Question

Bài 1, tìm x

a, x²+\(\frac{1}{4}\)=x

b, x²-4x=-4

c, x³+12x²+36x=0

d, (3x-1)²=(x-3)²

e, x³+3x²+3x+1=0

f, x³+1=0

Answer 1

f, \(x^3+1=0\)

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

Ta có 2 trương hợp:

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

(2) x² - x + 1 = 0

=> \(\Delta\) = (-1)² - 4.1.1 = 1 - 4 = -3 nhỏ hơn 0

=> Phương trình vô nghiệm.

Vậy x \(\in\) {-1}

Answer 2

a, \(x^2+\frac{1}{4}=x\)

\(x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\)

\(\left(x-\frac{1}{2}\right)^2=0\)

\(x-\frac{1}{2}=0\\ X=\frac{1}{2}\)

Answer 3

b, \(x^2-4x=-4\)

\(x^2-2.x.2+2^2=0\)

\(\left(x-2\right)^2=0\)

\(x-2=0\\ X=2\)

Answer 4

c, \(x^3+12x^2+36x=0\)

\(x\left(x^2+12x+36\right)=0\)

\(x\left(x+6\right)^2=0\)

Có: *x =0

*(x +6)² = 0 => x + 6 = 0 => x = -6

Vậy x ∈ {-6;0}

Answer 5

d, \(\left(3x-1\right)^2=\left(x-3\right)^2\)

\(\left(3x-1\right)^2-\left(x-3\right)^2=0\)

\(\left(3x-1-x+3\right)\left(3x-1+x-3\right)=0\)

\(\left(2x+2\right)\left(4x-4\right)=0\)

\(8\left(x+1\right)\left(x-1\right)=0\)

\(x^2-1=0\)

\(x^2=1\) => x = \(\pm\) 1

Answer 6

e, \(x^3+3x^2+3x+1=0\)

\(\left(x+1\right)^3=0\)

\(x+1=0\\ X=-1\)

Answer 7

b, x²-4x=-4

=>x²-4x+4=0

=>x²+2x+2x+4=0

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

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

=>x+2=0

x=-2

Vậy x=-2