Question

Giải phương trình

a) x² - 2x - 7 = 0

b) 4x² + 4x - 3 = 0

c) 3x² - 2x - 1 = 0

d) 4x² - 16x - 1 = 0

Giúp mk nha.mk tik

Answer 1

a) \(x^2-2x-7=0\)

\(\Rightarrow x^2-2.x.1+1^2-8=0\)

\(\Rightarrow\left(x-1\right)^2-8=0\)

\(\Rightarrow\left(x-1\right)^2-\sqrt{8}^2=0\)

\(\Rightarrow\left(x-1+\sqrt{8}\right)\left(x-1-\sqrt{8}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1+2\sqrt{2}\\x=-1-2\sqrt{2}\end{matrix}\right.\)

b) \(4x^2+4x-3=0\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{3}{2}\right)=0\) (tách hạng tử )

Rồi dùng máy tính nhấn. Mấy câu kia tương tự nhé!

Answer 2

3x² - 2x - 1 = 0

=>3x2+x-3x-1=0

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

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

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

vậy x=1 hoặc x=-\(\dfrac{1}{3}\)