a)x²-6x+9

=x²-2.x.3+3²

=(x-3)²

b)4x²+4x+1

=(2x)²+2.2x.1+1²

=(2x+1)²

c)4x²+12xy+9y²

=(2x)²+2.2x.3y+(3y)²

=(2x+3y)²

d)4x⁴-4x²+4

=(2x²)²-2.2x².2+2²

=(2x²-2)²

này mình có vài câu không làm được, xin lỗi bạn nha

\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)

Sửa lại đề ở câu 1: \(2ab\)chuyển thành \(2bx\)

1. \(2x^2+2b^2+2bx+2x+2b+2\)

\(=\left(x^2+2bx+b^2\right)+\left(x^2+2x+1\right)+\left(b^2+2b+1\right)\)

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

2. \(4x^2+4x+10+6y+y^2\)

\(=\left(4x^2+4x+1\right)+\left(y^2+6y+9\right)\)

\(=\left(2x+1\right)^2+\left(y+3\right)^2\)

a.

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

b.

\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

c.

\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)

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

d.

\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)

e.

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

\(=\left(x-y\right)\left(x+y-5\right)\)

f.

\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

g. 10x(x-y)-6y(y-x)

=10x(x-y)+6y(x-y)

=(x-y)(10x+6y)

h.x²-4x-5

=(x-5)(x+1)

i.x⁴-y⁴ = (x²-y²)(x²+y²)

B2.

a.5(x-2)=x-2

⇔5(x-2)-(x-2)=0

⇔4(x-2)=0

⇔x=2

b.3(x-5)=5-x

⇔3(x-5)+(x-5)=0

⇔4(x-5)=0

⇔x=5

c.(x+2)²-(x+2)(x-2)=0

⇔(x+2)[(x+2)-(x-2)]=0

⇔4(x+2)=0

⇔x=-2

a) Sửa đề: \(x^2+3x+1\rightarrow x^2+2x+1\)

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

b) \(x^2+y^2+2xy=\left(x+y\right)^2\)

c) \(9x^2+12x+4=\left(3x+2\right)^2\)

d) \(-4x^2-9-12x=-\left(4x^2+12x+9\right)=-\left(2x+3\right)^2\)

a) \(x^2+4x+4\)

\(=x^2+2\cdot2\cdot x+2^2\)

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

b) \(4x^2-4x+1\)

\(=\left(2x\right)^2-2\cdot2x\cdot1+1^2\)

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

c) \(x^2-x+\dfrac{1}{4}\)

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

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

d) \(4\left(x+y\right)^2-4\left(x+y\right)+1\)

\(=\left[2\left(x+y\right)\right]^2-2\cdot2\left(x+y\right)\cdot1+1^2\)

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

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

1)

\(=x^2-4x+4+y^2+2y+1\)

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

2)

\(=a^2+2ab+b^2+a^2-2ax+x^2\)

\(=\left(a+b\right)^2+\left(a-x\right)^2\)

3)

\(=x^2-2x+1+y^2+6y+9\)

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

4)

\(=x^2-2xy+y^2+x^2+10x+25\)

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

5)

\(=a^2+2ab+b^2+4b^2+4b+1\)

\(=\left(a+b\right)^2+\left(2b+1\right)^2\)

1/ x² - 4x + 5 + y² + 2y 

= ( x² - 4x + 4 ) + ( y² + 2y + 1 )

= ( x - 2 )² + ( y + 1 )²

2/ 2a² + 2ab - 2ax + x² + b²

= ( a² + 2ab + b² ) + ( x² - 2ax + a² )

= ( a + b )² + ( x - a )²

3/ x² - 2x + y² + 6y + 10

= ( x² - 2x + 1 ) + ( y² + 6y + 9 )

= ( x - 1 )² + ( y + 3 )²

4/ 2x² + y² - 2xy + 10x + 25

= ( x² - 2xy + y² ) + ( x² + 10x + 25 )

= ( x - y )² + ( x + 5 )²

5/ a² + 2ab + 5b² + 4b + 1

= ( a² + 2ab + b² ) + ( 4b² + 4b + 1 )

= ( a + b )² + ( 2b + 1 )²

a) x^2

a) hình như phải là 2x^2 chứ

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

     tách    2y^2 =y^2 +y^2    nha

c) \(z^2-6z+13+t^2+4t=\left(z^2-6z+9\right)+\left(t^2+4t+4\right)=\left(z-3\right)^2+\left(t+2\right)^2\)

tách 13 = 9+4 

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

cũng tách 2z^2 = z^2 + z^2

Lời giải:

Vì $x=9$ nên $x-9=0$
Ta có:

$F=(x^{2017}-9x^{2016})-(x^{2016}-9x^{2015})+(x^{2015}-9x^{2014})-....-(x^2-9x)+x-10$

$=x^{2016}(x-9)-x^{2015}(x-9)+x^{2014}(x-9)-....-x(x-9)+x-10$

$=x^{2016}.0-x^{2015}.0+x^{2014}.0-...-x.0+x-10$

$=x-10=9-10=-1$