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Đặt \(x+y-z=a;x-y+z=b;y+z-x=c\)
Ta có:\(A=\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(A=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(A=\left(a+b\right)^3+3\left(a+b\right)\cdot c\cdot\left(a+b+c\right)+c^3-a^3-b^3-c^3\)
\(A=a^3+b^3+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)+c^3-a^3-b^3-c^3\)
\(A=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(A=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Hay \(A=3\cdot2x\cdot2y\cdot2z\)
\(A=24xyz\)
\(4x^8+1=\)\(4x^8-4x^4+4x^4+1\)\(=\left(4x^8+4x^4+1\right)-4x^4\)
\(=\left(2x^4+1\right)^2-\left(2x^2\right)^2\)\(=\left(2x^4-2x^2+1\right)\left(2x^4-2x^2-1\right)\)
phần b em tự giải nhé
Mình nghĩ bạn ghi đề sai, đề đúng theo mình là:
\(x^2y^2\left(x-y\right)+y^2z^2\left(y-z\right)+z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(x-y\right)-y^2z^2\text{[}\left(x-y\right)+\left(z-x\right)\text{]}+z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(x-y\right)-y^2z^2\left(x-y\right)-y^2z^2\left(z-x\right)+z^2x^2\left(z-x\right)\)
\(=\left(x-y\right)\left(x^2y^2-y^2z^2\right)+\left(z-x\right)\left(z^2x^2-y^2z^2\right)\)
\(=\left(x-y\right).y^2\left(x+z\right)\left(x-z\right)+\left(z-x\right).z^2\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(x-z\text{ }\right)\text{[}y^2.\left(x+z\right)-z^2\left(x+y\right)\text{]}\)
\(=\left(x-y\right)\left(z-x\right)\left(y^2x+y^2z-z^2x-z^2y\right)\)
\(=\left(x-y\right)\left(z-x\right)\text{[}\left(y^2x-z^2x\right)+\left(y^2z-z^2y\right)\text{]}\)
\(=\left(x-y\right)\left(z-x\right)\text{[}x.\left(y-z\right)\left(y+z\right)+yz\left(y-z\right)\text{]}\)
\(=\left(x-y\right)\left(x-z\right)\left(y-z\right)\left(xy+x\text{z}+yz\right)\)
=8x3y + z3y + 8x3z -2xz3 - y3(z +2x)= y(8x3+z3) +2xz(4x2-z2) - y3(2x+z) = y(2x+z)(4x2 - 2xz + z2) +2xz(2x+z)(2x-z) - y3(2x+z)
=(2x+z)(4x2y -2xyz + z2y + 4x2z -2xz2 - y3) = (2x+z)( 4x2y+ 4x2z - 2xyx- 2xz2 +z2y - y3) = (2x+z)[ 4x2(y+z) -2xz(y+z) + y(z+y)(z-y)]
= (2x+z)(y+z)( 4x2- 2xz +yz- y2) = (2x+z)(y+z)(4x2 - y2 -2xz + yz) = (2x+z)(y+z)[(2x-y)(2x+y) - z(2x-y)]
= (2x+y)(y+z)(2x-y)(2x+y-z)