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Ta có (x^2 + y^2 )^3 + (z^2 – x^2 )^3 – (y^2 + z^2 )^3
= (x^2 + y^2 )^3 + (z^2 – x^2 )^3 + (-y^2 - z^2 )^3
Ta thấy x^2 + y^2 + z^2 – x^2 – y^2 – z^2 = 0
=> áp dụng nhận xét ta có: (x^2+y^2 )^3+ (z^2 -x^2 )^3 -y^2 -z^2 )^3
= 3(x^2 + y^2 ) (z^2 –x^2 ) (-y^2 – z^2 )
= 3(x^2+y^2 ) (x+z)(x-z)(y^2+z^2 )
\(6x^4-11x^2+3=6x^4-9x^2-2x^2+3\)
\(=3x^2\left(2x^2-3\right)-\left(2x^2-3\right)=\left(2x^2-3\right)\left(3x^2-1\right)\)
\(49.\left(y-4\right)^2-9y^2-36y-36\)
\(=7^2\left(y-4\right)^2-\left(9y^2+36y+36\right)\)
\(=\left(7y-28\right)^2-\left(3y+6\right)^2\)
\(=\left(7y-28+3y+6\right).\left(7y-28-3y-6\right)\)
\(=\left(10y-22\right).\left(4y-34\right)\)
\(=4.\left(5y-11\right).\left(2y-17\right)\)
\(49\left(y-4\right)^2-9y^2-36y-36\)
\(=\) \(4\left(2y-17\right)\left(5y-11\right)\)
(x -y)3 - 1 - 3(x -y)(x - y - 1)
= (x -y)3 - 3(x -y)(x - y - 1) - 1
Đặt x - y = t, khi đó ta có:
t3 - 3t. (t - 1) - 1
= t3 - 3t2 + 3t - 1
= (t - 1)3
Thay t = x - y vào (t - 1)3 , ta có: ( x - y - 1)3
Vậy (x -y)3 - 1 - 3(x -y)(x - y - 1) = ( x - y - 1)3
( x + 2 ) ( x + 3 ) ( x + 4 ) ( x + 5 ) - 24
= ( x2 + 7x + 10 ) ( x2 + 7x + 12 ) - 24
Đặt x2 + 7x + 10 = y
Ta có :
y2 + 2y - 24 = ( y - 4 ) ( y + 6 ) = ( x2 + 7x + 6 ) ( x2 + 7x + 16 )
= ( x + 1 ) ( x + 6 ) ( x2 + 7x + 16 )
Đặt x2+7x+10=t
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=t\left(t+2\right)-24=t^2+2t-24\)
\(=\left(t^2+2t+1\right)-25=\left(t+1\right)^2-5^2=\left(t-4\right)\left(t+6\right)\)=(x2+7x+6)(x2+7x+16)
=(x2+x+6x+6)(x2+7x+16)=[x(x+1)+6(x+1)](x2+7x+16)=(x+1)(x+6)(x2+7x+16)
\(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=\left(x-y\right)\left[3\left(x+y\right)-2\left(x-y\right)\right]\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)