Bài 4: Những hằng đẳng thức đáng nhớ (Tiếp)

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

\(A=3\left(x^2-2x+1\right)-\left(x^2+2x+1\right)+2\left(x^2-9\right)\)

\(A=3x^2-6x+3-x^2-2x-1+2x^2-18\)

\(A=4x^2-8x--16\)

\(B=5x\left(x-7\right)\left(x+7\right)-2\left(2x-1\right)^2\)

\(B=5x\left(x^2-49\right)-2\left(4x^2-4x+1\right)\)

\(B=5x^3-245x-8x^2+8x-2\)

\(B=5x^3-8x^2-253x-2\)

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

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

\(C=x^4-4-x^2+4\)

\(C=x^4-x^2\)

\(D=\left(4x-1\right)\left(1+16x^2+4x\right)-\left(4x+1\right)^3\)

\(D=\left(4x-1\right)\left(16x^2+4x+1\right)-\left(64x^3+48x^2+12x+1\right)\)

\(D=64x^3-1-64x^3-48x^2-12x-1\)

\(D=-48x^2-12x-2\)

\(A=3\left(x^2-2x+1\right)-\left(x^2+2x+1\right)+2\left(x^2-3^2\right)\\ =3x^2-6x+3-x^2-2x-1+2x^2-18\\ =4x^2-8x-16\)

\(B=5x\left(x^2-7^2\right)-2\left(4x^2-4x+1\right)\\ =5x^3-245x-8x^2+8x-2\\ =5x^3-8x^2-237x-2\)

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

\(D=\left(4x\right)^3-1^3-\left(64x^3+48x^2+12x+1\right)\\ =64x^3-1-64x^3-48x^2-12x-1\\ =-48x^2-12x-2\)

Bài 3:

\(I=64^2-72\cdot64+36^2\)

\(I=64^2-2\cdot36\cdot64+36^2\)

\(I=\left(64-36\right)^2\)

\(I=28^2\)

\(I=784\)

\(K=47\cdot53\)

\(K=\left(50-3\right)\left(50+3\right)\)

\(K=50^2-3^2\)

\(K=2500-9\)

\(K=2491\)

\(M=123^3-69\cdot123^2+369\cdot23^2-23^3\)

\(M=123^3-3\cdot23\cdot123^2+3\cdot123\cdot23^2-23^3\)

\(M=\left(123-23\right)^3\)

\(M=100^3\)

\(M=1000000\)

\(N=54^3+138\cdot54^2+162\cdot46^2+46^3\)

\(N=54^3+3\cdot46\cdot54^2+3\cdot54\cdot46^2+46^3\)

\(N=\left(54+46\right)^3\)

\(N=100^3\)

\(N=1000000\)

Bài 1

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

= 3x² - 6x + 1 - x² - 2x - 1 + 2x² - 18

= (3x² - x² + 2x²) + (-6x - 2x) + (1 - 1 - 18)

= 4x² - 8x - 18

B = 5x(x - 7)(x + 7) - 2(2x - 1)²

= 5x³ - 245x - 8x² + 8x - 2

= 5x³ - 8x² + (-245x + 8x) - 2

= 5x³ - 8x² - 237x - 2

C = (x² - 2)(x² + 2) - x² + 4

= x⁴ - 4 - x² + 4

= x⁴ - x²

D = (4x - 1)(1 + 16x + 4x) - (4x + 1)³

= 4x + 80x² - 1 - 20x - 64x³ - 48x² - 12x - 1

= -64x³ + (80x² - 48x²) + (4x - 20x - 12x) + (-1 - 1)

= -64x³ + 32x² - 28x - 2

Bài 3 

Sửa đề I = 64² + 72.64 + 36²

= 64² + 2.64.36 + 36²

= (64 + 36)²

= 100²

= 10000

K = 47.53

= (50 - 3)(50 + 3)

= 50² - 3²

= 2500 - 9

= 2491

M = 123³ - 69.123² + 369.23² - 23³

= 123³ - 3.123².23 + 3.123.23² - 23³

= (123 - 23)³

= 100³

= 1000000

N = 54³ + 138.54² + 162.46² + 46³

= 54³ + 3.54².46 + 3.54.46² + 46³

= (54 + 46)³

= 100³

= 1000000

Bài 2:

\(E=x^2-4x+6=x^2-4x+4+2\\ =\left(x-2\right)^2+2\ge2\)

GTNN của E đạt `2` khi và chỉ khi `x=2`

\(F=25x^2+20x+8=\left(5x\right)^2+2.5x.2+4+4\\ =\left(5x+2\right)^2+4\ge4\)

GTNN của F đạt `4` khi và chỉ khi `x=-2/5`

\(G=x^2-x=x^2-2.\dfrac{1}{2}.x+\dfrac{1}{4}-\dfrac{1}{4}\\ =\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

GTNN của G đạt \(-\dfrac{1}{4}\) khi và chỉ khi `x=1/2`

\(H=-4x^2-4x-3=-\left(4x^2+4x+3\right)\\ =-\left(\left(2x\right)^2+2.2x.1+1+2\right)\\ =-\left(2x+1\right)^2-2\le-2\)

