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23 tháng 9 2016

a) Hình như phân tích không được

b)  \(2x^4+5x^3+13x^2+25x+15\)

\(=x^3+1+2x^4+2x^3+13x^2+13x+12x+12+2+2x^3\)

\(=\left(x^3+1\right)+\left(2x^4+2x^3\right)+\left(13x^2+13x\right)+\left(12x+12\right)+2\left(1+x^3\right)\)

\(=\left(x+1\right)\left(x^2-x+1\right)+2x^3\left(x+1\right)+13x\left(x+1\right)+12\left(x+1\right)+2\left(x+1\right)\left(x^2-x+1\right)\)

\(=\left(x+1\right)\left(x^2-x+1+2x^3+13x+12+2x^2-2x+2\right)\)

\(=\left(x+1\right)\left(3x^2+10x+15+2x^3\right)\)

\(=\left(x+1\right)\left[x^2\left(2x+3\right)+5\left(2x+3\right)\right]\)

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

17 tháng 7 2023

a) 6x² + 7xy + 2y²

= 6x² + 4xy + 3xy + 2y²

= (6x² + 4xy) + (3xy + 2y²)

= 2x(3x + 2y) + y(3x + 2y)

= (3x + 2y)(2x + y)

b) x² - y² + 10x - 6y + 16

= x² + 10x + 25 - y² - 6y - 9

= (x² + 10x + 25) - (y² + 6y + 9)

= (x + 5)² - (y + 3)²

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

= (x - y + 2)(x + y + 8)

c) 4x⁴ + y⁴

= 4x⁴ + 4x²y² + y⁴ - 4x²y²

= (2x² + y²)² - (2xy)²

= (2x² + y² - 2xy)(2x² + y² + 2xy)

7 tháng 12 2021

a) 7x – 14 = 7(x - 2)

b) 5x3 - 10x2y +5xy2 = 5x(x2 - 2xy - y2) = 5x(x - y)2

c) 25 – x2= (x - 5)(x + 5)

7 tháng 12 2021

a, \(7\left(x-2\right)\)

c, \(\left(5-x\right)\left(5+x\right)\)

7 tháng 12 2021

a, 7x - 14 = 7 ( x - 2 )

b, 5x3 - 10x2y + 5xy2 

= 5x ( x2 - 2xy + y2 ) 

= 5x ( x - y )2

c, 25 - x2 = 52 - x2 = ( 5 - x ) ( 5 + x )

13 tháng 1

Bài 1:

\(a,x^4+5x^2+9\\=(x^4+6x^2+9)-x^2\\=[(x^2)^2+2\cdot x^2\cdot3+3^2]-x^2\\=(x^2+3)^2-x^2\\=(x^2+3-x)(x^2+3+x)\)

\(b,x^4+3x^2+4\\=(x^4+4x^2+4)-x^2\\=[(x^2)^2+2\cdot x^2\cdot2+2^2]-x^2\\=(x^2+2)^2-x^2\\=(x^2+2-x)(x^2+2+x)\)

\(c,2x^4-x^2-1\\=2x^4-2x^2+x^2-1\\=2x^2(x^2-1)+(x^2-1)\\=(x^2-1)(2x^2+1)\\=(x-1)(x+1)(2x^2+1)\)

13 tháng 1

Bài 2:

\(a,\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=120\)

\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\cdot\left[\left(x+2\right)\left(x+3\right)\right]=120\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=120\) (1)

Đặt \(x^2+5x+5=y\), khi đó (1) trở thành:

\(\left(y-1\right)\left(y+1\right)=120\)

\(\Leftrightarrow y^2-1=120\)

\(\Leftrightarrow y^2=121\)

\(\Leftrightarrow\left[{}\begin{matrix}y=11\\y=-11\end{matrix}\right.\)

+, TH1: \(y=11\Leftrightarrow x^2+5x+5=11\)

\(\Leftrightarrow x^2+5x-6=0\)

\(\Leftrightarrow x^2-x+6x-6=0\)

\(\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\left(\text{nhận}\right)\)

+, TH2: \(y=-11\Leftrightarrow x^2+5x+5=-11\)

\(\Leftrightarrow x^2+5x+16=0\)

\(\Leftrightarrow\left[x^2+2\cdot x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]-\dfrac{25}{4}+16=0\)

\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)

Ta thấy: \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}\ge\dfrac{39}{4}>0\forall x\)

Mà \(\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)

\(\Rightarrow\) loại

Vậy \(x\in\left\{1;-6\right\}\).

