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) \(8x^3+\frac{1}{27}\)
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TM
14 tháng 7 2017
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)\(8x^3+27y^3+36x^2y+54xy^2=\left(2x\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2+\left(3y\right)^3\)
\(=\left(2x+3y\right)^2\)
22 tháng 5 2022
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: \(D=\left(3x-2\right)^2+2\left(x+1\right)\left(3x-2\right)+\left(x+1\right)^2\)
\(=\left(3x-2+x+1\right)^2=\left(4x-1\right)^2=16x^2-8x+1\)
18 tháng 7 2021
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, \(8x^3+27y^3=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
18 tháng 7 2021
Câu 2 đề ko sai nha bạn.
2) x2 - (\(\sqrt{y^3}\))2 ( y>0)
= ( x -\(\sqrt{y^3}\)) ( x +\(\sqrt{y^3}\))
TD
1
19 tháng 10 2021
\(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{4}\right)^2\)
\(8x^3+27=\left(2x+3\right)\left(4x^2-6x+9\right)\)
\(-x^3+3x^2-3x+1=\left(-x+1\right)^3\)
H
4
13 tháng 7 2021
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`
N
31 tháng 7 2017
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
4 tháng 10 2021
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) \(=\left(x+3\right)^3\)
30 tháng 7 2020
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: \(x^6+1\)
\(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
TG
30 tháng 7 2020
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!
1) \(4x^2-y^2=\left(2x-y\right)\left(2x+y\right)\)
2) \(8x^3-27=\left(2x-3\right)\left(4x^2+6x+9\right)\)
3) \(x^3+27y^3=\left(x+3y\right)\left(x^2-3xy+9y^2\right)\)
4) \(x^2-25y^2=\left(x-5y\right)\left(x+5y\right)\)
5) \(8x^3+\frac{1}{27}=\left(2x+\frac{1}{3}\right)\left(4x^2-\frac{2}{3}x+\frac{1}{9}\right)\)