Vì mình mới họ định lí mới nên minhfm uốn làm thử nếu cậu không hiểu tì hỏi mình để mình làm cách bình thường .

a ) Áp dụng định lí Bezout :
Đặt \(f\left(x\right)=x^3-7x-6,\) ta thấy \(f\left(-1\right)=0\) nên \(-1\) là một ước của \(f\left(x\right)\).

Vậy \(f\left(x\right)\) chia hết cho \(\left(x+1\right)\). Ta có : \(f\left(x\right)=\left(x+1\right)\left(x^2-x-6\right)\)

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

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

b ) Áp dụng định lí Bezout :

Đặt \(f\left(x\right)=x^3-19x-30.\)Xét một số ước của 30 , ta được \(f\left(-2\right)=0\).

Ta chia \(f\left(x\right)\) cho \(\left(x+2\right);f\left(x\right)=\left(x+2\right)\left(x^2-2x-15\right)\)

\(x^2-2x-15\) nhận \(x=5\) làm nghiệm .

Do vậy \(f\left(x\right)=\left(x+2\right)\left(x+3\right)\left(x-5\right)\)

Chúc bạn học tốt

\(x^3-5x^2-14x\)

\(=x^3+2x^2-7x^2-14x\)

\(=x^2\left(x+2\right)-7x\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-7x\right)\)

\(=x\left(x+2\right)\left(x-7\right)\)

\(x^3-7x-6\)

\(=x^3+x^2-x^2-x-6x-6\)

\(=x^2\left(x+1\right)-x\left(x+1\right)-6\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-x-6\right)\)

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

\(=\left(x+1\right)\left[x\left(x+2\right)-3\left(x+2\right)\right]\)

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

\(x^3-19x-30\)

\(=x^3-5x^2+5x^2-25x+6x-30\)

\(=x^2\left(x-5\right)+5x\left(x-5\right)+6\left(x-5\right)\)

\(=\left(x-5\right)\left(x^2+5x+6\right)\)

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

\(=\left(x-5\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)

\(=\left(x-5\right)\left(x+3\right)\left(x+2\right)\)

a ) \(x^3-7x-6=x^3-x-6x-6=x^3-x-6\left(x+1\right)\)

\(=x\left(x^2-1\right)-6\left(x+1\right)=\left(x+1\right)\left[x\left(x-1\right)-6\right]\)

\(=\left(x+1\right)\left[\left(x^2-x-6\right)\right]=\left(x+1\right)\left[\left(x^2+2x-3x-6\right)\right]\)

\(=\left(x+1\right)\left[x\left(x+2\right)-3\left(x+2\right)\right]=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)

b )

\(x^3-19x-30=\left(x^3-9x\right)-\left(10x+30\right)=x\left(x^2-9\right)-10\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2-3x-10\right)=\left(x+2\right)\left(x+3\right)\left(x-5\right)\)

c )

\(a^3-6a^2+11a-6=\left(a-3\right)\left(a-2\right)\left(a-1\right).\)

a

\(x^2-3x-2=\left(x^2-3x+\frac{9}{4}\right)-\frac{17}{4}=\left(x-\frac{3}{2}\right)^2-\sqrt{\frac{17}{2}}^2\)

\(=\left(x-\frac{3}{2}-\sqrt{\frac{17}{2}}\right)\left(x-\frac{3}{2}+\sqrt{\frac{17}{2}}\right)\)

b

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

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

c

\(x^3-19x-30=x^3+2x^2-2x^2-4x-15x-30\)

\(=x^2\left(x+2\right)-2x\left(x+2\right)-15\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-2x-15\right)\)

Đề sai rồi nhé .... Bấm máy tính là biết.

Biết câu nào làm câu đấy thoy nha :))

3. \(x^4y^4+4\)

\(=\left(x^2y^2\right)^2+2\cdot x^2y^2\cdot2+2^2-2\cdot x^2y^2\cdot2\)

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

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

4. \(x^4+4y^4\)

\(=\left(x^2\right)^2+2\cdot x^2\cdot2y^2+\left(2y^2\right)^2-2\cdot x^2\cdot2y^2\)

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

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

2. \(x^4+x^2+1\)

\(=\left(x^2\right)^2+2\cdot x^2\cdot1+1^2-2x^2\)

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

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