Bài 9: Phân tích đa thức thành nhân tử bằng cách phối hợp nhiều phương pháp

a: \(2x^2+3xy-14y^2\)

\(=2x^2+7xy-4xy-14y^2\)

\(=\left(2x^2+7xy\right)-\left(4xy+14y^2\right)\)

\(=x\left(2x+7y\right)-2y\left(2x+7y\right)\)

\(=\left(2x+7y\right)\left(x-2y\right)\)

b: \(\left(x-7\right)\left(x-5\right)\left(x-3\right)\left(x-1\right)+7\)

\(=\left(x-7\right)\left(x-1\right)\left(x-5\right)\left(x-3\right)+7\)

\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)+7\)

\(=\left(x^2-8x\right)^2+15\left(x^2-8x\right)+7\left(x^2-8x\right)+105+7\)

\(=\left(x^2-8x\right)^2+22\left(x^2-8x\right)+112\)

\(=\left(x^2-8x\right)^2+8\left(x^2-8x\right)+14\left(x^2-8x\right)+112\)

\(=\left(x^2-8x\right)\left(x^2-8x+8\right)+14\left(x^2-8x+8\right)\)

\(=\left(x^2-8x+8\right)\left(x^2-8x+14\right)\)

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

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

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

\(=\left(x-3+6x-2\right)\left(x-3-3x+1\right)\)

\(=\left(7x-5\right)\left(-2x-2\right)\)

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

d: \(xy\left(x-y\right)+yz\left(y-z\right)+zx\left(z-x\right)\)

\(=x^2y-xy^2+y^2z-yz^2+zx\left(z-x\right)\)

\(=\left(x^2y-yz^2\right)-\left(xy^2-y^2z\right)+xz\left(z-x\right)\)

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

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

\(=\left(x-z\right)\left(xy+zy-y^2-xz\right)\)

\(=\left(x-z\right)\left[\left(xy-y^2\right)+\left(zy-zx\right)\right]\)

\(=\left(x-z\right)\left[y\cdot\left(x-y\right)-z\left(x-y\right)\right]\)

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

có `x-y=3`

`<=>x^2-2xy+y^2=9`

`<=>x^2-2*40+y^2=9`

`<=>x^2+y^2=89`

`x^3-y^3`

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

`=3*(89+40)`

`=3*129`

`=387`

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

=3^3+3*40*3

=27+360

=387

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

=5^3-3*6*5

=125-90

=35

Lời giải:
Ta thấy:

$2x^2+2x+5=2(x^2+x+\frac{1}{4})+\frac{9}{2}$

$=2(x+\frac{1}{2})^2+\frac{9}{2}\geq 0+\frac{9}{2}=\frac{9}{2}$

$\Rightarrow N=\frac{1}{2x^2+2x+5}\leq \frac{2}{9}$

Vậy $N_{\max}=\frac{2}{9}$. Giá trị này đạt tại $x+\frac{1}{2}=0\Leftrightarrow x=\frac{-1}{2}$

\(a,4x^2-4x+1\\ =\left(2x\right)^2-2.2x+1^2=\left(2x-1\right)^2\\ c,x^2-6xy-25z^2+9y^2\\ =\left(x^2-2.x.3y+9y^2\right)-\left(5z\right)^2\\ =\left(x-3y\right)^2-\left(5z\right)^2\\ =\left(x-3y-5z\right)\left(x-3y+5z\right)\)

Xem lại đề ý b

`#\text {Kr.Ryo}`

`a)`

`4x^2 - 4x + 1`

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

`= (2x - 1)^2`

`b)`

Xem lại đề

`c)`

`2x^2 + 7x + 5`

`= 2x^2 + 2x + 5x + 5`

`= (2x^2 + 2x) + (5x + 5)`

`= 2x(x + 1) + 5(x + 1)`

`= (2x + 5)(x + 1)`

`d)`

`x^2 - 6xy - 25z^2 + 9y^2`

`= (x^2 - 6xy + 9y^2) - 25z^2`

`= [ (x)^2 - 2*x*3y + (3y)^2] - (5z)^2`

`= (x + 3y)^2 - (5z)^2`

`= (x + 3y - 5z)(x + 3y + 5z)`

Em xem gõ lại đề cho đúng nha

a) \(C=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{16}+1\right)\)

\(C=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{16}+1\right)\)

\(C=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(C=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(C=\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(C=5^{32}-1\)

b) \(D=15\cdot\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)...\left(4^{64}+1\right)\)

.....

