1) x² + 10x + 26 + y² + 2y 

= (x² + 10x + 25) + (y² + 2y + 1)

= (x² + 5x + 5x + 25) + (y² + y + y + 1)

= x(x + 5) + 5(x + 5) + y(y +  1) + (y + 1)

= (x + 5)² + (y + 1)²

2) z² - 6z + 13 + t² + 4t 

= (z² - 6z + 9) + (t² + 4t + 4) 

= (z² - 3z - 3z + 9) + (t² + 2t + 2t + 4)

= z(z - 3) - 3(z - 3) + t(t + 2) + 2(t + 2)

= (z - 3)² + (t + 2)²

3) x² - 2xy + 2y² + 2y + 1

(x² - 2xy + y²) + (y² + 2y + 1)

= (x - xy - xy + y²) + (y² + y + y +1)

= x(x - y) - y(x - y) + y(y + 1) + (y + 1)

= (x - y)² + (y + 1)²

x² + 10x + 26 + y² + 2y
= x² + 10 + 25 + 1 + y² + 2y
= (x² + 10x + 25) + (y² + 2y + 1)
= (x + 5)² + (y + 1)²

x² - 2xy + 2y² + 2y + 1
= x² - 2xy + y² + y² + 2y + 1
= (x² - 2xy + y²) + (y² + 2y + 1)
= (x - y)² + (y + 1)²

4x² + 2z² - 4xz - 2z + 1
= 4x² + z² + z² - 4xz - 2z + 1
= (4x² - 4xz + z²) + (z² - 2z + 1)
= (2x + z)² + (z - 1)²

1)a)x²+10x+26+y²+2y

=(x²+10x+25)+(y²+2y+1)

=(x+5)²+(y+1)²

b)x²-2xy+2y²+2y+1

=(x²-2xy+y²)+(y²+2y+1)

=(x-y)²+(y+1)²

c)z²-6z+13+t²+4t

=(z²-6z+9)+(t²+4t+4)

=(z-3)²+(t+2)²

d)4x²+2z²-4xz-2z+1

=(4x²-4xz+z²)+(z²-2z+1)

=(2x-z)²+(z-1)²

2)a)(x-3)²-4=0

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

<=>(x-5)(x-1)=0

<=>x-5=0 hoặc x-1=0

<=>x=5 hoặc x=1

b)x²-2x=24

<=>x²-2x-24=0

<=>(x²-6x)+(4x-24)=0

<=>x(x-6)+4(x-6)=0

<=>(x-6)(x+4)=0

<=>x-6=0 hoặc x+4=0

<=>x=6 hoặc x=-4

a) x^2 + 10x + 26 + y^2 + 2y

=x²+10x+25+y²+2y+1

=x²+2.x.5+5²+y²+2.y.1+1²

=(x+5)²+(y+1)²

b)x^2 - 2xy + 2y^2 + 2y +1

=x²-2xy+y²+y²+2y+1

=(x-y)²+(y+1)²

c)z^2 - 6z + 13 + t^2 + 4t

=z²-6z+9+t²+4z+4

=z²-2.z.3+3²+t²+2.t.2+2²

=(z-3)²+(t+2)²

d)4x^2 + 2z^2 - 4xz - 2z + 1

=4x²-4xz+z²+z²-2z+1

=(2x)²-2.2x.z+z²+z²-2z.1+1²

=(2x-z)²+(z-1)²

a) x^2 + 10x + 26 + y^2 + 2y

=x²+10x+25+y²+2y+1

=x²+2.x.5+5²+y²+2.y.1+1²

=(x+5)²+(y+1)²

b)x^2 - 2xy + 2y^2 + 2y +1

=x²-2xy+y²+y²+2y+1

=(x-y)²+(y+1)²

c)z^2 - 6z + 13 + t^2 + 4t

=z²-6z+9+t²+4z+4

=z²-2.z.3+3²+t²+2.t.2+2²

=(z-3)²+(t+2)²

d)4x^2 + 2z^2 - 4xz - 2z + 1

=4x²-4xz+z²+z²-2z+1

=(2x)²-2.2x.z+z²+z²-2z.1+1²

=(2x-z)²+(z-1)²

a) (x-3)^2 - 4 =0

<=>(x-3)²-2²=0

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

<=>(x-5)(x-1)=0

<=>x-5=0 hoặc x-1=0

<=>x=5 hoặc x=1

b) x^2 - 2x = 24

<=>x²-2x-24=0

<=>x²-2x+1-25=0

<=>(x-1)²-25=0

<=>(x-1)²-5²=0

<=>(x-1-5)(x-1+5)=0

<=>(x-6)(x+4)=0

<=>x-6=0 hoặc x+4=0

<=>x=6 hoặc x=-4

1a/ z² - 6z + 5 - t² - 4t = z² - 2 . 3z + 3² - 4 - t² - 4t = (z² - 2 . 3z + 3²) - (2² + 2 . 2t + t²) = (z - 3)² - (2 + t)²

b/ x² - 2xy + 2y² + 2y² + 1 = x² - 2xy + y² + y² + 2y + 1 = (x² - 2xy + y²) + (y² + 2y + 1) = (x - y)² + (y + 1)²

c/ 4x² - 12x - y² + 2y + 8 = (2x)² - 12x - y² + 2y + 3² - 1 = [ (2x)² - 2 . 3 . 2x + 3² ] - (y² - 2y + 1) = (2x - 3)² - (y - 1)²

