Phân tích đa thức sau thành nhân tử
8+8\(\left(x-2\right)^3\)
phân tích các đa thức sau thành nhân tử8
7, 49 y mũ 2 - x mũ 2 + 6x - 9
8, 25x mũ 2 - 4y mũ 2 - 4y - 9
9, 4x mũ 2 -- y mũ 2 + 8y - 16
Trả lời:
7, 49y2 - x2 + 6x - 9
= 49y2 - ( x2 - 6x + 9 )
= ( 7y )2 - ( x - 3 )2
= ( 7y - x + 3 ) ( 7y - x - 3 )
8, sửa đề: 25x2 - 4y2 - 4y - 1
= 25x2 - ( 4y2 + 4y + 1 )
= ( 5x )2 - ( 2y + 1 )
= ( 5x - 2y - 1 ) ( 5x + 2y + 1 )
9, 4x2 - y2 + 8y - 16
= 4x2 - ( y2 - 8y + 16 )
= ( 2x )2 - ( y - 4 )2
= ( 2x - y + 4 ) ( 2x + y - 4 )
a, \(49y^2-x^2+6x-9=49y^2-\left(x-3\right)^2=\left(7y-x+3\right)\left(7y+x-3\right)\)
b, đề sai rồi bạn
c, \(4x^2-y^2+8y-16=4x^2-\left(y-4\right)^2=\left(2x-y+4\right)\left(2x+y-4\right)\)
Phân tích đa thức thành nhân tử:
\(x^3-8+2x\left(x-2\right)\)
\(x^3-8+2x\left(x-2\right)\\ =\left(x-2\right)\left(x^2+2x+4\right)+2x\left(x-2\right)\\ =\left(x-2\right)\left(x^2+2x+4+2x\right)=\left(x-2\right)\left(x^2+4x+4\right)\\ =\left(x-2\right)\left(x+2\right)^2\)
=\(\left(x-2\right)\left(x^2+2x+4\right)+2x\left(x-2\right)\)
=\(\left(x-2\right)\left(x^2+4x+4\right)\)
=\(\left(x-2\right)\left(x+2\right)^2\)
Phân tích đa thức sau thành nhân tử:
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+18\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
\(=\left(x^2+10x+20-4\right)\left(x^2+10x+20+4\right)-16\)
\(=\left(x^2+10x+20\right)^2-16+16=\left(x^2+10x+20\right)^2\)
Chúc bạn học tốt.
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(\Rightarrow\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+6\right)\left(x+8\right)\right]+16\)
\(\Rightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
\(\Rightarrow\left(x^2+10x+16\right)\left[\left(x^2+10x+16\right)+8\right]+16\)
\(\Rightarrow\left(x^2+10x+16\right)^2+8\left(x^2+10x+16\right)+4^2\)
\(\Rightarrow\left(x^2+10x+20\right)^2\)
PHÂN TÍCH ĐA THỨC THÀNH NHÂN TỬ:
\(\left(x^2+8x+8\right)\left(x^2+8x+15\right)+15\)
=(x^2+8x)^2+23(x^2+8x)+135
Cái này ko phân tích được nha bạn
\(\left(x^2+8x+8\right)\left(x^2+8x+15\right)+15\\ \Leftrightarrow\left(x^4+8x^3+15x^2+8x^3+64x^2+120x+8x^2+64x+120\right)+15\\ \Leftrightarrow x^4+16x^3+87x^2+184x+135\)
Gọi `A=(x^2+8x+8)(x^2+8x+15)+15`
Đặt `t=x^2+8x+11,5`
`=>A=(t-3,5)(t+3,5)+15=t^2-3,5^2+15=t^2-2,75=(t-sqrt(2,75))(t+sqrt(2,75))=(x^2+8x+11,5-(sqrt11)/2)(x^2+8x+11,5+(sqrt11)/2)=(x^2+8x+(23-\sqrt11)/2)(x^2+8x+(23+\sqrt11)/2)`
Phân tích đa thức thành nhân tử :\(\left(x^2-2x+3\right)\left(x^2-2x+5\right)-8\)
Đặt \(x^2-2x+4=a\)
