Phương pháp 2: dùng hằng đẳng thức
1) \(4x^2-y^2\)
2) \(8x^3-27\)
3) \(x^3+27y^3\)
4) \(x^2-25y^2\)
5) \(8x^3+\frac{1}{27}\)
dùng hằng đẳng thức để phân tích thành nhân tử:
1.1/ -x3+9x2-27x+27
1.2/ x4-2x3-x2+2x+1
1.3/ 8x3+27y3+36x2y+54xy2
a) \(-x^3+9x^2-27x+27=-\left(x^3-3.3.x^2+3.3^2.x-3^3\right)=-\left(x-3\right)^3\)
b)\(x^4-2x^3-x^2+2x+1=x^4+\left(-x\right)^2+\left(-1\right)^2+2x^2\left(-x\right)+2.\left(-x\right).\left(-1\right)+2x^2.\left(-1\right)\)
\(=\left(x^2-x-1\right)^2\)
c)\(8x^3+27y^3+36x^2y+54xy^2=\left(2x\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2+\left(3y\right)^3\)
\(=\left(2x+3y\right)^2\)
Bài 1 : Rút gọn đẳng thức :
C={x-3}{x+3}-{x+5}{x-1}
D={3x-2}^2+{x+1}^2+2{1+x}{3x-2}
Bài 2: Tính mỗi giá trị của mỗi đẳng thức :
D=-27-9/4x^2+1/8x^3+27/2x tại x =2
E=36x^4y+27y^3+8x^6+54x^2y^2 tại x=-2 ,y=2
Bài 3: Tìm x, bt:
c, {x^2+x+1}{x-1}-x{x+2}{x-2}=5
d, {x-1}^3-{x+3}{x^2+9}+3{x^2-4}=2
Bài 1:
a: \(C=\left(x-3\right)\left(x+3\right)-\left(x+5\right)\left(x-1\right)\)
\(=x^2-9-\left(x^2+4x-5\right)\)
\(=x^2-9-x^2-4x+5=-4x-4\)
b: \(D=\left(3x-2\right)^2+2\left(x+1\right)\left(3x-2\right)+\left(x+1\right)^2\)
\(=\left(3x-2+x+1\right)^2=\left(4x-1\right)^2=16x^2-8x+1\)
Bài 28 Khai triển hằng đẳng thức
1)x^3+3^3
2)x^2-y^3
3x^3+8
4)x^3-64
5) 1000 – y³
6) 125 – 8x³
7) x³ + 27y³
8)8) 8x³ + 27y³
1, \(x^3+3^3=\left(x+3\right)\left(x^2-3x+9\right)\)
2, đề sai
3, \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)
4, \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
5, \(1000-y^3=\left(10-y\right)=\left(100+10y+y^2\right)\)
tương tự ...
8, \(8x^3+27y^3=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
Hằng đẳng thức :x^2+x+1/4
27+8x^3
-x^3+3x^2-3x+1
8+36x+54x^2+27x^3
\(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{4}\right)^2\)
\(8x^3+27=\left(2x+3\right)\left(4x^2-6x+9\right)\)
\(-x^3+3x^2-3x+1=\left(-x+1\right)^3\)
phân tích ....
a, x^3 -1
b,8x^3-y^3
c,x^2-8x+16
d,25y^3-1
e,27-8y^3
2x^2-8x+8
Sửa lại câu d) là `25y^2`
`a)x^3-1`
`=(x-1)(x^2+x+1)`
`b)8x^3-y^3`
`=(2x)^3-y^3`
`=(2x-y)(4x^2+2xy+y^2)`
`c)x^2-8x+16`
`=x^2-2.x.4+4^2`
`=(x-4)^2`
`d)25y^2-1`
`=(5y)^2-1`
`=(5y-1)(5y+1(`
`e)27-8y^3`
`=3^3-(2y)^3`
`=(3-2y)(9+6y+4y^2)`
`f)2x^2-8x+8`
`=2(x^2-4x+4)`
`=2(x-2)^2`
a) x3 - 1 = x3 - 13
= (x - 1)(x2 - x + 1)
b) 8x3 - y3 = (2x)3 - y3
= (2x - y)(4x2 + 2xy + y2)
c) x2 - 8x + 16 = x2 - 2.4x + 42
= (x - 4)2
d) đề có bị sai không , nên mình sưa lại đề nhé :
25y2 - 1 = (5y)2 - 12
= (5y - 1)(5y + 1)
e) 27 - 8y3 = 33 - (2y)3
= (3 - 2y)(9 + 6y + 4y2)
f) 2x2 - 8x + 8 = 2(x2 - 4x + 4)
= 2(x - 2)2
Chúc bạn học tốt
Bài 1:Viết các hằng đẳng thức sau dưới dạng tích
A) x3-(3y)3
B) (xy)2-(x2y)3
C) 8x3-27y3
D) \(\frac{1}{64}\)x6y3-125
E) \(\frac{1}{27}\)-64x6
Bài 2: Viết các hằng đẳng thức sau dưới dạng tích
A) (x-1)(x2+x+1)
B) (2-x)(4+2x+x2)
C) (2x-1)(4x2+2x+1)
D) (x2-3)(x4+3x2+9)
E) (x2-y)(x4+x2y+y2)
MỌI NGƯỜI TRẢ LỜI GIÚP MÌNH VỚI MÌNH CẦN GẤP LẮP
Đề: Phân tích đa thức thành nhân tử bằng phương pháp dùng hằng đẳng thức:
1, x3+9x2+27x+27=
2, 8x6-27y3=
3, x6-y6=
4, 27a3-54ab+36ab2-8b2=
1. \(x^3+9x^2+27x+27=\left(x+3\right)^3\)
2.\(8x^6-27y^3=\left(2x\right)^3-\left(3y\right)^3=\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(8x^6+6xy+27y^3\right)\)
3.\(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right)\)4.câu cuối là \(8b^3\)bạn nhé !!
