\(c,P=2x-2x^2-5\)

      \(=-2\left(x^2-x+0,25\right)-4,5\) 

     \(=-2\left(x-0,5\right)^2-4,5\le-4,5\forall x\)

Dấu"=" xảy ra<=> \(-2\left(x-0,5\right)^2=0\Leftrightarrow x=0,5\)

Vậy...

\(4x^2-4x+3=0\)

\(\Rightarrow4x^2-4x+1+2=0\)

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

Vì \(\left(2x-1\right)^2\ge0\forall x\)\(\Rightarrow\left(2x-1\right)^2+2\ge2\)

\(\Rightarrow\left(2x-1\right)^2=0\)( vô lý )

\(\Rightarrow x\in\varnothing\)

\(4x^2-4x+3=0\)

\(\left(2x\right)^2-2.2x.1+1+2=0\)

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

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

\(\text{Vì }\left(2x-1\right)^2\ge0\forall x\)

\(\text{Mà}\left(2x-1\right)^2=-2\)

\(\Rightarrow\text{Ko có giá trị x thỏa mãn đề bài}\)

làm tắt ko hiểu thì hỏi 

a) \(=x^2+2.xy.\frac{1}{2}+\frac{1}{4}y^2-\frac{1}{4}y^2+y^2+1\)

\(=\left(x+\frac{1}{2}y\right)^2+\frac{3}{4}y^2+1>0\)

b) \(=\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6x+9\right)+1\)

\(=\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1>0\)

nhân ra mà tìm x 

a) 5x(x - 1) - (x + 2)(5x - 7) = 6

<=> 5x² - 5x - 5x² + 7x - 10 + 14 = 6

<=> -8x + 14 = 6

<=> -8x = 6 - 14

<=> -8x = -8

<=> x = 1

=> x = 1

b) 2(3x - 1)(2x + 5) - 6(2x - 1)(x + 2) = -6

<=> 12x² + 30x - 4x - 10 - 12x² - 24x + 6x + 12 = -6

<=> 8x + 2 = -6

<=> 8x = -6 - 2

<=> 8x = -8

<=> x = -1

=> x = -1

\(\left(x-2\right)\left(x-3\right)=\left(x-2\right)\left(x+1\right)\)

\(\Leftrightarrow x^2-5x+6=x^2-x-2\)

\(\Leftrightarrow-4x+8=0\)

\(\Leftrightarrow x=2\)

Vậy ...

(x-2)(x-3)=(x-2)(x+1)

\(x^2-5x+6=x^2-x-2\)

\(x^2-x^2-5x+x=-6-2\)

\(-4x=-8\)

\(x=2\)

\(\text{Vậy x=2}\)

\(2x^2-3x-2\)

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

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

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

2x^2-3x-2

=2x^2-4x+x-2

=2x(x-2)+(x-2)

=(x-2)(2x+1)

a) Đặt \(A=x^2-2x+3\)

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

Vì \(\left(x-1\right)^2\ge0;\forall x\)

\(\Rightarrow\left(x-1\right)^2+2\ge0+2;\forall x\)

Hay \(A\ge2>0;\forall x\)

b,c,d mình giúp biến đổi ra hằng đẳng thức còn tự làm

b) \(=\left(x-7\right)^2+5\)

c) giống b

d) \(=x^2+2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+2=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}\)

a) x² - 2x + 3 = (x² - 2x + 1) + 2 = (x - 1)² + 2 > 0

=> x² - 2x + 3 luôn dương \(\forall\)x

b) x²- 14x + 54 = (x² - 14x + 49) + 5 = (x - 7)² + 5 > 0

=> x² - 14x + 54 luôn dương \(\forall\)x

c) như trên

d) x² + x + 2 = (x² + x + 1/4) + 7/4 = (x + 1/2)² + 7/4 > 0

=> x²  + x + 2 luôn dương \(\forall\)x

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

Xét \(x+3=0\Rightarrow x=-3\)

Xét \(x+3\ne0\) ta có:

\(x+3=x-3\)

\(\Rightarrow0=6\left(VL\right)\)

Vậy \(x=-3\)

a) 

(x + 3)² = (x + 3)(x – 3)

⇔ (x + 3)² - (x + 3)(x - 3) = 0

⇔ (x + 3)(x + 3 - x + 3) = 0

⇔ 6(x + 3) = 0

⇔ x = -3

Vậy: x = -3

b) Ta có A = (x + 1)(x + 2)(x + 3)(x + 4) – 24

= (x + 1)(x + 4)(x + 2)(x + 3) - 24

= (x² + 5x + 4)(x2 + 5x + 6) - 24(*)

Đặt x² + 5x + 5 = t

Thay x² + 5x + 5 = t vào (*) ta được:

A = (t - 1)(t + 1) - 24

= t² - 25

= (t - 25)(t + 25)

= (x² + 5x + 5 + 5)(x² + 5x + 5 - 5)

= (x² + 5x + 10)(x² + 5x)

(x² + 5x + 10).x(x + 5) chia hết cho x (Với x ≠ 0)

Vậy: A chia hết cho x (Với x ≠ 0)