GTLN của H đạt `-2` khi và chỉ khi `x=-1/2`

$HaNa$☘

có `x-y=4`

`<=>x^2 -2xy+y^2 =16`

`<=>12-2xy+16`

`<=>-2xy=4`

`<=>xy=-2`

`x^3 -y^3`

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

`=4(12-2)`

`=4*10`

`=40`

Có: \(x-y=4\)
\(\Rightarrow\left(x-y\right)^2=16\)

\(\Rightarrow x^2-2xy+y^2=16\)

\(\Rightarrow x^2+y^2-2xy=16\)

\(\Rightarrow12-2xy=16\) \(\Leftrightarrow2xy=-4\Leftrightarrow xy=-2\)

Lại có: \(A=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\)

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

\(=4.\left(12-2\right)\) (do \(x^2+y^2=12;xy=-2\))

\(=4.10=40\)

                                                                            Vậy \(A=40\).

`@` `\text {Ans}`

`\downarrow`

`(2x - 3)^2 - (2x + 3)^2`

`= 4x^2 - 12x + 9 - (4x^2 + 12x + 9)`

`= 4x^2 - 12x + 9 - 4x^2 - 12x - 9`

`= (4x^2 - 4x^2) + (-12x - 12x) + (9-9)`

`= -24x`

____

`@` CT: 

`(A + B)^2 = A^2 + 2AB + B^2`

`(A - B)^2 = A^2 - 2AB + B^2`

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

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

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

\(=4x\cdot-6\)

\(=-24x\)

\(a,\left(2x+1\right)^2-4\left(x+2\right)^2=9\\ \Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)-9=0\\ \Leftrightarrow4x^2-4x^2+4x-16x+1-16-9=0\\ \Leftrightarrow-12x=24\\ \Leftrightarrow x=\dfrac{24}{-12}=-2\\ b,\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\\ \Leftrightarrow x^2+6x+9-\left(x^2+4x-32\right)=1\\ \Leftrightarrow x^2-x^2+6x-4x=1-9-32\\ \Leftrightarrow2x=-40\\ \Leftrightarrow x=-20\\ c,3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\\ \Leftrightarrow3\left(x^2+4x+4\right)+\left(4x^2-4x+1\right)-7\left(x^2-9\right)=36\\ \Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2+63=36\\ \Leftrightarrow3x^2+4x^2-7x^2+12x-4x=36-12-1-63\\ \Leftrightarrow8x=-40\\ \Leftrightarrow x=\dfrac{-40}{8}=-5\)

1: (4a-b)(4a+b)=(4a)^2-b^2=16a^2-b^2

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

=(x^2y)^2-4y^2

=x^4y^2-4y^2

(3/4x+3/5y)(3/5y-3/4x)=(3/5y)^2-(3/4x)^2=9/25y^2-9/16x^2

2: (x+2)(x^2-2x+4)=x^3+8

(3x+2y)(9x^2-6xy+4y^2)=27x^3+8y^3

3: (5-3x)(5+3x+9x^2)=125-27x^3

(1/2x-1/5y)(1/4x^2+1/10xy+1/25y^2)=1/8x^3-1/125y^3

a, \(25+10x+x^2=5^2+2.5x+x^2=\left(5+x\right)^2\)
b, \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3=\left(2x-\dfrac{1}{2}\right)\left[\left(2x\right)^2+2x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
c, \(x^2-10x+25=x^2-2.5x+5^2=\left(x-5\right)^2\)

1. \(25+10x+x^2\\ \Leftrightarrow5^2+2\cdot5\cdot x+x^2\\ \Leftrightarrow\left(5+x\right)^2\\ \Leftrightarrow\left(5+x\right)\left(5+x\right)\)

2. \(8x^3-\dfrac{1}{8}\\ \Leftrightarrow\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3\\ \Leftrightarrow\left(2x-\dfrac{1}{2}\right)\left[\left(2x\right)^2+2x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]\\ \Leftrightarrow\left(2x-\dfrac{1}{2}\right)\left[4x^2+x+\dfrac{1}{4}\right]\)

3. \(x^2-10x+25\\ \Leftrightarrow x^2-2\cdot5\cdot x+5^2\\ \Leftrightarrow\left(x-5\right)^2\\ \Leftrightarrow\left(x-5\right)\left(x-5\right)\)

`a, 4x^2 - 25y^2 = (2x-5y)(2x+5y)`.

`b, 8x^3 +27 = (2x+3)(4x^2 - 6x + 9)`.

`c, 125x^3 - 64y^3 = (5x)^3 - (4y)^3 = (5x-4y)(25x^2 + 20xy + 16y^2)`.

\(a,\\ 4x^2-25y^2=\left(2x\right)^2-\left(5y\right)^2=\left(2x-5y\right)\left(2x+5y\right)\\ b,\\ 8x^3+27=\left(2x\right)^3+3^3=\left(2x+3\right)\left(4x^2+6x+9\right)\\ c,\\ 125x^3-64y^3=\left(5x\right)^3-\left(4y\right)^3=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)

a: =>x^2-6x+9+y^2+8y+16=0

=>(x-3)^2+(y+4)^2=0

=>x=3 và y=-4