Question

Tìm x : 

a) 2( x + 3 ) - x^2 - 3x = 0

b) 2x( 3x - 5 ) = 10 - 6x

c) x^3 + x^2 + x + 1 = 0

Answer 1

a, 2x +6 -x²-3x=0

<=>(x+3)(x-2)=0

<=> x=2 hoặc x=-3

b, 6x² - 10x -10 + 6x =0

<=> (3x-5)(x+1) =0

<=> x= 5/3 hoặc x= -1

c, (x+1)(x²+1)=0

<=> x=-1

Answer 2

cho mình làm lại  hồi này làm sai 

b)\(2x\left(3x-5\right)=10-6x\)

\(\Rightarrow6x^2-10x=10-6x\)

\(\Rightarrow6x^2-10x-10+6x=0\)

\(\Rightarrow6x\left(x+1\right)-10\left(x+1\right)=0\)

\(\Rightarrow\left(6x-10\right)\left(x+1\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}6x-10=0\\x+1=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{3}\\x=-1\end{array}\right.\)

Answer 3

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

\(\Rightarrow2\left(x+3\right)-x\left(x+3\right)=0\)

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

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

b)\(2x\left(3x-5\right)=10-6x\)

\(\Rightarrow2x\left(3x-5\right)=2\left(5-3x\right)\)

\(\Rightarrow x=-1\)

c)\(x^3+x^2+x+1=0\)

\(\Rightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x^2+1\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x+1=0\\x^2+1=0\end{array}\right.\)

Vì \(x^2\ge0\Rightarrow x^2+1\ge1>0\Rightarrow x+1=0\)

=>x=-1