Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
BÀi 1 : xem lại đề
bài 2
a) 27 - x^3
= ( 3 -x )( 9 + 3x + x^2)
b) 8x^3 + 0,001
= (2x + 0,1) ( 4x^2 - 0,2x + 0,01)
\(\frac{x^3}{64}-\frac{y^3}{125}=\left(\frac{x}{4}-\frac{y}{5}\right)\left(\frac{x^2}{16}-\frac{xy}{20}+\frac{y^2}{25}\right)\)
a+b=7=>(a+b)2=49
=>a2+2ab+b2=49
Do ab=3
=>2ab=6
=>b2+a2=43
Ta có:a3+b3=(a+b)(a2-ab+b2)
Thay a2+b2=43 ab=3 a+b=7
=> a3+b3=7.(43-3)=7.40=280
a)27-x3=(3-x)(9+3x+x2)
b)8x3+0,001=(2x+0,1)(4x2-0,2x+0,01)
c)x3/64-y3/125=(x/4-y/5)(x2/16+xy/20+y2/25)
Làm bài 1 thôi !! Mấy bài kia tương tự . Tìm nhân tử chung ra .
a) \(m^2-n^2=\left(m-n\right)\left(m+n\right)\)
b) \(\left(x^2+x-1\right)^2-\left(x^2+2x+3\right)^2=\left(x^2+x-1+x^2+2x+3\right)\left(x^2+x-1-x^2-2x-3\right)\)
\(=\left(2x^2+3x+2\right)\left(-x-4\right)\)
c) \(-16+\left(x-3\right)^2=\left(x-3+4\right)\left(x-3-4\right)=x\left(x-7\right)\)
d) \(64+16y+y^2=\left(y+8\right)\left(y+8\right)\)
Bài 1:
\(B=\dfrac{1}{9}x^2-2x+9\)
\(=\left(\dfrac{1}{3}x\right)^2-2\cdot\dfrac{1}{3}x\cdot3+3^2=\left(\dfrac{1}{2}x-3\right)^2\)
\(C=x^3-9x^2+27x-27=\left(x-3\right)^3\)
\(D=27x^3+27x^2+9x+1=\left(3x+1\right)^3\)
\(E=\left(x-2y\right)^3\)
a) \(27x^3+8^3\)
\(=\left(3x\right)^3+2^3\)
\(=\left(3x+2\right)\left[\left(3x\right)^2+6x+2^2\right]\)
\(=\left(3x+2\right)\left(9x^2-6x+4\right)\)
b) \(8x^3-y^3\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
c) \(x^2+4xy+4y^2\)
\(=\left(x+2y\right)^2\)
\(27x^3+8\)
\(=\left(3x\right)^3+2^3\)
\(=\left(3x+2\right)\left(9x^2-6x+4\right)\)
\(8x^3-y^3\)
\(=\left(2x\right)^3-y^3\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(x^2+4xy+4y^2\)
\(=x^2+2.x.2y+\left(2y\right)^2\)
\(=\left(x+2y\right)^2\)
_Minh ngụy_
\(x^3-9x^2+27x-27\)
\(=x^3-3x^2.3+3.x.3^2-3^3\)
\(=\left(x-3\right)^3\)
\(27x^6-y^3\)
\(=\left(3x^2\right)^3-y^3\)
\(=\left(3x^2-y\right)\left[\left(3x^2\right)^2+3x^2y+y^2\right]\)
\(=\left(3x^2-y\right)\left(9x^4+3x^2y+y^2\right)\)
a) (x+2) \(\left(x^2-2x+4\right)\)
b) (3 - 2y) \(\left(9+6y+4y^2\right)\)
d) (4x - y) \(\left(16x^2+4xy+y^2\right)\)
bạn giải chi tiết hộ mình với nha.
mk sắp phải nộp bài r. huhuhuhu
a) Ta có: \(x^2+2x+1\)
\(=x^2+2\cdot x\cdot1+1^2\)
\(=\left(x+1\right)^2\)
b) Ta có: \(1-2y+y^2\)
\(=y^2-2\cdot y\cdot1+1^2\)
\(=\left(y-1\right)^2\)
c) Ta có: \(x^3-3x^2+3x-1\)
\(=x^3-x^2-2x^2+2x+x-1\)
\(=x^2\left(x-1\right)-2x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x+1\right)\)
\(=\left(x-1\right)^3\)
d) Ta có: \(27+27x+9x^2+x^3\)
\(=x^3+3x^2+6x^2+18x+9x+27\)
\(=x^2\left(x+3\right)+6x\left(x+3\right)+9\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+6x+9\right)\)
\(=\left(x+3\right)^3\)
e) Ta có: \(8-125x^3\)
\(=2^3-\left(5x\right)^3\)
\(=\left(2-5x\right)\left(4+10x+25x^2\right)\)
f) Ta có: \(64x^3+\frac{1}{8}\)
\(=\left(4x\right)^3+\left(\frac{1}{2}\right)^3\)
\(=\left(4x+\frac{1}{2}\right)\left(16x^2-2x+\frac{1}{4}\right)\)
g) Ta có: \(1-x^2y^4\)
\(=1^2-\left(xy^2\right)^2\)
\(=\left(1-xy^2\right)\left(1+xy^2\right)\)
a) \(x^2+2x+1=x^2+2x.1+1^2=\left(x+1\right)^2\)
b) \(1-2y+y^2=1^2-2y.1+y^2=\left(1-y\right)^2\)
c) \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
d) \(27+27x+9x^2+x^3=3^3+3.3^2x+3.3x^2+x^3=\left(3+x\right)^3\)
e) \(8-125x^3=2^3-\left(5x\right)^3=\left(2-5x\right)\left[2^2+2.5x+\left(5x\right)^2\right]=\left(2-5x\right)\left(4+10x+25x^2\right)\)
f) \(64x^3+\frac{1}{8}=\left(4x\right)^3+\left(\frac{1}{2}\right)^3=\left(4x+\frac{1}{2}\right)\left[\left(4x\right)^2-4x.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]=\left(4x+\frac{1}{2}\right)\left(16x^2-2x+\frac{1}{4}\right)\)
Ko chắc ạ!
\(a,=\left(3+x\right)\left(9-3x+x^2\right)\\ b,=\left(4x+0,1\right)\left(16x^2-0,4x+0,01\right)\\ c,=\left(2-3x\right)\left(4+6x+9x^2\right)\\ d,=\left(\dfrac{x}{5}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{25}+\dfrac{xy}{15}+\dfrac{y^2}{9}\right)\)
a) \(27+x^3=3^3+x^3=\left(3+x\right)\left(9-3x+x^2\right)\)
b) \(64x^3+0,001=\left(4x\right)^3+\left(\dfrac{1}{10}\right)^3=\left(4x+\dfrac{1}{10}\right)\left(16x^2-\dfrac{4x}{10}+\dfrac{1}{100}\right)\)