Rút gọn biểu thức:
(5x + 3)^2 - 2(5x + 3) . (x+3) + (x+3)^2
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=5x^2+5x-2x-2-(5x^2+x-15x-3)-17x-51
=5x^2-14x-53-5x^2+14x+3
=-50
c: C=125x^3+150x^2+60x+8+125x^3-150x^2+60x-8-2(x^2-4)
=250x^3+120x-2x^2+8
=250x^3-2x^2+120x+8
d: D=(4x)^3-3^3-(4x)^3-3^3
=64x^3-27-64x^3-27
=-54
c) \(C=\left(5x+2\right)^3+\left(5x-2\right)^3-2\left(x-2\right)\left(x+2\right)\)
\(=\left[\left(5x\right)^3+3\cdot\left(5x\right)^2\cdot2+3\cdot5x\cdot2^2+2^3\right]+\left[\left(5x\right)^3-3\cdot\left(5x\right)^2\cdot2+3\cdot5x\cdot2^2-2^3\right]-2\left(x^2-4\right)\)
\(=125x^3+150x^2+60x+8+125x^3-150x^2+60x-8-2x^2+8\)
\(=\left(125x^3+125x^3\right)+\left(150x^2-150x^2-2x^2\right)+\left(60x+60x\right)+\left(8-8+8\right)\)
\(=250x^3-2x^2+120x+8\)
d) \(D=\left(4x-3\right)\left(16x^2+12x+9\right)-\left(4x+3\right)\left(16x^2-12x+9\right)\)
\(=\left(4x\right)^3-3^3-\left[\left(4x\right)^3+3^3\right]\)
\(=64x^3-27-\left(64x^3+27\right)\)
\(=64x^3-27-64x^3-27\)
\(=-27-27\)
\(=-54\)
-(5x - 3)(2x + 4) - (x + 7)(x - 3) + (5x - 2)(3x + 4)
= -(10x2 + 20x - 6x - 12) - (x2 - 3x + 7x - 21) + (15x2 + 20x - 6x - 8)
= -10x2 - 20x + 6x + 12 - x2 + 3x - 7x + 21 + 15x2 + 20x - 6x - 8
= 4x2 - 4x + 25
\(a,x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^3-3x-5x^3-x^2+x^2\)
\(=-3x^3-3x\)
\(b,3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=-11x+24\)
=.= hok tốt!!
a)
\(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2=x\left(2x^2-3-5x^2-x+x\right)\)
=\(x\left(-3x^2-3\right)=-3x\left(x^2+1\right)\)
b)
Nhân ra:
\(3x^2-6x-5x+5x^2-8x^2+24=-11x+24\)
Chúc bạn học tốt!
a) \(2x\left(5-3x^2\right)-10\left(6+x\right)\)
\(=10x-6x^3-60-10x\)
\(=\) \(-6x^3-60\)
a) \(2x\left(5-3x^2\right)-10\left(6+x\right)\\ =2x.5-2x.3x^2-10.6-10.x\\ =10x-6x^3-60-10x\)
b) \(3\left(-x+2\right)-6\left(1-x+5x^{20}\right)\\ =-3.x+3.2-6.1+6.x-5.5x^{20}\\ =-3x+6-6+6x-25x^{20}=25x^{20}+3x\)
c) \(7x\left(2-5x^2+\dfrac{1}{2}x^3\right)-14x\left(1-2x^2\right)\\ =7x.2-7x.5x^2+7x.\dfrac{1}{2}x^3-14x.1+14x.2x^2\\ =14x-25x^3+\dfrac{7}{2}x^4-14x+28x^3=3x^2+\dfrac{7}{2}x^4\)
\(\left(5x+3\right)^2-2\left(5x+3\right)\left(x+3\right)+\left(x+3\right)^2\)
Dễ thấy đây là hằng đẳng thức thứ hai với 5x + 3 là A và x + 3 là B
Do đó : \(\left(5x+3\right)^2-2\left(5x+3\right)\left(x+3\right)+\left(x+3\right)^2\)
\(=\left(5x+3-x-3\right)^2\)
\(=\left(4x\right)^2\)
\(=16x^2\)