\(x^3+x^2+4=0\)

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

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

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

\(\Leftrightarrow\left(x+2\right)\left[\left(x^2-2.\frac{1}{2}x+\frac{1}{4}\right)+\frac{7}{4}\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left[\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\right]=0\)

\(\text{Vì}\)\(\left[\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\right]>0\)

\(\text{Nên x + 2 = 0 }\Leftrightarrow x=-2\)

\(\text{Vậy x = -2}\)

     \(x^3+x^2+4=0\)

\(\Rightarrow x^3+2x^2-x^2-2x+2x+4=0\)

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

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

Mà \(x^2-x+2=\left(x-\frac{1}{2}\right)^2+\frac{7}{4}>0\forall x\)

Do đó: \(x+2=0\Rightarrow x=-2\)

a) \(x^3+2x^2y+xy^2-4xz^2=x\left(x^2+2xy+y^2-4z^2\right)=x\left[\left(x+y\right)^2-\left(2z\right)^2\right]\)

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

b)\(-8x^3+12x^2y-6xy^2+y^3=y^3+3.y.\left(2x\right)^2-3.y^2.2x-\left(2x\right)^3\)\(=\left(y-2x\right)^3\)

c)\(6x^2+7x-5=2x\left(3x+5\right)-\left(3x+5\right)=\left(3x+5\right)\left(2x-1\right)\)

d)\(x^4+64y^4=\left(x^2\right)^2+2.x^2.8y^2+\left(8y^2\right)^2-16x^2y^2=\left(x^2+8y^2\right)-\left(4xy\right)^2\)

\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)

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

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

h)\(3x^2-6xy+3y^2-12z^2=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

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

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

B2: \(x^3-5x=0\Rightarrow x\left(x^2-5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x^2-5=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=5\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{5}\end{cases}}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=5\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\\orbr{\begin{cases}x=\sqrt{5}\\x=-\sqrt{5}\end{cases}}\end{cases}}\)

\(6a^2b-20ab^2+14ab\)

\(=2ab.\left(3a-10b+7\right)\)

Tham khảo nhé~

=a²b.(6-20+14)

k mik nha

Đặt thương khi chia x⁴+ax+b cho x² - 1 là Q(x), ta có:

\(x^4+ax+b=\left(x^2-1\right)Q\left(x\right)=\left(x-1\right)\left(x+1\right)Q\left(x\right)\)

Vì đẳng thức đúng với mọi x nên lần lượt cho x = 1 và  x = -1, ta được:

\(\hept{\begin{cases}1+a+b=0\\1-a+b=0\end{cases}\Leftrightarrow\hept{\begin{cases}a+b=-1\\a-b=1\end{cases}\Leftrightarrow}\hept{\begin{cases}a=0\\b=-1\end{cases}}}\)

Vậy với a = 0, b = -1 thì \(x^4+ax+b⋮x^2-1\)

P/s: đây là pp xét giá trị riêng

Đặt: \(A=\left(n^2+10\right)^2-36n^2\)

\(=\left(n^2+10\right)^2-\left(6n\right)^2\)

\(=\left(n^2-6n+10\right)\left(n^2+6n+10\right)\)

Vì \(n\in N\Rightarrow n^2+6n+10\ge10\)

Điều kiện cần để A là số nguyên tố:

     \(n^2-6n+10=1\)

\(\Rightarrow n^2-6n+9=0\)

\(\Rightarrow\left(n-3\right)^2=0\Rightarrow n=3\)

Ta phải thử lại:

\(A=\left(n^2+10\right)^2-36n^2=\left(3^2+10\right)^2-36.3^2=19^2-324=37\)

Vì 37 là số nguyên tố nên n = 3 thỏa mãn đề bài.