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a) 16(12 t 2 +1).
b) Gợi ý x 3 + y 3 = ( x + y ) 3 - 3xy(x + y)
(x + y - z)( x 2 + y 2 + z 2 - xy + xz + yz).
x 2 y + x y 2 + x 2 z + x z 2 + y 2 z + y z 2 + 3xyz.
= ( x 2 y + x 2 z + xyz) + (x y 2 + y 2 z + xyz) + (x z 2 + y z 2 + xyz)
= x(xy + xz + yz) + y(xy + yz + xz) + z(xz + yz + xy)
= (x + y + z)(xy + xz + yz).
\(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)
\(=\left(x^2y+x^2z+xyz\right)+\left(xz^2+yz^2+xyz\right)+\left(xy^2+y^2z+xyz\right)\)
\(=x\left(xy+xz+yz\right)+z\left(xz+yz+xy\right)+y\left(xy+yz+xz\right)\)
\(=\left(x+y+z\right)\left(xy+yz+xz\right)\)
Ta có : \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
Ta có: \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
Ta có : x4 + 2018x2 + 2017x + 2018
= x4 - x + 2018x2 + 2018x + 2018
= x(x3 - 1) + 2018(x2 + x + 1)
= x(x - 1)(x2 + x + 1) + 2018(x2 + x + 1)
= (x2 + x + 1)(x2 - x + 2018)
\(\text{a) }4x^{16}+81=4x^4+36x^2+81-36x^8\)
\(=\left(4x^{16}+36x^8+81\right)-36x^8\)
\(=\left[\left(2x^8\right)^2+2.2x^8.9+9^2\right]+\left(6x^4\right)^2\)
\(=\left(2x^8+9\right)^2-\left(6x^4\right)^2\)
\(=\left(2x^8+9-6x^4\right)\left(2x^8+9+6x^4\right)\)
\(\text{b) }x^4+2018x^2+2017x+2018\)
\(=x^4+2018x^2+2018x-x+2018\)
\(=\left(x^4-x\right)+\left(2018x^2+2018x+2018\right)\)
\(=x\left(x^3-1\right)-2018\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2018\left(x^2+x+1\right)\)
\(=\left(x^2-x\right)\left(x^2+x+1\right)+2018\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2018\right)\)
\(x^4+2x^3+2x^2+2x+1\\ =\left(x^4+x^3\right)+\left(x^3+x^2\right)+\left(x^2+x\right)+\left(x+1\right)\\ =x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\\ =\left(x^3+x^2+x+1\right)\left(x+1\right)\\ =\left[\left(x^3+x^2\right)+\left(x+1\right)\right]\left(x+1\right)\\ =\left[x^2\left(x+1\right)+\left(x+1\right)\right]\left(x+1\right)\\ =\left(x^2+1\right)\left(x+1\right)^2\)
\(x^4+8x=x\left(x^3+8\right)=x\left(x+2\right)\left(x^2-2x+4\right)\)
\(a,=\left(2x-5\right)\left(x+1\right)\\ b,=\left(x-10\right)\left(x+1\right)\\ c,=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
ai thông minh trả lời nhanh dùm mk dc ko vậy ạ
Ở đây ko có ai thông minh đâu bạn à. :)