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\(\left(x^2+2x-1\right)^2\)
\(=\left(x^2+2x\right)^2-2\left(x^2+2x\right)+1\)
\(=x^4+4x^3-2x^2+4x^2+4x+1\)
\(=x^4+4x^3+2x^2+4x+1\)
a) \(\left(x^2-2x-1\right)^2\)
\(=\left[x^2+\left(-2x\right)+\left(-1\right)\right]\left[x^2+\left(-2\right)+\left(-1\right)\right]\)
\(=\left(x^2\right)\left(x^2\right)+\left(x^2\right)\left(-2x\right)+\left(x^2\right)\left(-1\right)+\left(-2x\right)\left(x^2\right)+\left(-2x\right)\)
\(=x^4-2x^3-x^2-2x^3+4x^3+2x-x^2+2x+1\)
\(=x^4-4x^3+2x^2+4x+1\)
Mk ko chắc
a) \(\left(x^2-2x-1\right)^2\)
\(=\left(x^2-2x\right)^2-2\left(x^2+2x\right)-1\)
\(=x^4+4x^3-2x^2+4x^2+4x+1\)
\(=x^4+4x^3-2x^2+4x+1\)
b) Tương tự
2x y 2 + x 2 y 4 + 1 = x y 2 2 + 2.x y 2 .1 + 1 2 = x y 2 + 1 2
ta có :(2x-3)^2+(4x+6)(1-2x)+(1-2x)^2
=(2x-3+1-2x)^2
=(-2)^2=4
chúc bạn học tốt!!!
a: \(\left(a^2+2a+3\right)\left(a^2-2a-3\right)\)
\(=\left[a^2+\left(2a+3\right)\right]\left[a^2-\left(2a+3\right)\right]\)
\(=\left(a^2\right)^2-\left(2a+3\right)^2\)
\(=a^4-\left(2a+3\right)^2\)
b: \(\left(-a^2-2a+3\right)^2\)
\(=\left(a^2+2a-3\right)^2\)
\(=a^4+4a^2+9+4a^3-18a-6a^2\)
\(=a^4+4a^3-2a^2-18a+9\)
c: \(\left(x-y-z\right)^2\)
\(=x^2-2x\left(y+z\right)+\left(y+z\right)^2\)
\(=x^2-2xy-2xz+y^2+2yz+z^2\)
d: \(\left(x+y+z\right)\left(x-y-z\right)\)
\(=x^2-\left(y+z\right)^2\)
\(=x^2-y^2-2yz-z^2\)
`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}`
\(4\left(x^2+2x+1\right)-12x-3=4x^2+8x+4-12x-3\)
\(=4x^2-4x+1=\left(2x-1\right)^2\)
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)
(2x+1)3
= (2x)3+3.2x2 + 3.2x + 13
= 8x3 + 6x2 + 6x + 1
exin lỗi (2x+1)^3