Question

bài 2 tìm x

Answer 1

a.

<=> \(-3x^2+15x+5x-5+3x^2-4+x=0\)

<=> \(\left(-3x^2+3x^2\right)+\left(15x+5x+x\right)+\left(-5-4\right)=0\)

<=> \(21x-9=0\)

=> \(x=\dfrac{9}{21}=\dfrac{3}{7}\)

b.

<=> \(7x^2-14x-5x+5-21x-7x^2-3=0\)

<=> \(\left(7x^2-7x^2\right)+\left(-14x-5x-21x\right)+\left(5-3\right)=0\)

<=>  \(2-40x=0\)

<=> 40x = 2 

=> \(x=\dfrac{2}{40}=\dfrac{1}{20}\)

c.

<=> \(15x-3-x^2+2x+x^2-13x-7=0\)

<=> \(\left(-x^2+x^2\right)+\left(15x+2x-13x\right)+\left(-3-7\right)=0\)

<=> \(4x-10=0\)

=> 4x = 10 

=> \(x=\dfrac{10}{4}\)

d.

<=> \(2x^2-3x-2x^2+10x-12=0\)

<=> \(\left(2x^2-2x^2\right)+\left(-3x+10x\right)-12=0\)

<=> \(7x-12=0\)

=> 7x = 12 

=> \(x=\dfrac{12}{7}\)