Rút gọn biểu thức:
(x2+2)2-(x+2)(x-2)(x2+4)
Help me, pls!
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1/2 x 2 (6x – 3) – x( x 2 + 1/2) + 1/2.(x + 4)
= (3 x 3 – 3/2. x 2 ) – ( x 3 + 1/2.x) + (1/2.x + 2)
= 3 x 3 - 3/2 x 2 – x 3 - 1/2 x + 1/2 x + 2
= ( 3 x 3 – x 3 ) - 3/2. x 2 – (1/2 x - 1/2 x) + 2
= 2 x 3 - 3/2 x 2 + 2
a) A = (x - 5)(x² + 5x + 25) - (x - 2)(x + 2) + x(x² + x + 4)
= x³ - 125 - x² + 4 + x³ + x² + 4x
= (x³ + x³) + (-x² + x²) + 4x + (-125 + 4)
= 2x³ + 4x - 121
b) Tại x = -2 ta có:
A = 2.(-2)³ + 4.(-2) - 121
= 2.(-8) - 8 - 121
= -16 - 129
= -145
c) x² - 1 = 0
x² = 1
x = -1; x = 1
*) Tại x = -1 ta có:
A = 2.(-1)³ + 4.(-1) - 121
= 2.(-1) - 4 - 121
= -2 - 125
= -127
*) Tại x = 1 ta có:
A = 2.1³ + 4.1 - 121
= 2.1 + 4 - 121
= 2 - 117
= -115
\(a,=\left(x+8-x+2\right)^2=10^2=100\\ b,=x^2\left(x^2-16\right)-\left(x^4-1\right)=x^4-16x^2-x^4+1=1-16x^2\\ c,=x^3+1-x^3+1=2\)
a) Ta có: \(\left(x+2y\right)\left(x^2-2xy+4y^2\right)-\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3+\left(2y\right)^3-\left(x^3-y^3\right)\)
\(=x^3+8y^3-x^3+y^3\)
\(=9y^3\)
b) Ta có: \(\left(x+1\right)\left(x-1\right)^2-\left(x+2\right)\left(x^2-2x+4\right)\)
\(=\left(x+1\right)\left(x^2-2x+1\right)-\left(x+2\right)\left(x^2-2x+4\right)\)
\(=x^3-2x^2+x+x^2-2x+1-\left(x^3+8\right)\)
\(=x^3-x^2-x+1-x^3-8\)
\(=-x^2-x-7\)
\(a,x^2+4x-21-x^2-4x+5=-16\\ b,=\left(x+8-x+2\right)^2=10^2=100\\ c,=x^2\left(x^2-16\right)-\left(x^4-1\right)\\ =x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)
\(a,=x^2+4x-21-x^2-4x+5=-16\\ b,=\left(x+8-x+2\right)^2=10^2=100\\ c,=x^2\left(x^2-16\right)-\left(x^4-1\right)\\ =x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)
(x – 2) . (2x3 – x2 + 1) + (x – 2) x2(1 – 2x)
= (x – 2). [(2x3 – x2 + 1) + x2(1 – 2x)]
= (x – 2). [2x3 – x2 + 1 + x2 . 1 + x2 . (-2x)]
= (x – 2) . (2x3 – x2 + 1 + x2 – 2x3)
= (x – 2) .1
= x – 2
a. \(\left(x+2\right)^{^2}-\left(x-4\right)^{^2}+x^{^2}-3x+1=x^{^2}+4x+4-x^{^2}+8x-16+x^{^2}-3x+1=x^{^2}+9x-11\)
b. \(\left(2x+2\right)^{^2}-4x\left(x+2\right)=4x^{^2}+8x+4-4x^{^2}-8x=4\)
\(\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)
\(=x^4+4x^2+4-\left(x^2-4\right)\left(x^2+4\right)\)
\(=x^4+4x^2+4-x^4+16=4x^2+20=4\left(x^2+5\right)\)