\(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow x^2+8x+16-\left(x^2-1\right)=0\)

\(\Leftrightarrow x^2+8x+16-x^2+1=0\)

\(\Leftrightarrow8x=-17\)

\(\Leftrightarrow x=\frac{-17}{8}\)

b) \(x^2-2x=24\)

\(\Leftrightarrow x^2-2x-24=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=6\end{cases}}}\)

Vậy x = 6 hoặc x = -4

a) x^2 + 8x + 16 - x^2 + 1 = 0
<=> 8x = -17
<=> x = -17/8
b) x^2 - 2x - 24 = 0 
<=> x^2 - 2x + 1 - 25 = 0
<=> ( x -1)^2 -25 = 0
<=> (x - 1)^2 = 25
<=>  x - 1 = 5  hoặc  x - 1 = -5
<=>  x = 6       hoặc  x = -4

\(x^4+x^2+1\)

\(=x^4+2x^2+1-x^2\)

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

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

Tương tự như cách giải vừa nãy, ta có:  

\(P\left(x\right)=ax^3+bx^2+5x-50=Q\left(x\right).\left(x-2\right)\left(x+5\right)\)

  Thay lần lượt  \(x=2,x=-5\)(để vế phải = 0), ta được: 

\(\hept{\begin{cases}a.2^3+b.2^2+5.2-50=0\\a.\left(-5\right)^3+b.\left(-5\right)^2+5.\left(-5\right)-50=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}8a+4b+10-50=0\\-125a+25b-25-50=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}8a+4b=40\\-125a+25b=75\end{cases}}\Rightarrow\hept{\begin{cases}2a+b=10\\-5a+b=3\end{cases}\Rightarrow\hept{\begin{cases}2a+b-\left(-5a-b\right)=10-3\\-5a+b=3\end{cases}}}\) 

\(\Rightarrow\hept{\begin{cases}7a=7\\-5a+b=3\end{cases}\Rightarrow\hept{\begin{cases}a=1\\-5+b=3\end{cases}\Rightarrow}\hept{\begin{cases}a=1\\b=8\end{cases}}}\)

Vậy \(a=1,b=8\)

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

             \(=x^4+4x^2+4-4x^2\)

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

             \(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)⋮\left(x^2+ax+b\right)\)

Vậy \(\orbr{\begin{cases}a=-2,b=2\\a=2,b=2\end{cases}}\)

Chúc bạn học tốt.

\(\Leftrightarrow30xy-72x+55y-132-42x+16=0\)

\(\Leftrightarrow30xy-72x+55y-42x=0-16+132\)

........................................ Bạn tự làm tiếp nhé!!!

<=> x(x+2)²(x+4)=1.3².5
=> x=1
Vậy.............
Nghĩ v thui :)
 

ĐK: \(x\ne-4\)

\(x\left(x+2\right)^2=\frac{45}{x+4}\)

=>  \(x\left(x+2\right)^2\left(x+4\right)=45\)

<=> \(\left(x^2+4x\right)\left(x^2+4x+4\right)-45=0\)

Đặt:  \(x^2+4x+2=t\)

Khi đó pt trở thành:

\(\left(t-2\right)\left(t+2\right)-45=0\)

<=> \(t^2-49=0\)

<=> \(t=\pm7\)

đến đây thay trở lại đc pt bậc 2 ez bn lm nốt nhé

\(x^2+4x-y^2+4\)

\(=\left(x^2+2.x.2+2^2\right)-y^2\)

\(=\left(x+2\right)^2-y^2\)

\(=\left(x+2-y\right).\left(x+2+y\right)\)

Tham khảo nhé~

\(x^2+4x-y^2+4\)

\(=x^2+4x+4-y^2\)

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

\(=\left(x^2+2x.2+2^2\right)-y^2\)

\(=\left(x+2\right)^2-y^2\)

\(=\left[\left(x+2\right)+y^2\right].\left[\left(x+2\right)-y^2\right]\)

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