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\(\left(x-3\right)^2-\left(x+2\right)^2\)
\(=x^2-6x+9-x^2-4x-4\)
\(=-10x+5\)
\(\left(4x^2-2xy+y^2\right)\left(2x-y\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=\left(2x-y\right)\left(4x^2-2xy+y^2-4x^2-2xy-y^2\right)\)
\(=\left(2x-y\right)\cdot\left(-4xy\right)\)
a,\(\left(x-3\right)^2-\left(x+2\right)^2\)
\(=x^2-6x+9-x^2-4x-4\)
\(=-10x+5\)
b, \(\left(4x^2-2xy+y^2\right).\left(2x-y\right)-\left(2x-y\right).\left(4x^2+2xy+y^2\right)\)
\(=\left(2x-y\right).\left(4x^2-2xy+y^2-4x^2-2xy-y^2\right)\)
\(=\left(2x-y\right).\left(-4xy\right)\)
a) Theo tớ thì để phải là:
\(A=\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3-2\right)=x^3+8-x^3+2=10.\)
b) \(B=\left(x+3\right)\left(x^3-3x+9\right)-\left(54+x^3\right)=x^3+27-54-x^3=-27\)
c) \(C=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)=8x^3+y^3-8x^3+y^3=2y^3\)
Cả 3 bài đều áp dụng hằng đẳng thức: \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\) và \(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
a) \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
b) \(\left(x^2-1\right)^2-\left(x^4+x^2+1\right)\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-1-x^4-x^2-1\right)\)
\(=\left(x^2-1\right)\left(-x^4-2\right)\)
\(=-x^6+x^4-2x^2+2\).
a ) \(\left(x+3\right)\left(x^2-3x+9\right)-\left(5x+x^3\right)\)
\(=\left(x+3\right)\left(x^2-3x+3^2\right)-\left(54+x^3\right)\)
\(=x^3+3^3-\left(54+x^3\right)\)
\(=x^3+27-54-x^3\)
\(=-27\)
b ) \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=\left(2x+y\right)\left[\left(2x\right)^2-2.x.y+y^2\right]-\left(2x-y\right)\left[\left(2x\right)^2+2.x.y+y^2\right]\)
\(=\left[\left(2x\right)^3+y^3\right]-\left[\left(2x\right)^3-y^3\right]\)
\(=\left(2x\right)^3+y^3-\left(2x\right)^3+y^3\)
\(=2y^3\)
a ) (x+3)(x2−3x+9)−(5x+x3)(x+3)(x2−3x+9)−(5x+x3)
=(x+3)(x2−3x+32)−(54+x3)=(x+3)(x2−3x+32)−(54+x3)
=x3+33−(54+x3)=x3+33−(54+x3)
=x3+27−54−x3=x3+27−54−x3
=−27=−27
b ) (2x+y)(4x2−2xy+y2)−(2x−y)(4x2+2xy+y2)(2x+y)(4x2−2xy+y2)−(2x−y)(4x2+2xy+y2)
=(2x+y)[(2x)2−2.x.y+y2]−(2x−y)[(2x)2+2.x.y+y2]=(2x+y)[(2x)2−2.x.y+y2]−(2x−y)[(2x)2+2.x.y+y2]
=[(2x)3+y3]−[(2x)3−y3]=[(2x)3+y3]−[(2x)3−y3]
=(2x)3+y3−(2x)3+y3=(2x)3+y3−(2x)3+y3
=2y3
=(x+y)2-22/(x-y)(x+y)+4(x+y)
=(X+Y-2)(X+Y+2)/(X+Y)(X-Y+4)