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Bài 3:
b: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)+10=0\)
\(\Leftrightarrow6x^2+12-6x^2+12x-6=0\)
hay \(x=-\dfrac{1}{2}\)
Bài 2:
a: \(x^3+3x^2+3x+1=\left(x+1\right)^3\)
b: \(m^3+9m^2n+27mn^2+27n^3=\left(m+3n\right)^3\)
`a,-x^3/8 + 3/(4x^2) - 3/(2x) +1`
`=-(x^3/8 - 3/(4x^2) + 3/(2x) - 1)`
`=-(x/2 - 1)^3`
`b,x^6 - 3/(2x^{4} y) + 3/(4x^{2}y^{2}) - 1/(8y^{3})`
`=(x^3 - 1/(2y))^{3}`
Bài làm:
Ta có: \(\frac{x^3}{8}+\frac{3}{4}x^2y^2+\frac{3}{2}xy^4+y^6\)
\(=\left(\frac{x}{2}\right)^3+3.\left(\frac{x}{2}\right)^2.y^2+3.\frac{x}{2}.\left(y^2\right)^2+\left(y^2\right)^3\)
\(=\left(\frac{x}{2}+y^2\right)^3\)
này mình có vài câu không làm được, xin lỗi bạn nha
\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)
2:
-8x^6-12x^4y-6x^2y^2-y^3
=-(8x^6+12x^4y+6x^2y^2+y^3)
=-(2x^2+y)^3
3:
=(1/3)^2-(2x-y)^2
=(1/3-2x+y)(1/3+2x-y)
\(1,\\ a,=x^2+2xy+y^2\\ b,=x^2-4xy+4y^2\\ c,=x^2y^4-1\\ d,=\left[\left(x-y\right)\left(x+y\right)\right]^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\\ 2,\\ a,=\left(x+2\right)^2\\ b,=\left(3x-2\right)^2\\ c,=\left(\dfrac{x}{2}+1\right)^2\\ d,=\left(x+y-2\right)^2\)
a) \(x^4+4x^2+4=\left(x^2+2\right)^2\)
b) \(\left(2y-x\right)^2+2\left(2y-x\right)+1=\left(2y-x+1\right)^2\)
c) \(\left(2a-4b\right)^2+4a-8b+1=\left(2a-4b\right)^2+2\cdot\left(2a-4b\right)\cdot1+1^2=\left(2a-4b+1\right)^2\)
a) x 2 + y 2 3 . b) ( m + 3 n ) 3 .
c) ( 2 u + 4 v ) 3 . d) ( z – t + 5 ) 3 .