tik mik nhé
\(\dfrac{1}{27}+a^3\\ 8x^3+27y^3\\ \dfrac{1}{8}x^3+8y^3\\ x^6+1\\ x^9+1\\ x^3-64\\ x^3-125\\ 8x^6-27y^3\\ \dfrac{1}{64}x^6-125y^3\\ \dfrac{1}{8}x^3-8\\ x^3+6x^2+12x+8\\ x^3+9x^2+27x+27\) Giúp mình với mình cần gấp ;-;
1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)
2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)
4) \(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)
6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)
7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)
8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)
9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)
10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)
11) \(=\left(x+2\right)^3\)
12) \(=\left(x+3\right)^3\)
1) x^10-4x^8+4x^6
2) m ³+27
3) x ³+8
4) 1/27+a ³
5) 8x ³+27y ³
6) 1/8x ³+8y ³
7) 8x^6-27y ³
8) 1/8x ³-8
9) 1/64x^6-125y ³
10) (a+b) ³-c ³
11) x ³-(y-1) ³
12) x^6+1
1: Ta có: \(x^{10}-4x^8+4x^6\)
\(=x^6\left(x^4-4x^2+4\right)\)
\(=x^6\left(x-2\right)^2\left(x+2\right)^2\)
2: Ta có: \(m^3+27\)
\(=\left(m+3\right)\left(m^2-3m+9\right)\)
3: Ta có: \(x^3+8\)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
4: Ta có: \(\frac{1}{27}+a^3\)
\(=\left(\frac{1}{3}+a\right)\left(\frac{1}{9}-\frac{a}{3}+a^2\right)\)
5: Ta có: \(8x^3+27y^3\)
\(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
6: Ta có: \(\frac{1}{8}x^3+8y^3\)
\(=\left(\frac{1}{2}x+2y\right)\left(\frac{1}{4}x^2-xy+4y^2\right)\)
7: Ta có: \(8x^6-27y^3\)
\(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)
8: Ta có: \(\frac{1}{8}x^3-8\)
\(=\left(\frac{1}{2}x-2\right)\left(\frac{1}{4}x^2+x+4\right)\)
9: Ta có: \(\frac{1}{64}x^6-125y^3\)
\(=\left(\frac{1}{4}x^2-5y\right)\left(\frac{1}{16}x^4+\frac{5}{4}x^2y+25y^2\right)\)
10: Ta có: \(\left(a+b\right)^3-c^3\)
\(=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)\cdot c+c^2\right]\)
\(=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\)
11: Ta có: \(x^3-\left(y-1\right)^3\)
\(=\left[x-\left(y-1\right)\right]\cdot\left[x^2+x\left(y-1\right)+\left(y-1\right)^2\right]\)
\(=\left(x-y+1\right)\left(x^2+xy-x+y^2-2y+1\right)\)
12: Ta có: \(x^6+1\)
\(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
1) \(x^{10}-4x^8+4x^6\)
\(=x^6\left(x^4-4x^2+4\right)\)
2) \(m^3+27=m^3+3^3=\left(m+3\right)\left(m^2-3m+3^2\right)\)
3) \(x^3+8=x^3+2^3=\left(x+2\right)\left(x^2-2x+2^2\right)\)
4) \(\frac{1}{27}+a^3=\left(\frac{1}{3}\right)^3+a^3=\left(\frac{1}{3}+a\right)\left[\left(\frac{1}{3}\right)^2-\frac{1}{3}a+a^2\right]\)
5) \(8x^3+27y^3=\left(2x\right)^3+\left(3y\right)^3=\left(2x+3y\right)\left[\left(2x\right)^2-2x.3y+\left(3y\right)^2\right]=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
6) \(\frac{1}{8}x^3+8y^3=\left(\frac{1}{2}x\right)^3+\left(2y\right)^3=\left(\frac{1}{2}x+2y\right)\left[\left(\frac{1}{2}x\right)^2-\frac{1}{2}x.2y+\left(2y\right)^2\right]=\left(\frac{1}{2}x+2y\right)\left(\frac{1}{4}x^2-xy+4y^2\right)\)
8) \(\frac{1}{8}x^3-8=\left(\frac{1}{2}x\right)^3-2^3=\left(\frac{1}{2}x-2\right)\left[\left(\frac{1}{2}x\right)^2+\frac{1}{2}x.2+2^2\right]=\left(\frac{1}{2}x-2\right)\left(\frac{1}{4}x^2+x+4\right)\)
10) \(\left(a+b\right)^3-c^3=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]=\left(a+b-c\right)\left[\left(a^2+2ab+b^2\right)+ac+bc+c^2\right]=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\)11) \(x^3-\left(y-1\right)^3=\left(x-y+1\right)\left[x^2+x\left(y-1\right)+\left(y-1\right)^2\right]=\left(x-y+1\right)\left[x^2+xy-x+\left(y^2-2y+1\right)\right]=\left(x-y+1\right)\left(x^2+xy-x+y^2-2y+1\right)\)
P/s: Đăng ít thôi chớ bạn!