2a/ (x + y + 4)(x + y - 4) = x² + xy - 4x + xy + y² - 4y + 4x + 4y + 16 = x² + (xy + xy) + (-4x + 4x) + (-4y + 4y) + y² + 16

= x² + 2xy + y² + 4² = (x + y)² + 4²

b/ (x - y + 6)(x + y - 6) = x² + xy - 6x - xy - y² + 6y + 6x + 6y - 36 = x² + (xy - xy) + (-6x + 6x) + (6y + 6y) - y² - 36

= x² - y² + 12y - 6² = x² - (y² - 12y + 6²) = x² - (y² - 2 . 6y + 6²) = x² - (y - 6)²

c/ (y + 2z - 3)(y - 2z - 3) = y² -2yz - 3y + 2yz - 4z² - 6z - 3y + 6z + 9 = y² + (-2yz + 2yz) + (-3y - 3y) + (-6z + 6z) - 4z² + 9

= y² - 6y - 4z² + 9 = (y² - 6y + 9) - 4z² = (y - 3)² - (2z)²

d/ (x + 2y + 3z)(2y + 3z - x) = 2xy + 3xz - x² + 4y² + 6yz - 2xy + 6yz + 9z² - 3xz = (2xy - 2xy) + (3xz - 3xz) - x² + (6yz + 6yz) + 9z² + 4y²

= -x² + 4y² + 12yz + 9z² = (4y² + 12yz + 9z²) - x² = [ (2y)² + 2 . 2 . 3yz + (3z)² ] - x² = (2y + 3z)² - x²

:v dễ mà có trong nâng cao mới hc qua :3

a, x2+10x+26+y2+2y

=(x2+2.x.5+52)+(12+2.1.y+y2)

=(x+5)2+(y+1)2

b, x2−2xy+2y2+2y+1

=x2−2xy+y2+y2+2y+1

=(x2−2.x.y+y2)+(y2+2.y.1+12)

=(x−y)2+(y+1)2

c,z2−6z+5−t2−4t

=−(t2+4t−z2+6z−5)

=−(t2+2.t.2+22−z2+2.z.3−32)

=−((t2+2.t.2+22)−(z2−2.z.3+32))

=−((t+2)2−(z−3)2)

=(z−3)2−(t+2)2

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

b) \(\left(z^2-6z+9\right)+\left(t^2+4t+4\right)=\left(z-3\right)^2+\left(t+2\right)^2\)

c) \(\left(4x^2-4xz+z^2\right)+\left(z^2-2z+1\right)=\left(4x-z\right)^2+\left(z-1\right)^2\)

`B=(x/2+y)^3-6(x/2+y)^2z + 6(x+2y)z^2-8z^3`

`=(x/2+y)^3 - 3. (x/2+y)^2 . 2z + 3. (x/2+y) . (2z)^2 - (2z)^3`

`=(x/2+y-2z)^3`

Sửa đề: Δ\(B=\left(\dfrac{x}{2}+y\right)^3-6\left(\dfrac{x}{2}+y\right)^2z+12\left(x+2y\right)\cdot z^2-8z^3\)

Ta có: \(B=\left(\dfrac{x}{2}+y\right)^3-6\left(\dfrac{x}{2}+y\right)^2z+12\left(x+2y\right)\cdot z^2-8z^3\)

\(=\left(\dfrac{1}{2}x+y\right)^2-3\cdot\left(\dfrac{1}{2}x+y\right)^2\cdot2z+3\cdot\left(\dfrac{1}{2}x+y\right)\cdot\left(2z\right)^2-\left(2z\right)^3\)

\(=\left(\dfrac{1}{2}x+y-2z\right)^3\)

a) \(x^2+10x+26+y^2+2y\)

\(=\left(x^2+10x+25\right)+\left(y^2+2y+1\right)\)

\(\left(x+5\right)^2+\left(y+1\right)^2\)

b) \(x^2-2xy+2y^2+2y+1\)

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

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

c) \(z^2-6z+13+t^2+4t\)

\(=\left(z^2-6x+9\right)+\left(t^2+4t+4\right)\)

\(=\left(z-3\right)^2+\left(t+2\right)^2\)

d) \(4x^2-2z^2-2xz-2z+1\)

\(=\left(4x^2-4xz+z^2\right)+\left(z^2-2z+1\right)\)

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

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