Ta có:   x^3 + 6x^2 - 13x - 42 = 0

             x^3 - 3x^2 + 9x^2 - 27x + 14x - 42=0

             (x^3 - 3x^2)+ (9x^2 - 27x) + (14x - 42)=0

             x^2(x-3) + 9x(x-3) + 14(x-3) = 0

              (x-3)(x^2 + 9x + 14) =0

=> x-3=0

     x=3            (do đa thức x^2 + 9x + 14 không có nghiệm nên ta không lấy)

vc ban x²+9x+14 co nghiem ma

a)\(6x^2+5x-6=0\)

\(\Leftrightarrow6x^2-4x+9x-6=0\)

\(\Leftrightarrow2x\left(3x-2\right)+3\left(3x-2\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x+3=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=\frac{2}{3}\end{array}\right.\)

b)\(6x^2-13x+6=0\)

\(\Leftrightarrow6x^2-4x-9x+6=0\)

\(\Leftrightarrow2x\left(3x-2\right)-3\left(3x-2\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=\frac{2}{3}\end{array}\right.\)

c)\(10x^2-13x-3=0\)

\(\Leftrightarrow10x^2-15x+2x-3=0\)

\(\Leftrightarrow5x\left(2x-3\right)+\left(2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(5x+1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=0\\5x+1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-\frac{1}{5}\end{array}\right.\)

d)\(20x^2+19x-3=0\)

\(\Delta=19^2-\left(-4\left(20.3\right)\right)=601\)

\(\Rightarrow x_{1,2}=\frac{-19\pm\sqrt{601}}{40}\)

e)\(3x^2-x+6=0\)

\(\Delta=\left(-1\right)^2-4\left(3.6\right)=-71< 0\)

Suy ra vô nghiệm

\(x^3+6x^2-13x-42\)

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

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

\(=\left(x-3\right)\left(x+2\right)\left(x+7\right)\)

x3+6x2−13x−42

x3+6x2−13x−42

=(x+7)(x−3)(x+2)

a) 3x⁴ - 13x³ + 16x² - 13x + 3 = 0

(x - 3)(3x - 1)(x² - x + 1) = 0

nhưng vì x² - x + 1 # 0 nên:

x - 3 = 0 hoặc 3x - 1 = 0

x = 0 + 3         3x = 0 + 1

x = 3               3x = 1

                        x = 1/3

b) 6x⁴ + 5x³ - 38x² + 5x + 6 = 0

(x - 2)(x + 3)(3x + 1)(2x - 1) = 0

x - 2 = 0 hoặc x + 3 = 0 hoặc 3x + 1 = 0 hoặc 2x - 1 = 0

x = 0 + 2         x = 0 - 3           3x = 0 - 1          2x = 0 + 1

x = 2               x = -3               3x = -1              2x = 1

                                                x = -1/3             x = 1/2

Chắc là phân tích đa thức thành nhân tử hả bạn?

\(x^3+6x^2-13x-42\)

\(=x^3+2x^2+4x^2+8x-21x-42\)

\(=x^2\left(x+2\right)+4x\left(x+2\right)-21\left(x+2\right)\)

\(=\left(x^2+4x-21\right)\left(x+2\right)\)

\(x^3+6x^2-13x-42\)

\(=x^3+2x^2+4x^2+8x-21x-42\)

\(=x^2\cdot\left(x+2\right)+4x\cdot\left(x+2\right)-21\left(x+2\right)\)

\(=\left(x+2\right)\cdot\left(x^2+4x-21\right)\)

\(=\left(x+2\right)\cdot\left(x^2+7x-3x-21\right)\)

\(=\left(x+2\right)\cdot\left(x\cdot\left(x+7\right)-3\left(x+7\right)\right)\)

\(=\left(x+2\right)\left(x-3\right)\left(x+7\right)\)

a)\(5x^2=13x\Leftrightarrow5x^2-13x=0\Leftrightarrow x\left(5x-13\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\5x-13=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{13}{5}\end{array}\right.\)

b)\(6x^4=9x^3\Leftrightarrow6x^4-9x^3=0\Leftrightarrow3x^3\left(2x-3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}3x^3=0\\2x-3=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{3}{2}\end{array}\right.\)

c)\(\left(x-2\right)^2-4x^2-12x-9=0\)

\(\Leftrightarrow\left(x-2\right)^2=4x^2+12x+9\)

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

\(\Leftrightarrow x-2=2x+3\)

\(\Leftrightarrow-x=5\Leftrightarrow x=-5\)

\(x^3-2x^2+x-2=0\\ \Leftrightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ Vậy:x=2\\ ---\\ 2x\left(3x-5\right)=10-6x\\ \Leftrightarrow6x^2-10x-10+6x=0\\ \Leftrightarrow6x^2-4x-10=0\\ \Leftrightarrow6x^2+6x-10x-10=0\\ \Leftrightarrow6x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(6x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}6x-10=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)

\(4-x=2\left(x-4\right)^2\\ \Leftrightarrow4-x=2\left(x^2-8x+16\right)\\ \Leftrightarrow2x^2-16x+32+x-4=0\\ \Leftrightarrow2x^2-15x+28=0\\ \Leftrightarrow2x^2-8x-7x+28=0\\ \Leftrightarrow2x\left(x-4\right)-7\left(x-4\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\\ ---\\ 4-6x+x\left(3x-2\right)=0\\ \Leftrightarrow4-6x+3x^2-2x=0\\ \Leftrightarrow3x^2-8x+4=0\\ \Leftrightarrow3x^2-6x-2x+4=0\\ \Leftrightarrow3x\left(x-2\right)-2\left(x-2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

\(x^3+6x^2-13x-42\)

\(x^3+6x^2-13x-42\)

\(=\left(x+7\right)\left(x-3\right)\left(x+2\right)\)

b, \(2x^3-x^2+3x+6\)

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

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

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

\(x^3+6x^2-13x-42\)

\(=x^3+7x^2-x^2-7x-6x-42\)

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

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

\(=\left(x+7\right)\left(x-3\right)\left(x+2\right)\)

\(x^3+6x^2-13x-42\)

\(=x^3-3x^2+9x^2-27x+14x-42\)

\(=x^2\left(x-3\right)+9x\left(x-3\right)+14\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2+9x+14\right)\)

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

\(=\left(x-3\right)\left(x+7\right)\left(x+2\right)\)

Chúc bạn học tốt!!!