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Bài làm
a) 2(x + y)3 + 2(x - y)3
= 2[(x + y)3 + (x - y)3]
= 2[x3 + 3x2y + 3xy2 + y3 + x3 - 3x2y + 3xy2 - y3]
= 2[(x3 + x3) + (3x2y - 3x2y) + (3xy2 + 3xy2) + (y3 - y3)]
= 2[2x3 + 6xy2]
= 4x3 + 12xy2
b)uhm... Mình sửa đề chút, thay vì là -3(x + y)2(x - y) thì mình sẽ thành +3(x + y)2(x - y)
(x - y)3 - (x + y)3 + 3(x + y)2(x - y) - 3(x + y)(x - y)2
= -[(x + y)3 - 3(x + y)2(x - y) + 3(x + y)(x - y)2 - (x - y)3]
= -[(x + y) - (x - y)]3
= -[x + y - x + y ]3
= -[y]3
= -y
a)\(9x^2+30x+25+9x^2-30x+25-\left(9x^2-2^2\right)\)
=\(9x^2+54\)=\(9\left(x^2+6\right)\)
b)\(2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)\)
=\(8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x\)
=\(x^3-16x^2+25x\)
c)\(\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2\)
=\(\left(x+y-z-\left(x+y\right)\right)^2\)=\(\left(-z\right)^2\)
a) \(5x^2-2x\left(3x+\frac{3}{2}\right)=-x^2-3x=-x\left(x+3\right)=-3\left(3+3\right)=-18\)
b) \(3x\left(x-4y\right)-\frac{12}{5}y\left(y-5x\right)=3x^2-\frac{12}{5}y^2=3\left(x^2-\frac{4}{5}y^2\right)\)
\(=3\left(4^2-\frac{4}{5}.5^2\right)=3.\left(-4\right)=-12\)
c) \(\left(x-2\right)^2-\left(x+7\right)\left(x-7\right)=x^2-4x+4-x^2+49=-4x+53=-4.3+53=41\)
d) \(x^2+12x+36=\left(x+6\right)^2=\left(64+6\right)^2=70^2=4900\)
e) \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)=x^2-6x+9-x^2+16=-6x+25=-6\left(-1\right)+25\)
= 31
f) \(\left(3x+2y\right)^2-4y\left(3x+y\right)=9x^2+12xy+4y^2-12xy-4y^2=9x^2=9\left(-\frac{1}{3}\right)^2=1\)
a, \(A=\left(3x-2\right)^2+\left(3x+2\right)^2+2\left(9x^2-4\right)\)
\(=\left(3x-2\right)^2+\left(3x+2\right)^2+2\left(3x-2\right)\left(3x+2\right)\)
\(=\left(3x-2+3x+2\right)^2\)
\(=36x^2=36.\left(-\frac{1}{3}\right)^2=4\)
b, \(B=\left(x+y-7\right)^2-2\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)
\(=\left[\left(x+y-7\right)-\left(y-6\right)\right]^2\)
\(=\left(x-1\right)^2\)
\(=\left(101-1\right)^2=10000\)
c, \(C=4x^2-20x+27\)
\(=\left(2x\right)^2-2.2x.5+5^2+2\)
\(=\left(2x-5\right)^2+2\)
\(=\left(52,5.2-5\right)^2+2\)
\(=100^2+2=10002\)
Bài này dễ mà chỉ dùng hằng đẳng thức thôi. Chúc bạn học tốt.