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a) Ta có: \(P=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(\dfrac{x^2-x-2}{x^2}\right)\)
\(=\dfrac{x\left(x-2\right)^2+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x^2-x-2\right)}{x^2}\)
\(=\dfrac{x\left[x^2-4x+4+4x\right]}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x+1}{2x}\)
b) Thay \(x=\dfrac{1}{2}\) vào P, ta được:
\(P=\dfrac{1}{2}+1=\dfrac{3}{2}\)
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b)Q = (x - 1)3 - 4x(x + 1)(x - 1) + 3(x - 1)(x2 + x + 1)
Q = (x - 1)3 - 4x(x2 - 1) + 3(x3 - 1)
Thay x = -2 vào Q ta dc :
(-3)3 - 4 . (-2) . 3 + 3 . (-9) = -27 + 24 - 27 = -30
a/ \(\left(8-5x\right)\left(x-2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)+10\)
\(=8\left(x-2\right)-5x\left(x-2\right)+\left(4x-8\right)\left(x+1\right)+2\left(x^2-2^2\right)\)
\(=8x-16-5x^2+10x+4x\left(x+1\right)-8\left(x+1\right)+2x^2-2.2^2\)
\(=8x-16-5x^2+10x+4x^2+4x-8x-8+2x^2-8\)
\(=14x+x^2-32\)
b/ \(4\left(x-1\right)\left(x+5\right)-\left(x+2\right)\left(x+5\right)-3\left(x-1\right)\left(x+2\right)\)
\(=\left(x+5\right)\left[4\left(x-1\right)-\left(x+2\right)\right]-\left(3x-3\right)\left(x+2\right)\)
\(=\left(x+5\right)\left(4x-4-x-2\right)-\left[3x\left(x+2\right)-3\left(x+2\right)\right]\)
\(=\left(x+5\right)\left(3x-6\right)-\left(3x^2+6x-3x-6\right)\)
\(=x\left(3x-6\right)+5\left(3x-6\right)-\left(3x^2+3x-6\right)\)
\(=3x^2-6x+15x-30-3x^2-3x+6\)
\(=9x-24\)
\(a,x^2+4x-21-x^2-4x+5=-16\\ b,=\left(x+8-x+2\right)^2=10^2=100\\ c,=x^2\left(x^2-16\right)-\left(x^4-1\right)\\ =x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)
\(a,=x^2+4x-21-x^2-4x+5=-16\\ b,=\left(x+8-x+2\right)^2=10^2=100\\ c,=x^2\left(x^2-16\right)-\left(x^4-1\right)\\ =x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)
\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)
\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)
\(=6x^2y\)
\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)
\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)
\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)
1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy
2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3
=6x^2y
3: =(x+y-x+y)^2=(2y)^2=4y^2
4: =(2x+3-2x-5)^2=(-2)^2=4
5: =18^8-18^8+1=1
a: \(P=\left(\dfrac{x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)\cdot\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)\cdot\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{1}{x+2}\cdot\dfrac{x^3-x-2x+2}{x^2+x+1}\right)\)
\(=\left(\dfrac{x}{x+2}-\dfrac{x^2-2x+4}{\left(x+2\right)^2}\right):\left(\dfrac{1}{x+2}\cdot\dfrac{x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)}{x^2+x+1}\right)\)
\(=\dfrac{x^2+2x-x^2+2x-4}{\left(x+2\right)^2}:\left(\dfrac{1}{x+2}\cdot\dfrac{\left(x-1\right)\left(x^2+x-2\right)}{x^2+x+1}\right)\)
\(=\dfrac{4x-4}{\left(x+2\right)^2}:\left(\dfrac{1}{x+2}\cdot\dfrac{\left(x-1\right)\left(x+2\right)\left(x-1\right)}{x^2+x+1}\right)\)
\(=\dfrac{4\left(x-1\right)}{\left(x+2\right)^2}\cdot\dfrac{x^2+x+1}{\left(x-1\right)^2}=\dfrac{4\left(x^2+x+1\right)}{\left(x+2\right)^2\left(x-1\right)}\)
b: Để P>0 thì x-1>0
hay x>1
a) Ta có: 2x2 + 8 = 2(x2 + 4).
8 – 4x + 2x2 – x3
= (8 – x3) - ( 4x - 2x2)
= (2 – x).(4 + 2x + x2) - 2x.(2 - x)
= (2 – x).(4 + 2x + x2 – 2x)
= (2 - x). (4 + x2 )
* Do đó:
b) Tại x = 1 2 hàm số đã cho xác định nên thay x = 1 2 vào biểu thức rút gọn của P ta được: