)^2\left(3x^2-6x-3\right)\)

\(\Leftrightarrow\left(x^2+4x+4\right)\left(3x^2-6x-3\right)-x^2=0\)

\(\Leftrightarrow\left(x^2+4x+4\right)\left(3x^2+6x+1\right)+\left(x^2+4x+4\right)\left(-12x-5\right)-x^2=0\)

\(\Leftrightarrow\left(x^2+4x+4\right)\left(3x^2+6x+2\right)-\left(12x^3+48x^2+48x+5x^2+20x+20+x^2\right)=0\)

\(\Leftrightarrow\left(x^2+4x+4\right)\left(3x^2+6x+2\right)-2\left(2x+5\right)\left(3x^2+6x+2\right)=0\)

\(\Leftrightarrow\left(3x^2+6x+2\right)\left(x^2-6\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\pm\sqrt{6}\\x=-1\pm\frac{1}{\sqrt{3}}\end{matrix}\right.\)

Vậy ..............

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 \(x=\dfrac{5}{2}\)

4.

\((2x+7)(x+3)^2(2x+5)=18\)

\(\Leftrightarrow [(2x+7)(2x+5)](x+3)^2=18\)

\(\Leftrightarrow (4x^2+24x+35)(x^2+6x+9)=18\)

\(\Leftrightarrow [4(x^2+6x+9)-1](x^2+6x+9)=18\)

\(\Leftrightarrow (4a-1)a=18\) (đặt \(x^2+6x+9=a\) )

\(\Leftrightarrow 4a^2-a-18=0\)

\(\Leftrightarrow (4a-9)(a+2)=0\Rightarrow \left[\begin{matrix} a=\frac{9}{4}\\ a=-2\end{matrix}\right.\)

Nếu \(a=x^2+6x+9=\frac{9}{4}\Leftrightarrow (x+3)^2=\frac{9}{4}\)

\(\Rightarrow \left[\begin{matrix} x+3=\frac{3}{2}\\ x+3=\frac{-3}{2}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-3}{2}\\ x=\frac{-9}{2}\end{matrix}\right.\)

Nếu \(a=x^2+6x+9=-2\Leftrightarrow (x+3)^2=-2< 0\) (vô lý)

Vậy ............

5.

PT \(\Leftrightarrow (x-1)(x-2)(2x-3)(2x-5)=30\)

\(\Leftrightarrow [(x-1)(2x-5)][(x-2)(2x-3)]=30\)

\(\Leftrightarrow (2x^2-7x+5)(2x^2-7x+6)=30\)

Đặt \(2x^2-7x+5=a\) thì:

PT \(\Leftrightarrow a(a+1)=30\)

\(\Leftrightarrow a^2+a-30=0\)

\(\Leftrightarrow (a-5)(a+6)=0\Rightarrow \left[\begin{matrix} a-5=0\\ a+6=0\end{matrix}\right.\)

Nếu \(a-5=0\Leftrightarrow 2x^2-7x=0\Leftrightarrow x(2x-7)=0\)

\(\Rightarrow \left[\begin{matrix} x=0\\ x=\frac{7}{2}\end{matrix}\right.\)

Nếu \(a+6=0\Leftrightarrow 2x^2-7x+11=0\)

\(\Leftrightarrow 2(x-\frac{7}{4})^2+\frac{39}{8}=0\Leftrightarrow 2(x-\frac{7}{4})^2=-\frac{39}{8}<0\) (vô lý)

Vậy...........

6.

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

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

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

\(\Leftrightarrow (a+4)(a+6)=24\) (đặt \(x^2+5x=a\) )

\(\Leftrightarrow a^2+10a=0\Leftrightarrow a(a+10)=0\Rightarrow \left[\begin{matrix} a=0\\ a+10=0\end{matrix}\right.\)

Nếu \(a=0\Leftrightarrow x^2+5x=0\Leftrightarrow x(x+5)=0\Leftrightarrow \left[\begin{matrix} x=0\\ x=-5\end{matrix}\right.\)

Nếu \(a+10=0\Leftrightarrow x^2+5x+10=0\Leftrightarrow (x+\frac{5}{2})^2+\frac{15}{4}=0\)

\(\Leftrightarrow (x+\frac{5}{2})^2=-\frac{15}{4}\) (vô lý)

Vậy...........

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