Question

Bài 1: giải các phương trình sau:

a) 2(x+5) - x² - 5x = 0                        b) 2x² + 3x - 5 = 0                   c) ( x - 1)² + 4(x+2) - (x² - 3 ) = 0

Answer 1

a) 2 (x + 5) - x² - 5x = 0

=> 2 (x + 5) - (x² + 5x) = 0

=> 2 (x + 5) - x (x + 5) = 0

=> (2 - x) (x + 5) = 0

Có 2 TH xảy ra :

TH1 : 2 - x = 0 => x = 2

TH2 : x + 5 = 0 => x = -5

 

Answer 2

a, 2\((\)x +5\()\) - x² - 5x =0

\(\Leftrightarrow\) 2x² +10-x² - 5x=0

\(\Leftrightarrow\)x² - 5x +10=0

\(\Delta'\) = \((\) -5\()\)² - 1. 10=15 \(\Rightarrow\) \(\sqrt{\Delta'}\) = \(\sqrt{15}\)

\(\Rightarrow\) x₁ = 5 + \(\sqrt{15}\) ; x₂ = 5- \(\sqrt{15}\)

pt có 2 nghiệm ........

b, 2x² + 3x -5 =0

có a+b+c= 2+3+ \((\) -5\()\) =0

\(\Rightarrow\) x₁=1 , x₂ =\(\dfrac{-5}{2}\)

c, \((\) x-1\()\)² + 4.\((x+2)\) - \((x^2-3)\)=0

\(\Rightarrow x^2-2x+1+4x+8-x^{2^{ }}+3=0\)

\(\Leftrightarrow\) -2x +12 =0

\(\Leftrightarrow\)-2x=-12\(\Leftrightarrow\) x= 6