Question

[hbvvyv]

Bài 1: Giải các phương trình sau 

Chú ý b chẵn thì dùng công thức nghiệm thu gọn nhé 

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(2x^2-7x+3=0\)

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

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

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

=>\(\left[{}\begin{matrix}x-3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

b: \(6x^2+x+5=0\)

\(\text{Δ}=1^2-4\cdot6\cdot5=1-24\cdot5=1-120=-119< 0\)

=>Phương trình vô nghiệm

c: \(6x^2+x-5=0\)

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

=>6x(x+1)-5(x+1)=0

=>(x+1)(6x-5)=0

=>\(\left[{}\begin{matrix}x+1=0\\6x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{6}\end{matrix}\right.\)

d: \(3x^2+5x+2=0\)

=>\(3x^2+3x+2x+2=0\)

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

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

=>\(\left[{}\begin{matrix}x+1=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{2}{3}\end{matrix}\right.\)

e: \(y^2-8y+16=0\)

=>\(\left(y-4\right)^2=0\)

=>y-4=0

=>y=4

f: \(16z^2+24z+9=0\)

=>\(\left(4z\right)^2+2\cdot4z\cdot3+3^2=0\)

=>\(\left(4z+3\right)^2=0\)

=>4z+3=0

=>4z=-3

=>\(z=-\dfrac{3}{4}\)