\(\left(x-1\right)\left(x+5\right)=x^2+4x-5\)(1)

\(x\left(x+4\right)=x^2+4x\)(2)

Lấy (1) nhân (2) \(\Leftrightarrow y.\left(y-5\right)=84\Leftrightarrow y^2-5y+\left(\frac{5}{2}\right)^2=84+\frac{25}{4}=\left(\frac{19}{2}\right)^2\)

\(\orbr{\begin{cases}y=\frac{5-19}{2}=-7\left(loai\right)\\y=\frac{5+19}{2}=12\end{cases}}\) 

\(x^2+4x=12\Leftrightarrow\left(x+2\right)^2=16\Rightarrow\orbr{\begin{cases}x=-6\\x=2\end{cases}}\)

 x(x - 1)(x + 4)(x + 5) = 84 

<=> x(x + 4)(x - 1)(x + 5) = 84 

<=> (x² + 4x)(x² + 4x - 5) - 84 = 0 

Đặt t = x² + 4x ta có 

t(t - 5) - 84 = 0 

<=> t² - 5t - 84 = 0 

<=> t² + 7t - 12t - 84 = 0 

<=> t(t + 7) - 12(t + 7) = 0 

<=> (t - 12)(t + 7) = 0 

<=> t = 12 hoặc t = -7 

Với t = 12 ta có 

x² + 4x = 12 

<=> x² + 4x - 12 = 0 

<=>x² - 2x + 6x - 12 = 0 

<=> x(x - 2) + 6(x - 2) = 0 

<=> (x + 6)(x - 2) = 0 

<=> x = -6 hoặc x = 2 

Với x = - 7 ta có 

x² + 4x = -7 

<=> x² + 4x + 7 = 0 

<=> x² + 4x + 4 + 3 =0 

<=> (x + 2)² + 3 = 0 

Lại có (x + 2)² + 3 > 0 với mọi x 

=> pt vô nghiệm 

Kết luận nghiêm x = - 6 ; x = 2

\(Tacó\)

\(x\left(x-1\right)\left(x+4\right)\left(x+5\right)=\left[\left(x-1\right)\left(x+5\right)\right]\left[x\left(x+4\right)\right]\)

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

\(Đặt:x^2+4x=t\)pt trở thành:

\(\left(t-5\right)t=84=7.12\Leftrightarrow t=12\)

\(\Leftrightarrow x^2+4x=12\Leftrightarrow x\left(x+4\right)=12=2.6\Leftrightarrow x=2\)

\(Vậy:x=2\)

\(x\left(x-1\right)\left(x+4\right)\left(x+5\right)=84\)

\(\Leftrightarrow x^4+8x^3+11x^2-20x=84\)

\(\Leftrightarrow x^4+8x^3+11x^2-20x=84-84\)

\(\Leftrightarrow x^4+8x^3+11x^2-20x=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-6\right)\left(x^2+4x+7\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=4\\x=-6\end{cases}}\)

\(\dfrac{x+1}{98}+\dfrac{x+2}{97}+\dfrac{x+90}{9}+\dfrac{x+84}{15}>-4\\ \Leftrightarrow\left(\dfrac{x+1}{98}+1\right)+\left(\dfrac{x+2}{97}+1\right)+\left(\dfrac{x+90}{9}+1\right)+\left(\dfrac{x+84}{15}+1\right)>0\\ \Leftrightarrow\dfrac{x+99}{98}+\dfrac{x+99}{97}+\dfrac{x+99}{9}+\dfrac{x+99}{15}>0\\ \Leftrightarrow\left(x+99\right)\left(\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{9}+\dfrac{1}{15}\right)>0\)

Vì \(\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{9}+\dfrac{1}{15}>0\Rightarrow x+99>0\Rightarrow x>-99\)

cho tớ hỏi ông  

bn ilovemath ơi

\(a,x\left(x-1\right)\left(x+4\right)\left(x+5\right)=84\)

\(\Leftrightarrow\left[x\left(x+4\right)\right]\left[\left(x-1\right)\left(x+5\right)\right]=84\)

\(\Leftrightarrow\left(x^2+4x\right)\left(x^2+4x-5\right)=84\)

Đặt \(x^2+4x=a\)

Ta có : \(a=x^2+4x+4-4=\left(x+2\right)^2-4\ge-4\)

\(\Rightarrow a\ge-4\)

\(Ta\text{ }co'\text{ }pt:a\left(a-5\right)=84\)

\(\Leftrightarrow a^2-5a-84=0\)

\(\Leftrightarrow\left(a-12\right)\left(a+7\right)=0\)

Mà \(a\ge-4\Rightarrow a=12\)

                       \(\Rightarrow x^2+4x=12\)

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

                        \(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)

\(b,x^3-5x^2+8x-4=0\)

\(\Leftrightarrow\left(x-2\right)^2\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=1\end{cases}}\)

dong ho chi may gio

a: \(\Leftrightarrow5\sqrt{x+3}-4\sqrt{x+3}=3\sqrt{x-2}-3\sqrt{x-2}+2\)

\(\Leftrightarrow\sqrt{x+3}=2\)

=>x+3=4

hay x=1

c: \(\Leftrightarrow\left(x^2+4x\right)\left(x^2+4x-5\right)=84\)

\(\Leftrightarrow\left(x^2+4x\right)^2-5\left(x^2+4x\right)-84=0\)

\(\Leftrightarrow\left(x^2+4x\right)^2-12\left(x^2+4x\right)+7\left(x^2+4x\right)-84=0\)

\(\Leftrightarrow x^2+4x-12=0\)

=>(x+6)(x-2)=0

=>x=-6 hoặc x=2

Đáp án B

Đáp án B

Đáp án B