Question

[nam hai]

ai bt lm thì giúp e vs ạ

Answer 1

[Nguyễn Lê Phước Thịnh]

1: \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)

=>\(\left(2x-1\right)^2-\left(2x-1\right)\left(2x+1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-2x-1\right)=0\)

=>-2(2x-1)=0

=>2x-1=0

=>\(x=\dfrac{1}{2}\)

2: \(\left(x+2\right)^2-x\left(x-3\right)=2\)

=>\(x^2+4x+4-x^2+3x=2\)

=>7x+4=2

=>7x=-2

=>\(x=-\dfrac{2}{7}\)

3: \(\left(x-5\right)^2-x\left(x+2\right)=5\)

=>\(x^2-10x+25-x^2-2x=5\)

=>-12x+25=5

=>-12x=5-25=-20

=>\(x=\dfrac{20}{12}=\dfrac{5}{3}\)

4: \(\left(x-1\right)^2+x\left(4-x\right)=11\)

=>\(x^2-2x+1+4x-x^2=11\)

=>2x+1=11

=>2x=10

=>x=5

5: \(\left(x-3\right)\left(x+3\right)=\left(x-5\right)^2\)

=>\(x^2-9=x^2-10x+25\)

=>-10x+25=-9

=>-10x=-25-9=-34

=>\(x=\dfrac{34}{10}=\dfrac{17}{5}\)

6: \(\left(2x+1\right)^2-4x\left(x-1\right)=17\)

=>\(4x^2+4x+1-4x^2+4x=17\)

=>8x+1=17

=>8x=16

=>x=2

7: \(\left(3x+1\right)^2-9x\left(x-2\right)=25\)

=>\(9x^2+6x+1-9x^2+18x=25\)

=>24x+1=25

=>24x=24

=>x=1

8: \(\left(3x-2\right)\left(3x+2\right)-9x\left(x-1\right)=0\)

=>\(9x^2-4-9x^2+9x=0\)

=>9x-4=0

=>9x=4

=>\(x=\dfrac{4}{9}\)

9: \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

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

=>4(x+2)=0

=>x+2=0

=>x=-2

10: \(\left(x+2\right)^2-\left(x-3\right)\left(x+3\right)=-3\)

=>\(x^2+4x+4-\left(x^2-9\right)+3=0\)

=>\(x^2+4x+7-x^2+9=0\)

=>4x+16=0

=>4x=-16

=>x=-4

11: \(\left(3x+2\right)^2-\left(3x-5\right)\left(3x+2\right)=0\)

=>(3x+2)(3x+2-3x+5)=0

=>7(3x+2)=0

=>3x+2=0

=>3x=-2

=>\(x=-\dfrac{2}{3}\)

12: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

=>\(x^2+6x+9-x^2+4=4x+17\)

=>6x+13=4x+17

=>2x=4

=>x=2

13: \(3\left(x-1\right)^2+\left(x+5\right)\left(-3x+2\right)=-25\)

=>\(3\left(x^2-2x+1\right)+2x-3x^2+10-15x=-25\)

=>\(3x^2-6x+3-3x^2-13x+10=-25\)

=>-19x+13=-25

=>-19x=-38

=>x=2

14: \(\left(x+3\right)^2+\left(x-2\right)^2=2x^2\)

=>\(x^2+6x+9+x^2-4x+4=2x^2\)

=>2x=-13

=>\(x=-\dfrac{13}{2}\)