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26 tháng 10 2023

1)

\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)

Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:

\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)

2) Bạn xem lại đề!

AH
Akai Haruma
Giáo viên
26 tháng 10 2023

Lời giải:

1. 
$x^3+3x^2-16x-48=(x^3+3x^2)-(16x+48)=x^2(x+3)-16(x+3)$

$=(x+3)(x^2-16)=(x+3)(x-4)(x+4)$

2.

$4x(x-3y)+12y(3y-x)=4x(x-3y)-12y(x-3y)=(x-3y)(4x-12y)=4(x-3y)(x-3y)=4(x-3y)^2$

3.

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

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

17 tháng 3 2020

bạn ơi hình như sai đề thì phải a bạn mình nghĩ phải là \(\left(x^2-x+2\right)^2\)

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

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

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

\(=2x^2+2-x=2x^2+2.1-x=2\left(x^2+1\right)-x\)

22 tháng 9 2021

2. \(2xy-4y+2y^2\)

 \(=2y\left(x-2+y\right)\)

11 tháng 7 2016

1/ \(3x^2+6x+3-3y^2=3x^2+3x+3x+3-3y^2\)

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

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

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

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

2/ \(25-x^2-y^2+2xy=5^2-\left(x^2+y^2-2xy\right)\)

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

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

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

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

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

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

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

30 tháng 4 2017

 1)    \(25x^4-10x^2y+y^2\)

\(\Leftrightarrow\left(5x^2\right)^2+2\cdot\left(5x^2\right)\cdot y+y^2\)

\(\Leftrightarrow\left(5x^2+y\right)^2\)

 2)   \(x^4+2x^3-4x-4\)

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

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

 3)  \(x^4+x^2+1\)

\(\Leftrightarrow x^4+x^2-x+x+1\)

 \(\Leftrightarrow\left(x^4-x\right)+\left(x^2+x+1\right)\)

\(\Leftrightarrow x\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2-x+1\right)\)

 4)    \(x^3-5x^2-14x\)\(\Leftrightarrow x^3-7x^2+2x^2-14x\)

\(\Leftrightarrow x^2\left(x-7\right)+2x\left(x-7\right)\)\(\Leftrightarrow x\left(x+2\right)\left(x-7\right)\)

 5)  \(x^2yz+5xyz-14yz\)\(\Leftrightarrow yz\left(x^2+5x-14\right)\)

\(\Leftrightarrow yz\left(x^2+7x-2x-14\right)\)

\(\Leftrightarrow yz\left[x\left(x+7\right)-2\left(x+7\right)\right]\) 

\(\Leftrightarrow yz\left(x+7\right)\left(x-2\right)\)

1 tháng 5 2017

Cảm ơn bạn Nguyễn Kim Thương :))

DD
30 tháng 11 2021

\(x^2\left(x+1\right)-\left(x+1\right)\left(3x+1\right)+7x-x^2\)

\(=x^3+x^2-3x^2-4x-1+7x-x^2\)

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

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

2 tháng 10 2016

a) \(4a^3b^3c^2x+12a^3b^4c^2-16a^4b^5cx\)

\(=4a^3b^3c\left(cx+3bc-4ab^2x\right)\)

b) \(\left(b-2c\right)\left(a-b\right)-\left(a+b\right)\left(2c-b\right)\)

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

c) \(3a\left(a+5\right)-2\left(5+a\right)=\left(a+5\right)\left(3a-2\right)\)

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

15 tháng 10 2020

Bài 1:

a) \(3x^2-9x=3x\left(x-3\right)\)

b) \(x^2-4x+4=\left(x-2\right)^2\)

c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)

Bài 2: 

a) \(101^2-1=\left(101-1\right)\left(101+1\right)=102.100=10200\)

b) \(67^2+66.67+33^2=67^2+2.33.67+33^2\)

\(=\left(67+33\right)^2=100^2=10000\)

Bài 3:

\(x\left(x-3\right)+2\left(x+3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

Vậy \(x=-2\)hoặc \(x=3\)

15 tháng 10 2020

B1:

a) \(3x^2-9x=3x.\left(x-3\right)\)

b) \(x^2-4x+4=\left(x-2\right)^2\)

c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3+y\right).\left(x+3-y\right)\)

B2:

a) \(101^2-1=\left(101+1\right).\left(101-1\right)=102.100=10200\)

b) \(67^2+66.67+33^2=67^2+2.33.67+33^2=\left(67+33\right)^2=100^2=10000\)

B3:

\(x\left(x-3\right)+2\left(x-3\right)=0\)

\(\left(x-3\right).\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

a) 2x + 2y - x2 - xy

= 2(x + y) + x(x + y)

= (x + y) (x + 2)

mk ko bít phân tích đúng ko đúng thì t i c  k nhé!! 245433463463564564574675687687856856846865855476457

12 tháng 7 2016

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

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

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

\(=\left(x+3\right)\left(8-x\right)\)

c)\(\left(3x+2\right)^2+\left(3x-2\right)^2-2\left(9x^2-4\right)\)

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

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

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

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

=4.4=16