Question

Tìm x

a) 3x² = 12 = 0

b) 2 (x+5) - x² - 5x =0

c) 2x² + 3x - 5 = 0

Answer 1

Sửa: a)\(3x^2-12=0\)

\(\Rightarrow3x^2=12\)

\(\Rightarrow x^2=\frac{12}{3}=4\)

\(\Rightarrow x=\sqrt{4}=2\)

Vậy: x=2

b)\(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=\left(x+5\right)\left(2-x\right)=0\)

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

Vậy: \(x=-5;2\)

c)\(\Rightarrow2x^2-2x+5x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)=0\)

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

Vậy: \(x=1;-\frac{5}{2}\)

 

 

Answer 2

ủa bạn xem lại đề đi làm gì có chuyện 12 = 0 đc