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YQ
2
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)
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
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 6 2017
Chắc là phân tích đa thức thành nhân tử hả bạn?
\(x^3+6x^2-13x-42\)
\(=x^3+2x^2+4x^2+8x-21x-42\)
\(=x^2\left(x+2\right)+4x\left(x+2\right)-21\left(x+2\right)\)
\(=\left(x^2+4x-21\right)\left(x+2\right)\)
Q
5 tháng 6 2017
\(x^3+6x^2-13x-42\)
\(=x^3+2x^2+4x^2+8x-21x-42\)
\(=x^2\cdot\left(x+2\right)+4x\cdot\left(x+2\right)-21\left(x+2\right)\)
\(=\left(x+2\right)\cdot\left(x^2+4x-21\right)\)
\(=\left(x+2\right)\cdot\left(x^2+7x-3x-21\right)\)
\(=\left(x+2\right)\cdot\left(x\cdot\left(x+7\right)-3\left(x+7\right)\right)\)
\(=\left(x+2\right)\left(x-3\right)\left(x+7\right)\)
VM
2
3 tháng 8 2016
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)\(6x^4=9x^3\Leftrightarrow6x^4-9x^3=0\Leftrightarrow3x^3\left(2x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}3x^3=0\\2x-3=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{3}{2}\end{array}\right.\)
3 tháng 8 2016
c)\(\left(x-2\right)^2-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x-2\right)^2=4x^2+12x+9\)
\(\Leftrightarrow\left(x-2\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow x-2=2x+3\)
\(\Leftrightarrow-x=5\Leftrightarrow x=-5\)
BT
3
1 tháng 8 2016
\(x^3+6x^2-13x-42\)
\(x^3+6x^2-13x-42\)
\(=\left(x+7\right)\left(x-3\right)\left(x+2\right)\)
2 tháng 8 2016
b, \(2x^3-x^2+3x+6\)
\(=2x^3+2x^2-3x^2-3x+6x+6\)
\(=2x^2\left(x+1\right)-3x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(2x^2-3x+6\right)\)
LP
1
DH
4 tháng 9 2017
\(x^3+6x^2-13x-42\)
\(=x^3-3x^2+9x^2-27x+14x-42\)
\(=x^2\left(x-3\right)+9x\left(x-3\right)+14\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+9x+14\right)\)
\(=\left(x-3\right)\left(x^2+7x+2x+14\right)\)
\(=\left(x-3\right)\left(x+7\right)\left(x+2\right)\)
Chúc bạn học tốt!!!
KN
25 tháng 8 2020
1. a . 3x2 - 6x = 0
\(\Leftrightarrow3x\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}3x=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
b. x3 - 13x = 0
\(\Leftrightarrow x\left(x^2-13\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-13=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{13}\end{cases}}\)
c. 5x ( x - 2001 ) - x + 2001 = 0
<=> 5x ( x - 2001 ) - ( x - 2001 ) = 0
\(\Leftrightarrow\left(5x-1\right)\left(x-2001\right)=0\Leftrightarrow\orbr{\begin{cases}5x-1=0\\x-2001=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=2001\end{cases}}\)
\(x^3+6x^2-13x-42=0\)
\(\Leftrightarrow\left(x^3-3x^2\right)+\left(9x^2-27x\right)+\left(14x-42\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)+9x\left(x-3\right)+14\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)+\left(x^2+9x+14\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+7x+2x+14\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[x\left(x+7\right)+2\left(x+7\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(x+7\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\x+2=0\\x+7=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=-2\\x=-7\end{array}\right.\)
x3 + 6x2 - 13x - 42 = 0
=> x3 - 3x2 + 9x2 - 27x + 14x - 42 = 0
=> x2 ( x - 3 ) + 9x ( x - 3 ) + 14 ( x - 3 ) = 0
=> ( x - 3 ) ( x2 + 9x + 14) = 0
=> ( x - 3 ) ( x2 + 2x + 7x + 14 ) = 0
=> ( x - 3 ) [ x ( x + 2 ) + 7 ( x + 2 ) ] = 0
=> ( x - 3 ) ( x + 2 ) ( x + 7 ) = 0
=> x - 3 = 0 => x = 3
=> x + 2 = 0 => x = -2
=> x + 7 = 0 => x = -7