a: =x^3+8-1+27x^3=28x^3+7

b: Sửa đề: (2+y)(y^2-2y+4)+(5-y)(25+5y+y^2)

=8+y^3+125-y^3

=133

Bài 12:

a) \(\left(\dfrac{1}{2}x+4\right)^2\)

\(=\left(\dfrac{1}{2}x\right)^2+2\cdot\dfrac{1}{2}x\cdot4+4^2\)

\(=\dfrac{1}{4}x^2+4x+16\)

b) \(\left(7x-5y\right)^2\)

\(=\left(7x\right)^2-2\cdot7x\cdot5y+\left(5y\right)^2\)

\(=49x^2-70xy+25y^2\)

c) \(\left(6x^2+y^2\right)\left(y^2-6x^2\right)\)

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

\(=y^4-36x^4\)

d) \(\left(x+2y\right)^2\)

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

\(=x^2+4xy+4y^2\)

e) \(\left(x-3y\right)\left(x+3y\right)\)

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

\(=x^2-9y^2\)

f) \(\left(5-x\right)^2\)

\(=5^2-2\cdot5\cdot x+x^2\)

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

\(11,\)

\(a,\left(7x+4\right)^2-\left(7x+4\right)\left(7x-4\right)\)

\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)

\(=\left(7x+4\right).8=56x+32\)

\(b,\left(x+2y\right)^2-6xy\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x+2y-6xy\right)\)

Bài `12`

`(1/2x+4)^2`

`=(1/2x)^2 + 2 . 1/2x.4 + 4^2`

`= 1/4 x^2 +4x + 16`

__

`(7x-5y)^2`

`=(7x)^2-2.7x.5y+(5y)^2`

`= 49x^2 -  70xy   + 25y^2`

__

`(6x^2+y^2)(y^2-6x^2)`

`=(y^2+6x^2)(y^2-6x^2)`

`=(y^2)^2 - (6x^2)^2`

`=y^4-36x^4`

__

`(x+2y)^2`

`=x^2+ 2.x.2y+(2y)^2`

`= x^2 + 4xy +4y^2`

__

`(x-3y)(x+3y)`

`=x^2 - (3y)^2`

`=x^2 - 9y^2`

__

`(5-x)^2`

`=5^2 -2.5.x+x^2`

`=25 - 10x+x^2`

Bài `11`

`(7x+4)^2 -(7x+4)(7x-4)`

`= (7x+4)(7x+4) -(7x+4)(7x-4)`

`=(7x+4)(7x+4-7x+4)`

`=8(7x+4)`

`= 56x+32`

__

`(x+2y)^2-6xy (x+2y)`

`= (x+2y) (x+2y-6xy)`

a) \(P=x\left(5x+15y\right)-5y\left(3x-2y\right)-5\left(y^2-2\right)=5x^2+15xy-15xy+10y^2-5y^2+10=5x^2+5y^2+10\)

b) P = 0

=> \(5x^2+5y^2+10=0\)

\(\Rightarrow x^2+y^2=-2\)

Mà: \(x^2+y^2\ge0\)

=> Ko có cặp (x; y) nào thỏa mãn P = 0

P = 10

=> \(5x^2+5y^2+10=10\)

=> \(x^2+y^2=0\)

Mà: \(x^2+y^2\ge0\)

=> x = 0; y = 0

a) Ta có: \(P=x\left(5x+15y\right)-5y\left(3x-2y\right)-5\left(y^2-2\right)\)

\(=5x^2+15xy-15xy+10y^2-5y^2+10\)

\(=10\)

\(B=4x^2-y^4-y^4+4xy^2-4x^2-4xy^2=-2y^4=-32\)

Ta có:

a) Ta có: \(\left(x+2y\right)\left(x^2-2xy+4y^2\right)-\left(x-y\right)\left(x^2+xy+y^2\right)\)

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

\(=x^3+8y^3-x^3+y^3\)

\(=9y^3\)

b) Ta có: \(\left(x+1\right)\left(x-1\right)^2-\left(x+2\right)\left(x^2-2x+4\right)\)

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

\(=x^3-2x^2+x+x^2-2x+1-\left(x^3+8\right)\)

\(=x^3-x^2-x+1-x^3-8\)

\(=-x^2-x-7\)

2(x – y)(x + y) +  x + y 2  +  x - y 2

=  x + y 2  +2( x+ y).(x- y) + x - y 2

(áp dụng hằng đẳng thức thứ 1với A = x+ y, B = x- y)

= x + y + x - y 2 = 2 x 2 = 4 x 2