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

Trả lời:

1, x³ - x² - 4

= x³ - x² - 4 + 2x - 2x

= x³ - 2x² + x² - 4 + 2x - 2x

= ( x³ + x² + 2x ) - ( 2x² + 2x + 4 )

= x ( x² + x + 2 ) - 2 ( x² + x + 2 )

= ( x - 2 )( x² + x + 2 )

2, x³ - 2x - 4 

= x³ - 2x - 4 + 2x² - 2x²

= x³ - 4x + 2x - 4 + 2x² - 2x²

= ( x³ + 2x² + 2x ) - ( 2x² + 4x + 4 )

= x ( x² + 2x + 2 ) - 2 ( x² + 2x + 2 )

= ( x - 2 )( x² + 2x + 2 )

giúp mình đi gấp lm rùi

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a. $6x^2-11x=x(6x-11)$
b. $x^7+x^5+1=(x^7-x)+(x^5-x^2)+x+x^2+1$

$=x(x^6-1)+x^2(x^3-1)+(x^2+x+1)$
$=x(x^3-1)(x^3+1)+x^2(x^3-1)+(x^2+x+1)$
$=(x^3-1)(x^4+x+x^2)+(x^2+x+1)$

$=(x-1)(x^2+x+1)(x^4+x^2+x)+(x^2+x+1)$
$=(x^2+x+1)[(x-1)(x^4+x^2+x)+1]$

$=(x^2+x+1)(x^5-x^4+x^3-x+1)$

c.

$x^8+x^4+1=(x^4)^2+2.x^4+1-x^4$

$=(x^4+1)^2-(x^2)^2$

$=(x^4+1-x^2)(x^4+1+x^2)$

$=(x^4+1-x^2)(x^4+2x^2+1-x^2)$

$=(x^4-x^2+1)[(x^2+1)^2-x^2]$

$=(x^4-x^2+1)(x^2+1-x)(x^2+1+x)$

d.

$x^3-5x+8-4=x^3-5x+4$

$=x^3-x^2+x^2-x-(4x-4)$

$=x^2(x-1)+x(x-1)-4(x-1)=(x-1)(x^2+x-4)$

e.

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

$=x^2(x^3-1)+x(x^3-1)+x^2+x+1$

$=(x^3-1)(x^2+x)+(x^2+x+1)$
$=(x-1)(x^2+x+1)(x^2+x)+(x^2+x+1)$

$=(x^2+x+1)[(x-1)(x^2+x)+1]$

$=(x^2+x+1)(x^3-x+1)$

\(x^4+2x^2-24\)

\(=x^4-4x^2+6x^2-24\)

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

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

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

a, \(a^4+64=\left(a^2\right)^2+8^2=\left(a^2\right)^2+16a^2+8^2-16a^2\)

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

b, \(a^4+4b^2=\left(a^2\right)^2+\left(2b\right)^2=\left(a^2\right)^2+4a^2b+4b^2-4a^2b\)ĐK : b >= 0 

\(=\left(a+2b\right)^2-\left(2a\sqrt{b}\right)^2=\left(a+2b-2a\sqrt{b}\right)\left(a+2b+2a\sqrt{b}\right)\)

p/s : mình nghĩ phần b bạn nên để đề là \(a^4+4b^4\)sẽ hợp lí hơn 

Anh nhận bú lồn hoàn toàn miễn phí nha mấy em

\(2x^4+128y^4\)

\(=2x^4+2.\left(8y^2\right)^2\)

\(=2\left[x^4+\left(8y^2\right)^2\right]\)

\(=2\left[x^4+2x^28y^2+\left(8y^2\right)^2-2x^28y^2\right]\)

\(=2\left[\left(x^2+8y^2\right)^2-\left(4xy\right)^2\right]\)

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

2x4+128x^4

2x^4+2.(8y^2)^2

2.(x^4+(8y^2)^2)

2.((x^2)^2+2.x^2.8y^2+(8y^2)^2-2x^2.8y^2)

2.(x^2+8y^2)-(4.x.y)^2

2.((x^2+8y^2)-4xy).((x^2+8y^2)+4xy)

2.(x^2+8y^2-4xy).(x^2+8y^2+4xy)

a, 3x^2 + 13x + 10  

= 3x^2 + 3x + 10x + 10 

= 3x(x + 1) + 10(x + 1)

= (3x + 10)(x + 1)

b, x^2 - 10x + 21

= x^2 - 3x - 7x + 21

= x(x - 3) - 7(x - 3)

= (x - 7)(x - 3)

c, 6x^2 - 5x + 1

= 6x^2 - 3x - 2x + 1

= 3x(2x - 1) - (2x - 1)

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

Bạn đăng 1 lần nhiều bài như vậy làm người khác nản lắm đấy =) đơn giản bài rất dài mà mik cx ko chắc là bản thân mik có đc k hay ko nên phải nản vậy thôi :)

1a)\(3x^2+13x+10=3x^2+3x+10x+10\)

\(3x\left(x+1\right)+10\left(x+1\right)=\left(3x+10\right)\left(x+1\right)\)

b)\(x^2-10x+21=x^2-3x-7x+21\)

\(=x\left(x-3\right)-7\left(x-3\right)=\left(x-7\right)\left(x-3\right)\)

c)\(6x^2-5x+1=6x^2-3x-2x+1\)

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

1,

\(a,3x^2+13x+10=3x^2+3x+10x+10\)

\(=3x\left(x+1\right)+10\left(x+1\right)\)

\(=\left(x+1\right)\left(3x+10\right)\)

\(b,x^2-10x+21=x^2-7x-3x+21\)

\(=x\left(x-7\right)-3\left(x-7\right)\)

\(=\left(x-7\right)\left(x-3\right)\)

\(c,6x^2-5x+1=6x^2-2x-3x+1\)

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

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

2,

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

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

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

\(b,x^4+18^2=\left(x^2\right)^2+36x^2+18^2-36x^2\)

\(=\left(x^2+18\right)-36x^2\)

\(=\left(x^2+18-6x\right)\left(x^2+18+6x\right)\)

\(c,x^4+3x^2+4=x^4+4x^2+4-x^2\)

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

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

Toán Học Team