a: \(x^2+4x+4=x^2+2\cdot x\cdot2+2^2=\left(x+2\right)^2\)

b: \(4x^2-4x+1=\left(2x\right)^2-2\cdot2x\cdot1+1^2=\left(2x-1\right)^2\)

c: \(2x-1-x^2\)

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

d: \(x^2+x+\dfrac{1}{4}=x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)

e: \(9-x^2=3^2-x^2=\left(3-x\right)\left(3+x\right)\)

g: \(\left(x+5\right)^2-4x^2=\left(x+5+2x\right)\left(x+5-2x\right)\)

\(=\left(5-x\right)\left(5+3x\right)\)

h: \(\left(x+1\right)^2-\left(2x-1\right)^2\)

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

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

i: \(=x^2y^2-4xy+4-3\)

\(=\left(xy-2\right)^2-3=\left(xy-2-\sqrt{3}\right)\left(xy-2+\sqrt{3}\right)\)

k: \(=y^2-\left(x-1\right)^2\)

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

l: \(=x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=\left(x+2\right)^3\)

m: \(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2-y^3=\left(2x-y\right)^3\)

Ta có

8 x 3   +   12 x 2 y   +   6 x y 2   +   y 3     =   ( 2 x ) 3   +   3 . ( 2 x ) 2 y   +   3 . 2 x . y 2   +   y 3   =   ( 2 x   +   y ) 3

Đáp án cần chọn là: B

a) Áp dụng HĐT 5 thu được ( 2 a   -   3 b ) 3 .

b) Ta có  8 x 3   +   12 x 2 y   +   6 xy 2   +   y 3  = ( 2 x   +   y ) 3 .

Áp dụng HĐT 7 với A = 2x + y; B = z

( 2 x   +   y ) 3 - z 3 = (2x + y - z)(4 x 2   +   y 2   +   z 2  + 4xy + 2xz + zy).

\(8x^3+12x^2y+6xy^2+y^3-z^3\)

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

\(=\left(2x+y-z\right)\left[4x^2+z\left(2x+y\right)+z^2\right]\)

a, 8a3 - 36a2 +54ab2 - 27b3

=(8a3-36a2b +54ab2 - 27b3)

=(2a-3b)2

=(2a-3b)(2a-3b)(2a-3b)

b, 8x3 + 12x2y + 6xy2 + y3 - z 3

=(8x3 + 12x2y + 6xy2 + y3) - z3

=(2x + y)3 - y3

=(2x + y +z) . [ (2x + Y)2 + 2(2x + y)+ z2

= (2x + y + z)(4x2 + 4xy + y2 + 4x + 2y + z2

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

Chúc bạn học tốt và nhớ click cho mình với nhá!

= (5x-25) + (5x - x²)

= 5(x-5) + x(5-x)

= 5(x-5) - x(x-5)

= (5 - x)(x - 5)

\(10x-25-x^2=-\left(x^2-2\cdot x\cdot5+5^2\right)=-\left(x-5\right)^2\)

\(a,x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right).\)

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

\(b,9x^2+y^2+6xy=\left(3x\right)^2+2.3x.y+y^2=\left(3x+y\right)^2\)

\(c,6x-9-x^2=-\left(x^2-6x+9\right)=-\left(x^2-2.x.3+3^2\right)=-\left(x-3\right)^2\)

\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2=\left(x+2y\right)^2\)

a/ x² + 6x + 9 = (x + 3)² = (x + 3)(x + 3)

b/ 10x - 25 - x² = -x² + 10x - 25 = -(x² -10x + 25) = -(x - 5)² = -(x - 5)(x - 5)

c/ \(8x^3+\frac{1}{8}=\left(2x\right)^3+\left(\frac{1}{2}\right)^3=\left(2x+\frac{1}{2}\right)\left(4x^2-x+\frac{1}{4}\right)\)

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

= 2x(4x + 2)

Em áp dụng hđt số 3 trong sgk nhé.

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

                                      = 4x(2x+1)