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a) (x+3)(x^2-3x+9)-(54+x^3)
= x^3- 3x^2+9x+3x^2-9x+27-54-x63
= -27
b) (2x + y)(4x^2 – 2xy + y^2) – (2x – y)(4x^2+ 2xy + y^2)
= (2x + y)[(2x)^2 – 2x.y + y^2] – (2x – y)[(2x)^2 + 2x.y + y^2]
= [(2x)3^3+ y^3] – [(2x)^3 – y^3]
= (2x)^3 + y^3 – (2x)^3 + y^3
= 2y^3
a)(x+3)(X^2-3x+9)-(54+x^3)
= \(x^3\)+ \(3^3 \) - 54 -\(x^3\)
= 27- 54
= -27
b)(2x+y)(4x^2-2xy+y^2)-(2x-y)(4x^2+2xy+y^2)
= \((2x)^3\) + \(y^3\) - [\((2x)^3\) - \(y^3\) ]
= \(8x^3\) + \(y^3\) - \(8x^3\) + \(y^3\)
= \(2y^3\)
b) \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=\left(2x+y\right)\left(4x^2-2xy+y^2\right)+\left(2x+y\right)\left(4x^2+2xy+y^2\right)\)
\(=\left(2x+y\right)\left(4x^2-2xy+y^2+4x^2+2xy+y^2\right)\)
\(=\left(2x+y\right)\left(8x^2+2y^2\right)\)
\(=\left(2x+y\right)\left(4x+y\right).2xy\)
a: \(N=\left(2x-3y\right)\left(2x+3y\right)=\left(2x\right)^2-\left(3y\right)^2\)
\(=4x^2-9y^2\)
Thay x=1/2 và y=1/3 vào N, ta được:
\(N=4\cdot\left(\dfrac{1}{2}\right)^2-9\left(\dfrac{1}{3}\right)^2\)
\(=4\cdot\dfrac{1}{4}-9\cdot\dfrac{1}{9}\)
=1-1
=0
b: \(N=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=\left(2x-y\right)\left[\left(2x\right)^2+2x\cdot y+y^2\right]\)
\(=\left(2x\right)^3-y^3=8x^3-y^3\)
Khi x=1 và y=3 thì \(N=8\cdot1^3-3^3=8-27=-19\)
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
Bài 1:
- a,(2+xy)^2=4+4xy+x^2y^2
- b,(5-3x)^2=25-30x+9x^2
- d,(5x-1)^3=125x^3 - 75x^2 + 15x^2 - 1
\(\left(x-3\right)^2-\left(x+2\right)^2\)
\(=x^2-6x+9-x^2-4x-4\)
\(=-10x+5\)
\(\left(4x^2-2xy+y^2\right)\left(2x-y\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=\left(2x-y\right)\left(4x^2-2xy+y^2-4x^2-2xy-y^2\right)\)
\(=\left(2x-y\right)\cdot\left(-4xy\right)\)
a,\(\left(x-3\right)^2-\left(x+2\right)^2\)
\(=x^2-6x+9-x^2-4x-4\)
\(=-10x+5\)
b, \(\left(4x^2-2xy+y^2\right).\left(2x-y\right)-\left(2x-y\right).\left(4x^2+2xy+y^2\right)\)
\(=\left(2x-y\right).\left(4x^2-2xy+y^2-4x^2-2xy-y^2\right)\)
\(=\left(2x-y\right).\left(-4xy\right)\)
\(a,\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x\left(x-3\right)\)
\(=x^3-6x^2+12x-27-x^3+x+6x^2-18x\)
\(=-5x-27\)
\(b,\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-\left(8x^3-y^3\right)\)
\(=8x^3+y^3-8x^3+y^3=2y^3\)
\(\left(x+y+z\right)^2-2\left(x+y+z\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x+y+z-x-y\right)^2\)
\(=z^2\)
a)
=\(x^3-6x^2+12x+8-27-x^3+x+6x^2-18x\)
=-5x-19
b)
=\(8x^3+y^3-8x^3+y^3\)
=\(2y^3\)
c)
=(x+y+z-x-y)\(^2\) +x+y
=\(z^2+x+y\)
hc tốt
a) Ta có: \(\left(y+3\right)\left(y^2-3y+9\right)-\left(60-y^3\right)\)
\(=y^3+27-60+y^3\)
\(=2y^3-33\)
b) Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)