a)
<=> (2x+1).(3x-2) - (5x-8 ) . (2x+1 ) =0
<=> (2x+1 ) ( 3x-2-5x+8) = 0
<=> (2x+1)(6-2x)=0
<=> 2x + 1 = 0
6 - 2x = 0
<=> 2x = -1
2x = 6
=> x = -1/2 ; x = 3
b)
<=> (2x)^2 - 1 = (2x+1)(3x-5)
<=> (2x-1 ) ( 2x + 1 ) - (2x+1) ( 3x-5)=0
<=> (2x+1 ) (2x-1-3x+5)=0
=>
2x + 1 = 0
4 - x =0
=> x = -1/2 ; x = 4
c)
<=> x^2 + 2x+1 = 4x^2 -8x+4
<=>
x^2 + 2x + 1 - 4x^2 + 8x - 4 = 0
<=>
-3x^2 +10x - 3=0
Δ = b^2 - 4ac = \(10^2-4.-3.-3=64\)
\(\sqrt{\Delta}=\sqrt{64}=8>0\)
=> pt có 2 nghiệm pb
\(x_1=\dfrac{-10+8}{-6}=\dfrac{1}{3};x_2=\dfrac{-10-8}{-6}=3\)
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