phân tích đa thức thành nhân tử
a, x^4 - x^5
b, -8x^2y^2 - 12xy^3 - 4xy^2
c,(x-y)^3 - x^3 + y^3
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![](https://rs.olm.vn/images/avt/0.png?1311)
a) 6x - x2 - 5
= -x2 + 6x - 9 + 4
= -( x2 - 6x + 9 ) + 4
= -( x - 3 )2 + 4 ≤ 4 ∀ x ( chưa kl luôn âm được :)) )
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{7}{x+4}+\frac{8}{x-4}+\frac{14x}{x^2-16}=\frac{7}{x+4}+\frac{8}{x-4}+\frac{14x}{\left(x-4\right)\left(x+4\right)}\)
\(=\frac{7\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}+\frac{8\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\frac{14x}{\left(x-4\right)\left(x+4\right)}\)
\(=\frac{7x-28+8x+32+14x}{\left(x-4\right)\left(x+4\right)}=\frac{29x+4}{\left(x-4\right)\left(x+4\right)}\)
\(B=\frac{x^2-2x+1}{x-1}+\frac{x^2-9}{x+3}=\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x-3\right)\left(x+3\right)}{x+3}\)
\(=x-1+x-3=2x-4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) A=x3+3x2+3x
A=x3+3x2.1+3x.12+13
A=(x+1)3
b)A=x3-3x2+3x-1
A=x3-3x2.1+3x.12-13
A=(x-1)3
c)A=x3+6x2+12x
A=x3+3.2x2+3.22x+13
A=(x+1)3
![](https://rs.olm.vn/images/avt/0.png?1311)
1, \(A=\frac{9}{x+1}-\frac{8}{1-x}-\frac{16}{x^2-1}\)
\(=\frac{9}{x+1}-\frac{8}{1-x}-\frac{16}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{9\left(1-x\right)\left(x-1\right)}{\left(x+1\right)\left(1-x\right)\left(x-1\right)}-\frac{8\left(x+1\right)\left(x-1\right)}{\left(1-x\right)\left(x+1\right)\left(x-1\right)}-\frac{16\left(1-x\right)}{\left(1-x\right)\left(x+1\right)\left(x-1\right)}\)
\(=\frac{9\left(1-x\right)\left(x-1\right)-8\left(x+1\right)\left(x-1\right)-16\left(1-x\right)}{\left(x+1\right)\left(x-1\right)\left(1-x\right)}\)
\(=\frac{18x-9-9x^2-8x^2+8-16+16x}{\left(x+1\right)\left(x-1\right)\left(1-x\right)}=\frac{-17x^2+34x-17}{\left(x+1\right)\left(x-1\right)\left(1-x\right)}\)
\(=\frac{-17\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)\left(1-x\right)}=\frac{-17\left(x-1\right)}{\left(x+1\right)\left(1-x\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có x = 100
=> x + 1 = 101
Khi đó A = x15 - 101x14 + 101x13 - 101x12 + ... + 101x3 - 101x2 + 101x + 2020
= x15 - (x + 1)x14 + (x + 1)x13 - (x + 1)x12 + ... + (x + 1)x3 - (x + 1)x2 + (x + 1)x + 2020
= x15 - x15 - x14 + x14 + x13 - x13 - x12 + ... + x4 + x3 - x3 - x2 + x2 + x + 2020
= x + 2020
= 101 + 2020 (Vì x = 100)
= 2121
Vậy A = 2121 khi x = 100
A = x15 - 101x14 + 101x13 - ... - 101x2 + 101x + 2020 tại x = 100
x = 100 => 101 = x + 1
Thế vào A ta được
A = x15 - ( x + 1 )x14 + ( x + 1 )x13 - ... - ( x + 1 )x2 + ( x + 1 )x + 2020
= x15 - ( x15 + x14 ) + ( x14 + x13 ) - ... - ( x3 + x2 ) + ( x2 + x ) + 2020
= x15 - x15 - x14 + x14 + x13 - ... - x3 - x2 + x2 + x + 2020
= x + 2020
= 100 + 2020 = 2120
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa: \(A=\frac{x^2+6x+9}{x+3}+\frac{x^2-16}{x-4}=\frac{\left(x+3\right)^2}{x+3}+\frac{\left(x-4\right)\left(x+4\right)}{x-4}=x+3+x+4=2x+7\) (đk: \(x\ne-3;x\ne4\))
\(B=\frac{5}{x+2}+\frac{6}{x-2}-\frac{10x}{x^2-4}\)(đk x\(\ne\)\(\pm\)2)
\(B=\frac{5\left(x-2\right)+6\left(x+2\right)-10x}{\left(x-2\right)\left(x+2\right)}\)
\(B=\frac{5x-10+6x+12-10x}{\left(x-2\right)\left(x+2\right)}=\frac{x+2}{\left(x-2\right)\left(x+2\right)}=\frac{1}{x-2}\)
Gọi 2 số tự nhiên cần tìm là a,b (a,b \(\in\)N)
Theo bài ra, ta có: \(\hept{\begin{cases}a+b=13\\ab=36\end{cases}}\)
<=> \(\hept{\begin{cases}a=13-b\\ab=36\end{cases}}\) <=> \(\hept{\begin{cases}a=13-b\\\left(13-b\right)b=36\end{cases}}\) <=> \(\hept{\begin{cases}a=13-b\\b^2-13b+36=0\end{cases}}\)
<=> \(\hept{\begin{cases}a=13-b\\b^2-4x-9x+36=0\end{cases}}\) <=> \(\hept{\begin{cases}a=13-b\\\left(b-4\right)\left(b-9\right)=0\end{cases}}\)
<=> a = 13 - b và b = 4 hoặc b = 9
Với b = 4 => a = 13 - 4 = 9
Với b = 9 => a = 13 - 9 = 4
![](https://rs.olm.vn/images/avt/0.png?1311)
a) vật thể nhân tạo
b) vật thể tự nhiên
c) vật thể nhân tạo
d) vật thể tự nhiên
a) x4 - x5 = x4( x - 1 )
b) -8x2y2 - 12xy3 - 4xy2
= -4xy( 2xy + 3y2 + y )
c) ( x - y )3 - x3 + y3
= x3 - 3x2y + 3xy2 - y3 - x3 + y3
= 3xy2 - 3x2y
= 3xy( y - x )