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8 tháng 7 2018
\(1)\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)\left(x-1\right)\\ =x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x-1\right)^2\\ =6x^2+2-6\cdot\left(x^2-2x+1\right)\\ =6x^2+2-6x^2+12x-6\\ =12x-4\)
\(2)x\left(x-1\right)\left(x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\\ =x\left(x^2-1\right)-\left(x^3+1\right)\\ =x^3-x-x^3-1\\=-x-1\)
\(3)\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x-4\right)\left(x+4\right)\\ =x^3-3x^2+3x-1-(x^3+8)+3\cdot\left(x^2-16\right)\\ =x^3-3x^2+3x-1-x^3-8+3x^2-48\\ =3x-55\)
TH
0
6 tháng 12 2022
a: =x^2+2x-15-x^2+4
=2x-11
b: =x^2-4x+4+x^2+6x+9-2(x^2-1)
=2x^2+2x+13-2x^2+2
=2x+15
c: \(=x^2-4x+4+x^3-1-x^3+4x\)
=x^2+3
d: \(=\left(2x+5-2x+1\right)^2=6^2=36\)
e: \(=x^3+1-x^3+1-x^2=2-x^2\)
27 tháng 11 2022
a: \(=\dfrac{4x^2+4x+1-4x^2+4x-1}{\left(2x+1\right)\left(2x-1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x\cdot5}{4x\left(2x+1\right)}=\dfrac{10}{2x+1}\)
b: \(=\left(\dfrac{1}{x^2+1}+\dfrac{x-2}{x+1}\right):\dfrac{1+x^2-2x}{x}\)
\(=\dfrac{x+1+x^3+x-2x^2-2}{\left(x+1\right)\left(x^2+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x^3-2x^2+2x-1}{\left(x+1\right)\left(x^2+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x\left(x^2-x+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}\)
c: \(=\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}\cdot\left(\dfrac{1}{\left(x-1\right)^2}-\dfrac{1}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}\)
\(=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
Y
2 tháng 8 2018
\(P=\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+12\right)\)
\(P=\left[\left(x+2\right)+\left(x-2\right)\right]\left[\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)+\left(x-2\right)^2\right]-2x^3-24x\)
\(P=2x\left(x^2+4x+4-x^2+4+x^2-4x+4\right)-2x^3-24x\)
\(P=2x\left(x^2+12\right)-2x^3-24x\)
\(P=2x^3+24x-2x^3-24x\)
\(P=0\)
=> P không phụ thuộc vào biến x
\(Q=\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)
\(Q=\left[\left(x-1\right)-\left(x+1\right)\right]\left[\left(x-1\right)^2+\left(x-1\right)\left(x+1\right)+\left(x+1\right)^2\right]+6\left(x^2-1\right)\)
\(Q=-2\left(x^2-2x+1+x^2-1+x^2+2x+1\right)+6x^2-6\)
\(Q=-2\left(3x^2+1\right)+6x^2-6\)
\(Q=-6x^2-2+6x^2-6\)
\(Q=-8\)
=> Q không phụ thuộc vào biến x
\(N=y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)
\(N=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)\)
\(N=0\)
=> N không phụ thuộc vào biến y
\(M=\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3\left(1-x\right)x\)
\(M=x^3-3x^2+3x-1-x^3+1-3x+3x^2\)
\(M=0\)
=> M không phụ thuộc vào biến x
\(H=\left(x+1\right)^3-\left(x-1\right)^3-3\left[\left(x-1\right)^2+\left(x+1\right)^2\right]\)
\(H=x^3+3x^2+3x+1-x^3+3x^2-3x+1-3\left(x^2-2x+1+x^2+2x+1\right)\)
\(H=6x^2+2-3\left(2x^2+2\right)\)
\(H=6x^2+2-6x^2-6\)
\(H=-4\)
=> H không phụ thuộc vào biến x
NT
0
\(\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=x^3+1-\left(x^3-1\right)\)
\(=x^3+1-x^3+1\)
\(=2\)