Khi đó \(\left(x^2-2x+3\right)\left(x^2-2x+5\right)-8=\left(a-1\right)\left(a+1\right)-8\)
\(=a^2-1-8\)
\(=a^2-9\)
\(=\left(a-3\right)\left(a+3\right)\)
\(=\left(x^2-2x+4-3\right)\left(x^2-2x+4+3\right)\)
\(=\left(x^2-2x+1\right)\left(x^2-2x+7\right)\)
\(=\left(x-1\right)^2\left(x^2-2x+7\right)\)
1.Phân tích đa thức thành nhân tử:
d. \(\text{2(3-x)}^{\text{2}}-\dfrac{1}{8}\left(x-1\right)^2\)
\(=2\left[\left(x-3\right)^2-\dfrac{1}{16}\left(x-1\right)^2\right]\\ =2\left(x-3-\dfrac{1}{4}x+\dfrac{1}{4}\right)\left(x-3+\dfrac{1}{4}x-\dfrac{1}{4}\right)\\ =2\left(\dfrac{3}{4}x-\dfrac{11}{4}\right)\left(\dfrac{5}{4}x-\dfrac{13}{4}\right)\)
Phân tích đa thức thành nhân tử
\(x^2+27+\left(x+3\right)\left(x-9\right)\)
x3+27+(x+3)(x+9)
= (x+3)(x2-3x+9)+(x+3)(x+9)
= (x+3)(x2-3x+9+x+9)
=(x+3)(x2-2x+18)
\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\\ =\left(x+3\right)\left(x^2-3x+9+x-9\right)\\ =\left(x+3\right)\left(x^2-2x\right)=x\left(x-2\right)\left(x+3\right)\)
x3+27+(x+3)(x+9)
= (x+3)(x2-3x+9)+(x+3)(x+9)
= (x+3)(x2-3x+9+x+9)
=(x+3)(x2-2x+18)
Phân tích đa thức thành nhân tử:
\(\left(x+3\right)^4+\left(x+5\right)^4-2\)
`(x+3)^4+(x+5)^4-2`
`={[(x+3)^2]^2-1^2}+{[(x+5)^2]^2 -1^2}`
`=[(x+3)^2-1^2][(x+3)^2+1]+[(x+5)^2-1^2][(x+5)^2+1]`
`=(x+3-1)(x+3+1)[(x+3)^2+1]+(x+5-1)(x+5+1)[(x+5)^2+1]`
`=(x+2)(x+4)[(x+3)^2+1]+(x+4)(x+6)[(x+5)^2+1]`
`=(x+4){(x+2)[(x+3)^2+1]+(x+6)[(x+5)^2+1]}`
`=(x+4)(2x^3+24x^2+108x+176)`
\(\left(x+3\right)^4+\left(x+5\right)^4-2\)
\(=\left[\left(x+3\right)^4-1\right]+\left[\left(x+5\right)^4-1\right]\)
\(=\left[\left(x^2+6x+9-1\right)\left(x^2+6x+9+1\right)\right]+\left[\left(x^2+10x+25-1\right)\left(x^2+10x+25+1\right)\right]\)
\(=\left(x^2+6x+8\right)\left(x^2+6x+10\right)+\left(x^2+10x+24\right)\left(x^2+10x+26\right)\)
\(=\left(x+2\right)\left(x+4\right)\left(x^2+6x+10\right)+\left(x+4\right)\left(x+6\right)\left(x^2+10x+26\right)\)
\(=\left(x+4\right)\left[\left(x+2\right)\left(x^2+6x+10\right)+\left(x+6\right)\left(x^2+10x+26\right)\right]\)
\(=\left(x+4\right)\left(x^3+6x^2+10x+2x^2+12x+20+x^3+10x^2+26x+6x^2+60x+156\right)\)
\(=\left(x+4\right)\left(2x^3+24x^2+108x+176\right)\)
\(=2\left(x+4\right)\left(x^3+12x^2+54x+88\right)\)
Phân tích các đa thức sau thành nhân tử:
a) \({x^3} + 4x\)
b) \(6ab - 9a{b^2}\)
c) \(2a\left( {x - 1} \right) + 3b\left( {1 - x} \right)\)
d) \({\left( {x - y} \right)^2} - x\left( {y - x} \right)\)
`a, x^3 + 4x = x(x^2+4)`
`b, 6ab - 9ab^2 = 3ab(2-b)`
`c, 2a(x-1) + 3b(1-x)`
`= (2a-3b)(x-1)`
`d, (x-y)^2 - x(y-x)`
`= (x-y+x)(x-y)`
`= (2x-y)(x-y)`