Bài 5: Những hằng đẳng thức đáng nhớ (Tiếp)

a)

\(8x^3-64=\left(2x-4\right)\left(4x^2+8x+16\right)=8\left(x-2\right)\left(x^2+2x+4\right)\)

b)

\(1+8x^6y^3=\left(1+2x^2y\right)\left(1-2x^2y+4x^4y^2\right)\)

c)

\(=\left(3x+\dfrac{y}{2}\right)\left(9x^2-\dfrac{3}{2}xy+\dfrac{y^2}{4}\right)\)

d)

\(=\left(5x+3y\right)\left(25x^2-15xy+9y^2\right)\)

a)

\(=\left(3x-1-16\right)\left(3x-1+16\right)=\left(3x-17\right)\left(3x+15\right)\)

b)

\(=\left(5x-4-7x\right)\left(5x-4+7x\right)=\left(-2x-4\right)\left(12x-4\right)\)

c)

\(\left(2x+5-x+9\right)\left(2x+5+x-9\right)=\left(x+14\right)\left(3x-4\right)\)

d)

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

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

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

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

e)

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

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

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

\(=\left(4x-7\right)\left(8x+11\right)\)

f)

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

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

a) =(3x -1-16)(3x-1 +16)

=( 3x-17)(3x+ 15)

b) =(5x- 4- 7x)(5x -4 +7x)

=(-2x-4)(12x-4)

c) =(2x+5-x-9)(2x+5+x-9)

=(x-4)(3x-4)

a)

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

b)

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

c)

\(1+12x+36x^2=\left(1+6x\right)^2\)

d)

\(9x^2-24xy+16y^2=\left(3x-4y\right)^2\)

e)

\(\dfrac{x^2}{4}+2xy+4y^2=\left(\dfrac{x}{2}+2y\right)^2\)

f)

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

a)

\(2xy^2-10xy=2xy\left(y-5\right)\)

b)

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

c)

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

a) `2xy^2-10xy=2xy(y-5)`

b) `x^2+xy-5x-5y`

`=x(x+y)-5(x+y)`

`=(x-5)(x+y)`

c) `25-x^2-y^2+2xy`

`= 25-(x^2-2xy+y^2)`

`= 5^2-(x-y)^2`

`= (5-x+y)(5+x-y)`

a:=x^2-2xy+y^2+x^2+2xy+y^2

=2x^2+2y^2

b: =4a^2+4ab+b^2-4a^2+4ab-b^2

=8ab

c: =x^2+2xy+y^2-x^2+2xy-y^2

=4xy

d: =4x^2-4x+1-2(4x^2-12x+9)+4

=4x^2-4x+5-8x^2+24x-18

=-4x^2+20x-13

a)

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

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

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

b)

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

\(=2b\cdot4a=8ab\)

c)

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

\(=2y\cdot2x=4xy\)

d)

\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)

\(=4x^2-4x+1-8x^2+24x-18+4\)

\(=4x^2-8x^2-4x+24x+1-18+4\)

\(=-4x^2+20x-13\)

bn tách từng bài 1 ra

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

\(x^2-8x+16=\left(x-4\right)^2\)

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

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

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

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

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

a, x^2 + 4x +4= x^2 + 2*x*2 + 2^2 = (x+2)^2

b,x^2 + 8x + 16 = x^2 + 2*x*4 + 4^2 = (x+4)^2

c, = x^2 - 25

d, = x^2 + 1x +1x +1 = x(x+1)+(x + 1)= ( x+1)(x+1)=(x+1)^2

e,= (2x)^2 - 3^2 = (2x+3)(2x-3)

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

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

\(c,\left(x^2+y^2\right)^2=x^4+2x^2y^2+y^4\)

\(d,\left(x^3+2y^2\right)=x^6+4x^3y^2+4y^4\)

\(e,\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\)

\(f,\left(x-y^2\right)^2=x^2-2xy^2+y^4\)

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

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

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

\(\left(x^3+2y^2\right)^2=x^6+4x^3y^2+4y^4\)

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

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

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

`( x -1) ( x^2 + x +1 ) - x^2 ( x + 1 )`

`=x^2 -1 - x^3 + x^2`

`= 2x^2 -1-x^3`

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

\(x^2-4+\left(x-2\right)^2=0\)
\(\Rightarrow x^2-4+x^2+4x+4=0\)
\(\Rightarrow2x^2+4x=0\)
\(\Rightarrow2x\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x+2=0\end{matrix}\right.\)            \(\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
                Vậy \(x\in\left\{0;-2\right\}\)

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

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

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

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

\(\Rightarrow\left(x-2\right)2x=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-2=0\\2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

Vậy ...