\(D=\left(4^{32}-1\right)\left(4^{32}+1\right)\left(4^{64}+1\right)\)

\(D=\left(4^{64}+1\right)\left(4^{64}-1\right)\)

\(D=4^{128}-1\)

c) \(E=24\left(5^2+1\right)\left(5^4+1\right)...\left(5^{256}+1\right)\)

\(E=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{256}+1\right)\)

\(E=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{256}+1\right)\)

.....

\(E=\left(5^{128}-1\right)\left(5^{128}+1\right)\left(5^{256}+1\right)\)

\(E=\left(5^{256}-1\right)\left(5^{256}+1\right)\)

\(E=5^{512}-1\)

a: \(C=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^{16}-1\right)\left(5^{16}+1\right)=5^{32}-1\)

b: \(D=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\cdot...\cdot\left(4^{64}+1\right)\)

\(=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\left(4^{64}+1\right)\)

\(=\left(4^8-1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\left(4^{64}+1\right)\)

\(=\left(4^{16}-1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\left(4^{64}+1\right)\)

\(=\left(4^{32}-1\right)\left(4^{32}+1\right)\left(4^{64}+1\right)\)

\(=\left(4^{64}-1\right)\left(4^{64}+1\right)=4^{128}-1\)

c: \(E=\left(5^2-1\right)\left(5^2+1\right)\cdot...\cdot\left(5^{128}+1\right)+5^{256}-1\)

\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\cdot...\cdot\left(5^{128}+1\right)+5^{256}-1\)

\(=\left(5^8-1\right)\left(5^8+1\right)\cdot...\cdot\left(5^{128}+1\right)+5^{256}-1\)

\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\left(5^{128}+1\right)+5^{256}-1\)

\(=\left(5^{32}-1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\left(5^{128}+1\right)+5^{256}-1\)

\(=\left(5^{64}-1\right)\left(5^{64}+1\right)\left(5^{128}+1\right)+5^{256}-1\)

\(=5^{256}-1+5^{256}-1=2\cdot5^{256}-2\)

a) C = (5 - 1)(5 + 1)(5² + 1)(5⁴ + 1)...(5¹⁶ + 1)

= (5² - 1)(5² + 1)(5⁴ + 1)...(5¹⁶ + 1)

= (5⁴ - 1)(5⁴ + 1)...(5¹⁶ + 1)

= (5⁸ - 1)...(5¹⁶ + 1)

= 5³² - 1

b) D = 15(4² + 1)(4⁴ + 1)...(4⁶⁴ + 1)

= (4² - 1)(4² + 1)(4⁴ + 1)...(4⁶⁴ + 1)

= (4⁴ - 1)(4⁴ + 1)...(4⁴⁶ + 1)

= (4¹⁶ - 1)...(4⁶⁴ + 1)

= 4¹²⁸ - 1

c) E = 24(5² + 1)(5⁴ + 1)...(5¹²⁸ + 1)(5²⁵⁶ + 1)

= (5² - 1)(5² + 1)(5⁴ + 1)...(5¹²⁸ + 1)(5²⁵⁶ + 1)

= (5⁴ - 1)(5⁴ + 1)...(5¹²⁸ + 1)(5²⁵⁶ + 1)

= (5⁸ - 1)...(5¹²⁸ + 1)(5²⁵⁶ + 1)

= (5²⁵⁶ - 1)(5²⁵⁶ + 1)

= 5⁵¹² - 1

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

=10^2-4*30

=100-120=-20

=>Vô lý