\(b,\) Đề thiếu vế phải rồi bạn.

a: =(x^2+x)^2+4(x^2+x)-12

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

=(x^2+x+6)(x+2)(x-1)

b: =(x^2+8x)^2+22(x^2+8x)+120

=(x^2+8x+12)(x^2+8x+10)

=(x+2)(x+6)(x^2+8x+10)

c: =8x^2+12x-2x-3

=(2x+3)(4x-1)

a: (x^2+x)^2+4x^2+4x-12

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

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

=(x^2+x+6)(x+2)(x-1)

b: =(x^2+8x)^2+22(x^2+8x)+105+15

=(x^2+8x)^2+22(x^2+8x)+120

=(x^2+8x+10)(x^2+8x+12)

=(x^2+8x+10)(x+2)(x+6)

c: =8x^2+12x-2x-3

=(2x+3)(4x-1)

30 tháng 10 2021

a/ \(=5x\left(x^2-2x+3\right)\)

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

30 tháng 10 2021

a) \(5x^3-10x^2+15x=5x\left(x^2-2x+3\right)\)

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

AH
Akai Haruma
Giáo viên
25 tháng 10 2021

a. 

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

b.

$x^2(x-1)+16(1-x)=x^2(x-1)-16(x-1)=(x-1)(x^2-16)=(x-1)(x-4)(x+4)$

c.

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

d.

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

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

AH
Akai Haruma
Giáo viên
25 tháng 10 2021

e.

$x^4+4y^4=(x^2)^2+(2y^2)^2+2.x^2.2y^2-4x^2y^2$

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

f.

$x^4-13x^2+36=(x^4-4x^2)-(9x^2-36)$

$=x^2(x^2-4)-9(x^2-4)=(x^2-9)(x^2-4)=(x-3)(x+3)(x-2)(x+2)$

g.

$(x^2+x)^2+4x^2+4x-12=(x^2+x)^2+4(x^2+x)-12$

$=(x^2+x)^2-2(x^2+x)+6(x^2+x)-12$

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

$=[x(x-1)+2(x-1)](x^2+x+6)=(x-1)(x+2)(x^2+x+6)$

h.

$x^6+2x^5+x^4-2x^3-2x^2+1$

$=(x^6+2x^5+x^4)-(2x^3+2x^2)+1$

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

25 tháng 10 2021

a) \(3x\left(2x-y\right)+5y\left(y-2x\right)\)

\(=3x\left(2x-y\right)-5y\left(2x-y\right)\)

\(=\left(3x-5y\right)\left(2x-y\right)\)

b) \(\left(x-5\right)^2-9\left(x+y\right)^2\)

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

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

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

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

25 tháng 10 2021

a: \(3x\left(2x-y\right)+5y\left(y-2x\right)=\left(2x-y\right)\left(3x-5y\right)\)

e: \(x^2-10x+24=\left(x-4\right)\left(x-6\right)\)

26 tháng 7 2021

a, \(x^2-5x+6=x^2+x-6x+6=x\left(x-1\right)-6\left(x-1\right)=\left(x-1\right)\left(x-6\right)\)

b, \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x^2-2x+5x-10\right)\)

\(=3\left[x\left(x-2\right)+5\left(x-2\right)\right]=3\left(x-2\right)\left(x+5\right)\)

c, \(x^2+7x+10=x^2+2x+5x+10=x\left(x+2\right)+5\left(x+2\right)=\left(x+2\right)\left(x+5\right)\)

26 tháng 7 2021

a) x2 - 5x + 6 = (x2-2x)-(3x-6)=x(x-2)-3(x-2)=(x-3)(x-2)

b) 3x2 + 9x -30= 3(x2+3x-10) = 3((x2+5x)-(2x+10)) = 3(x(x+5)-2(x+5)) = 3(x-2)(x+5)

c) x2 + 7x + 10 =( x2+5x)+(2x+10)=x(x+5)+2(x+5)=(x+2)(x+5)