Bài 1 yêu cầu gì em?

Bài 2:

\(a,x\left(x-1\right)+5\left(x-1\right)=\left(x+5\right)\left(x-1\right)\\ b,3x\left(x+1\right)+3\left(x+1\right)=\left(3x+3\right)\left(x+1\right)=3\left(x+1\right)\left(x+1\right)=3\left(x+1\right)^2\\ c,x\left(x-3\right)+xy\left(x-3\right)=\left(x+xy\right)\left(x-3\right)=x\left(y+1\right)\left(x-3\right)\\ d,2x\left(x-2\right)-6\left(x-2\right)=\left(2x-6\right)\left(x-2\right)=2\left(x-3\right)\left(x-2\right)\)

Bài 1:

a) \(3xy+6y\)

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

b) \(3x^2+9x\)

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

c) \(6x-9y^2\)

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

d) \(10xy^2-6x^2y\)

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

Bài 2:

a) \(x\left(x-1\right)+5\left(x-1\right)\)

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

b) \(3x\left(x+1\right)+3\left(x+1\right)\)

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

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

\(=3\left(x+1\right)^2\) 

c) \(x\left(x-3\right)+xy\left(x-3\right)\)

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

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

d) \(2x\left(x-2\right)-6\left(x-2\right)\)

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

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

Bài `1`

`a,3xy +6y`

`= 3y(x+2)`

`b,3x^2+9x`

`= 3x(x+3)`

`c,6x-9y^2`

`= 3(2x- 3y^2)`

`d,10xy^2-6x^2y`

`= 2xy(5y-3x)`

Bài 1:

\(a,=3x\left(3xy+5y-1\right)\\ b,=\left(z-2\right)\left(3z-5\right)\\ c,=\left(x+2y\right)^2-4z^2=\left(x+2y+2z\right)\left(x+2y-2z\right)\\ d,=x^2-3x+5x-15=\left(x-3\right)\left(x+5\right)\)

Bài 2:

\(a,\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x+2-4x^2-12x=9\\ \Leftrightarrow4x^2+10x+7=0\\ \Leftrightarrow4\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{3}{4}=0\\ \Leftrightarrow4\left(x+\dfrac{5}{6}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\ \Leftrightarrow x\in\varnothing\\ c,\Leftrightarrow x^2-12x+36=0\\ \Leftrightarrow\left(x-6\right)^2=0\\ \Leftrightarrow x=6\)

tìm có mà link https://h7.net/hoi-dap/toan-8/phan-h-da-thuc-x-1-x-3-x-5-x-7-15-thanh-nhan-tu-faq257547.html

tí mình gửi qua cho 

học tốt

\(B=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(=\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)(1)

Đặt \(x^2+8x+11=t\)thay vào (1) ta được : 

\(\left(t-4\right)\left(t+4\right)+15\)

\(=t^2-16+15\)

\(=t^2-1\)

\(=\left(t-1\right)\left(t+1\right)\)Thay \(t=x^2+8x+11\)vào bt ta được:

\(\left(x^2+8x+11-1\right)\left(x^2+8x+11+1\right)\)

\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)

\(=\left(x^2+8x+10\right)\left(x^2+2x+6x+12\right)\)

\(=\left(x^2+8x+10\right)\left[x\left(x+2\right)+6\left(x+2\right)\right]\)

\(=\left(x^2+8x+10\right)\left(x+2\right)\left(x+6\right)\)

Bài làm

B = ( x + 1 )( x + 3 )( x + 5 )( x + 7 ) + 15 

B = [ ( x + 1 ) ( x + 7 ) ] [ ( x + 3 ) ( x + 5 ) ] + 15

B = [ x² + 7x + x + 7 ] [ x² + 5x + 3x + 15 ] + 15

B = [ x² + 8x + 7 ] [ x² + 8x + 15 ] + 15

Đặt [ x² + 8x + 7 ] [ x² + 8x + 15 ] + 15 = k

=> B = k . ( k + 8 ) + 15

=> B = k² + 8k + 15

=> B = k² + 3k + 5k + 15

=> B = ( k² + 5k ) + ( 3k + 15 )

=> B = [ k( k + 5 ) ] + [ 3( k + 5 ) ]

=> B = ( k + 5 ) ( k + 3 )

Hay B = ( x² + 8x + 7 + 3 ) ( x² + 8x + 7 + 5 )

=> B = ( x² + 8x + 10 ) ( x² + 8x + 12 )

=> B = ( x² + 8x + 10 ) ( x² + 2x + 6x + 12 )

=> B = ( x² + 8x + 10 ) [ ( x² + 6x ) + ( 2x + 12 )]

=> B = ( x² + 8x + 10 ) [ x( x + 6 ) + 2( x + 6 ) ]

=> B = ( x² + 8x + 10 ) ( x + 2 ) ( x + 6 )

# Học tốt #

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Bài làm:

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

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

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

Đặt \(x^2+5x+5=t\)\(\Rightarrow\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2\)

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

b) Tương tự như a phân tích và đặt ra được: \(t^2-1-24=t^2-25=\left(t-5\right)\left(t+5\right)\)

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

c) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)

