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5 tháng 10 2017

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

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

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

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

5 tháng 10 2017

x3+6x2−13x−42

x3+6x2−13x−42

=(x+7)(x−3)(x+2)

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

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

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

2 tháng 8 2016

b, \(2x^3-x^2+3x+6\)

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

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

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

20 tháng 11 2016

a) \(\left(x+8\right)^2-2\left(x+8\right)\left(x-2\right)+\left(x-2\right)^2\)

\(=\left[\left(x+8\right)-\left(x-2\right)\right]^2\)

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

\(=10^2\)

\(=2^2.5^2\)

b)\(x^3-4x^2-12x+27=\left(x^3+27\right)-\left(4x^2+12x\right)\)

\(=\left(x+3\right)\left(x^2-3x+9\right)-4x\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2-3x+9-4x\right)\)

\(=\left(x+3\right)\left(x^2-7x+9\right)\)

c)\(x^3+6x^2+11x+6=x^3+x^2+5x^2+5x+6x+6\)

\(=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+5x+6\right)\)

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

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

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

d)\(x^3+6x^2-13x-42=x^3-3x^2+9x^2-27x+14x-42\)

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

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

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

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

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

27 tháng 10 2016

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)

28 tháng 10 2016

vc ban x2+9x+14 co nghiem ma

31 tháng 10 2018

a) \(x^3+x^2+5x^2+5x+6x+6=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\)

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

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

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

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

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

c) \(\left(x^2+x+4\right)^2+3x\left(x^2+x+4\right)+5x\left(x^2+x+4\right)+15x^2\)

\(=\left(x^2+x+4\right)\left(x^2+x+4+3x\right)+5x\left(x^2+x+4+3x\right)\)

\(=\left(x^2+6x+4\right)\left(x^2+4x+4\right)\)

\(=\left(x^2+6x+4\right)\left(x+2\right)^2\)

d) \(\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(=\left(x^2+10x\right)^2+40\left(x^2+10x\right)+16.24+16\)

\(=\left(x^2+10x\right)^2+40\left(x^2+10x\right)+400\)

\(=\left(x^2+10x+20\right)^2\)

30 tháng 5 2018

a) 4x3 - 13x2 + 9x - 18

= 4x3 - 12x2 - x2 + 3x + 6x - 18

= 4x2( x - 3) - x( x - 3) + 6( x - 3)

= ( x - 3)( 4x2 - x + 6)

b) - x3 - 6x2 + 6x + 1

= 6x( 1 - x) + 1 - x3

= 6x( 1 - x) + ( 1 - x )( x2 + x + 1)

= ( 1 - x)( x2 + 7x + 1)

c) x3 + 3x2 + 3x + 2

= x3 + 2x2 + x2 + 2x + x + 2

= x2( x + 2) + x( x + 2) + x + 2

= ( x + 2)( x2 + x + 1)

a) \(4x^3-13x^2+9x-18\)

\(=4x^3-12x^2-x^2+3x+6x-18\)

\(=4x^2\left(x-3\right)-x\left(x-3\right)+6\left(x-3\right)\)

\(=\left(x-3\right)\left(4x^2-x+6\right)\)

25 tháng 11 2016

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

26 tháng 11 2016

ơn pạn nhìu nha