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~

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

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

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

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

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

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

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

\(x^3+2x^2+2x+1=\left(x^3+1\right)+\left(2x^2+2x\right)\)

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

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

\(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27\)

\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-x+9\right)\)

\(x^4+2x^3+2x^2+2x+1=x^4+x^2+2x^3+x^2+2x+1\)

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

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

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

\(x^4-2x^3+2x-1=\left(x^4-1\right)-2x\left(x^2-1\right)\)

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

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

\(x^3+2x^2+2x+1=\left(x^3+x^2\right)+\left(x^2+x\right)+\left(x+1\right)\)

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

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

\(x^3-4x^2+12x-27\)

\(=\left(x^3-x^2\right)-\left(3x^2-3x\right)+\left(9x-27\right)\)

\(=x^2.\left(x-1\right)-3x.\left(x-1\right)+9.\left(x-3\right)\)

\(=\left(x-1\right).\left(x^2-3x\right)+9.\left(x-3\right)\)

\(=x.\left(x-1\right).\left(x-3\right)+9.\left(x-3\right)\)

\(=\left(x-3\right)\left[x.\left(x-1\right)+9\right]\)

a, \(\left(5n+2\right)^2-4=\left(5n+2-2\right)\left(5n+2+2\right)=5n\left(5n+4\right)⋮5\)

b, \(n^3-n=n\left(n^2-1\right)=\left(n-1\right)n\left(n+1\right)\)

Vì (n-1)n(n+1) là tích 3 số nguyên liên tiếp

=>(n-1)n(n+1) chia hết cho 6 hay n^3-n chia hết cho 6

c, \(a+b+c=0\Rightarrow a+b=-c\)

\(\Rightarrow\left(a+b\right)^3=\left(-c\right)^3\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\Rightarrow a^3+b^3-3abc=-c^3\)

=>a^3+b^3+c^3=3abc