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

Đặt \(x^2+8x+11=t\)\(\Rightarrow\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1\)

\(=\left(t-1\right)\left(t+1\right)=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)

\(=\left(x^2+8x+10\right)\left(x+2\right)\left(x+6\right)\)

d) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x+11=t\)\(\Rightarrow\left(t-1\right)\left(t+1\right)-24=t^2-1-24=t^2-25\)

\(=\left(t-5\right)\left(t+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

Làm mẫu cho 1 vd:

a, (x+1)(x+2)(x+3)(x+4)+1

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

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

Đặt \(y=x^2+5x+5\)

Khi đó ::

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

\(=y^2-1+1=y^2\)

Thay vào ta được: \(\left(x^2+5x+5\right)^2\)

a) (x+1)(x+2)(x+3)(x+4)+1=[(x+1)(x+4)].[(x+2)(x+3)]+1=(x²+5x+4)(x²+5x+6)+1

đặt t=x²+5x+5 ta có đa thức (t-1)(t+1)+1=t²-1+1=t². mà t=x²+5x+5

=> (x+1)(x+2)(x+3)(x+4)+1=(x²+5x+5)²

b) (x+1)(x+2)(x+3)(x+4)-24. theo kết quả câu (a) ta được (x+1)(x+2)(x+3)(x+4)=(x²+5x+4)(x²+5x+6)

đặt t=x²+5x+5 ta có đa thức (t-1)(t+1)-24=t²-1-24=t²-25=(t-5)(t+5)

mà t=x²+5x+5 => (t-5)(t+5)=(x²+5x)(x²+5x+10)

c) (x+1)(x+3)(x+5)(x+7)+15=[(x+1)(x+7)].[(x+3)(x+5)]+15=(x²+8x+7)(x²+8x+15)+15

đặt x²+8x+11=t ta có đa thức (t-4)(t+4)+15=t²-16+15=t²-1=(t-1)(t+1)

mà t=x²+8x+11 => (t-1)(t+1)=(x²+8x-10)(x²+8x+12)

d) (x+2)(x+3)(x+4)(x+5)-24=[(x+2)(x+5)][(x+3)(x+4)]-24=(x²+7x+12)(x²+7x+10)-24

đặt t=x²+7x+11 ta có đa thức (t-1)(t+1)-24=t²-1-24=t²-25=(t+5)(t-5)

mà t=x²+7x+11 => (t-5)(t+5)=(x²+7x+6)(x²+7x+16)

Cưu là mình vs (x^2+x)^2-2(x^2+x)-15

1)(x^2+3x+1)(x^2+3x+2)-6

Đặt t = x²  + 3x + 1

Khi đó PT có dạng:

t.(t + 1) - 6

= t² + t - 6

= t² - 2t - 3t - 6

= t.(t - 2) + 3.(t - 2)

= (t + 3).(t - 2)

= (x² + 3x + 1 + 3).(x² + 3x + 1 - 2)

= (x² + 3x + 4).(x² + 3x - 1)

\(1\hept{\begin{cases}\left(x^2+3x+2-1\right)\left(x^2+2x+2\right)-6\\\left(t-1\right)\left(t\right)-6\\t^2-t-6\end{cases}}.\) " đặt x^2+3x+2 = t

\(\hept{\begin{cases}t^2-\frac{2t.1}{2}+\frac{1}{4}-\left(\frac{24+1}{4}\right)\\\left(t-\frac{1}{2}\right)^2-\frac{25}{4}\\\left(t-\frac{1}{2}\right)^2-\frac{25}{4}\end{cases}}\)

\(\hept{\begin{cases}\left(t-\frac{1}{2}-\frac{5}{2}\right)\left(t-\frac{1}{2}+\frac{5}{2}\right)\\\left(t-\frac{7}{2}\right)\left(t+\frac{4}{2}\right)\\\left(t-\frac{7}{2}\right)\left(t+\frac{4}{2}\right)\end{cases}}\)

2)  \(\hept{\begin{cases}\left\{\left(x+1\right)\left(x+7\right)\right\}\left\{\left(x+5\right)\left(x+3\right)\right\}+15\\\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\\t\left(t+8\right)+15\end{cases}}\)  

\(\hept{\begin{cases}t^2+8t+15\\\left(t^2+8t+16\right)-1\\\left(t+4\right)^2-1\end{cases}}\Leftrightarrow\left(t+5\right)\left(t+4\right)\)

\(\hept{\begin{cases}a^3\left(b-c\right)+b^3\left(c-a+b-b\right)+c^3\left(a-b\right)\\a^3\left(b-c\right)-b^3\left(-c+a-b+b\right)+c^3\left(a-b\right)\\a^3\left(b-c\right)-b^3\left(a-b\right)-b^3\left(b-c\right)+c^3\left(a-b\right)\end{cases}\Leftrightarrow\hept{\begin{cases}\left(b-c\right)\left(a^3-b^3\right)-\left(a-b\right)\left(b^3-c^3\right)\\\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)-\left(a-b\right)\left(b-c\right)\left(b^2+ab+c^2\right)\\\left(a-b\right)\left(b-c\right)\left(a^2+2ab+2b^2+c^2\right)\end{cases}}}\)