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

\(a,A=x^3-6x^2+12x-8\)

\(=x^3-3x^2.2+3.2^2x-2^3=\left(x-2\right)^3\)

\(b,B=1-\dfrac{3x}{2}+\dfrac{3x^2}{4}-\dfrac{x^3}{8}\)

\(=\dfrac{8-3x.4+3x^2.2-x^3}{8}\)

\(=\dfrac{-\left(x^3-3x^2.2+3x.4-2^3\right)}{8}\)

\(=\dfrac{-\left(x-2\right)^3}{8}\)

\(C=\left(2x+y\right)^3-6\left(2x+y\right)^2x+12\left(2x+y\right)x^2-8x^3\)

\(=\left(2x+y\right)^3-3.2x\left(2x+y\right)^2+3.\left(2x\right)^2\left(2x+y\right)-\left(2x\right)^3\)

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

\(=y^3\)

a) \(x^2+8x+16\)

\(=x^2+2.4.x+4^2\)

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

b) \(9x^2-24x+16\)

\(=\left(3x\right)^2-2.4.3x+4^2\)

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

c) \(x^2-3x+\dfrac{9}{4}\)

\(=x^2-2.\dfrac{3}{2}.x+\left(\dfrac{3}{2}\right)^2\)

\(=\left(x-\dfrac{3}{2}\right)^2\)

d) \(4x^2y^4-4xy^3+y^2\)

\(=\left(2xy^2\right)^2-2.2xy^2.y+y^2\)

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

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

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

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

f) \(\left(x+3y\right)^2-12xy\)

\(=x^2+6xy+9y^2-12xy\)

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

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

lú quá

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

`b,=(3x-4)^2`

`c,=(x-3/2)^2`

`d,=(2xy^2-y)^2`

`e,=(x-2y-2)^2`

`f,=x^2-6xy+9y^2=(x-3y)^2`

HĐT số 1,2: `A^2+-2AB+B^2=(A-B)^2`

`x^2-2x+1=25`

`<=>(x-1)^2=25=5^2`

- TH1: `x-1=5<=>x=6`

- TH2: `x-1=-5<=>x=-4`

Vậy `S={6;-4}`

\(x^2-2x+1=25\)

\(\Leftrightarrow x^2-2.1.x+1^2=5^2\)

\(\Leftrightarrow\left(x-1\right)^2=5^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

Vậy \(S=\left\{6;-4\right\}\)

a: =-(x^2-4x-2)

=-(x^2-4x+4-6)

=-(x-2)^2+6<=6

Dấu = xảy ra khi x=2

b: =-(x^2-x-2)

=-(x^2-x+1/4-9/4)

=-(x-1/2)^2+9/4<=9/4

Dấu = xảy ra khi x=1/2

\(=\dfrac{\left(4x^2+6x\right)\left(6x+3\right)-5x-5}{\left(6x+3\right)\left(5x+5\right)}\)

\(=\dfrac{24x^3+12x^2+36x^2+18x-5x-5}{\left(6x+3\right)\left(5x+5\right)}\)

\(=\dfrac{24x^3+48x^2+13x-5}{\left(6x+3\right)\left(5x+5\right)}\)

a: =x^2-6x+9+x^2-6x+9

=2(x-3)^2>=0

Dấu = xảy ra khi x=3

b: =-(x^2+4x+y^2-2y)

=-(x^2+4x+4+y^2-2y+1-5)

=-(x+2)^2-(y-1)^2+5<=5

Dấu = xảy ra khi x=-2 và y=1

=x^2+5x+25/4-25/4

=(x+5/2)^2-25/4>=-25/4

Dấu = xảy ra khi x=-5/2

=ab(a-b)(a+b)

=ab(a-1)(a+1)-ab(b-1)(b+1)

Vì a;a-1;a+1là ba số liên tiếp

nên a(a-1)(a+1) chia hết cho 6

Vì b;b-1;b+1 là ba số liên tiếp

nen b(b-1)(b+1) chia hết cho 6

=>A chia hết cho 6

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

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

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

b.

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

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

c.

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

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

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

d.

\(4\left(x^2-y^2\right)-8\left(x-ay\right)-4\left(a^2-1\right)\)

\(=4x^2-8x+4-\left(4y^2-8ay+4a^2\right)\)

\(=4\left[\left(x-1\right)^2-\left(y-a\right)^2\right]\)

\(=4\left(x-1+y-a\right)\left(x-1-y+a\right)\)

a)

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

b)

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

\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)

\(=x^3-x^3+3x^2y+3x^2y+3xy^2-3xy^2+y^3+y^3\)

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

c)

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

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

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

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

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

d)

\(=4x^2-4y^2-8x+8ay-4a^2+4\)

\(=\left(4x^2-8x+4\right)-\left(4y^2-8ay+4a^2\right)\)

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

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

\(=4\left(x-1-y+a\right)\left(x-1+y-a\right)\)

a, (xy+1) mũ 2 -(x+y) mũ 2 =(xy+1-x-y).(xy+1+x+y) =[(xy-x)+(1-y)].[(xy+x)+(1+y)]=[x.(y-1)-(y-1)].[x.(y+1)-(y+1)]=(y-1).(x-1).(y+1).(x-1)=(x-1).[(y-1).(x-1)] dài thế :))

\(\left(3x-1\right)^2-4^2=\left(3x-1-4\right)\left(3x-1+4\right)=\left(3x-5\right)\left(3x+3\right)=3\left(x+1\right)\left(3x-5\right)\)

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

\(\left(3x+1\right)^2-\left(2x-4\right)^2=\left(3x+1-2x+4\right)\left(3x+1+2x-4\right)=\left(x+5\right)\left(5x-3\right)\)

\(\left(6x+9\right)^2-\left(2x+2\right)^2=\left(6x+9-2x-2\right)\left(6x+9+2x+2\right)=\left(4x+7\right)\left(8x+11\right)\)

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

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

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