2a) 3x\(^2\)-7x=0
⟺ x(3x-7)=0
⟺ x=0 hoặc 3x-7=0
⟺ x=0 hoặc 3x=7
⟺ x=0 hoặc x= \(\dfrac{7}{3}\)
b)2x\(^2\)-x-5=0
a=2 ; b= -1 ; c= -5
=>Δ=b\(^2\) - 4ac =(-1)\(^2\) - 4×2×(-5)
=1 - (-40) = 1+40 = 41>0
Vậy PT có hai nghiệm phân biệt.
x\(_1\)=\(\dfrac{-b+\sqrt{\text{Δ}}}{2a}\) = \(\dfrac{-\left(-1\right)+\sqrt{41}}{2\text{×}2}\) = \(\dfrac{1+\sqrt{41}}{4}\)
x\(_2\)= \(\dfrac{-b+\sqrt{\text{Δ}}}{2a}\) = \(\dfrac{-\left(-1\right)-\sqrt{41}}{2\text{×}2}\) = \(\dfrac{1-\sqrt{41}}{4}\)
c) \(\left(x+1\right)^2\)\(=4\left(x+4\right)\)
⟺ \(x^2+2x+1\) \(=4x+16\)
⟺ \(x^2+2x+1\)\(-4x-16\)\(=0\)
⟺ \(x^2-2x-15=0\)
⟺ \(x^2-2x=15\)
⟺ \(x\left(x-2\right)=0\)
⟺ \(x=0\) hoặc \(x-2=0\)
⟺ \(x=0\) hoặc \(x=2\)
Vậy S= \(\left\{0;2\right\}\)
c) (x+1)2(x+1)2=4(x+4)=4(x+4)
⟺ x2+2x+1x2+2x+1 =4x+16=4x+16
⟺ x2+2x+1x2+2x+1−4x−16−4x−16=0=0
⟺ x2−2x−15=0x2−2x−15=0
⟺ x2−2x=15x2−2x=15
⟺ x(x−2)=0x(x−2)=0
⟺ x=0x=0 hoặc x−2=0x−2=0
⟺ x=0x=0 hoặc x=2x=2
Vậy S= {0;2}
