(x – 2)(x + 2)(x2 + 4) Thực hiện các phép nhân
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\(a,=12x^2-4x-6x-2-x-3=12x^2-11x-5\\ b,=12x^2-9x-12x^2-4x+5=5-13x\\ c,=12x^3-4x^2-12x^3-12x^2+7x-3=-16x^2+7x-3\\ d,=\left(x^2-4\right)\left(x^2+4\right)=x^4-16\)
Bài 1:
b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)
Bài 2:
a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)
d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)
\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)
e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
2.
a. 3x(12x - 4) - 9x(4x - 3) = 30
<=> 36x2 - 12x - 36x2 + 27x = 30
<=> 36x2 - 36x2 - 12x + 27x = 30
<=> 15x = 30
<=> x = 2
b. x(5 - 2x) + 2x(x - 1) = 15
<=> 5x - 2x2 + 2x2 - 2x = 15
<=> -2x2 + 2x2 + 5x - 2x = 15
<=> 3x = 15
<=> x = 5
a) x2 ( 5x3 - x - 1212)= 5x5-x3-1212x
b) ( 3xy - x2 + y ) 2323x2y= 6969x3y2- 2323x4y+ 2323x2y2
c) x2 ( 4x3 - 5xy + 2x ) ( -1212 xy )=(4x5-5x3y+2x3).(-1212xy)
= -4848x6y +6060x4y2-2424x4y
2/ Tìm x, biết
a) 3x( 12x - 4 ) - 9x (4x - 3 ) = 30
=> 36x2-12x-36x2+27x=30
=> -12x +27x=30
=> 15x = 30
=>x =2
b ) x( 5 - 2x ) + 2x ( x - 1 )= 15
=> 5x-2x2+2x2-2x=15
=> 3x=15
=>x=5
\(a,\left(x^3+5x^2-2x+1\right)\left(x-7\right)\\ =x^4-7x^3+5x^3-35x^2-2x^2+14x+x-7\\ =x^4-2x^3-37x^2+15x-7\\ b,\left(2x^2-3xy+y^2\right)\left(x+y\right)\\ =2x^3+2x^2y-3x^2y-3xy^2+xy^2+y^3\\ =2x^3-x^2y-2xy^2+y^3\\ c,\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\\ =x^3-5x^2+x-2x^2+10x--x^3-11x\\ =x^3-7x^2\\ d,x\left(1-3x\right)\left(4-3x\right)-\left(x-4\right)\left(3x+5\right)\\ =x\left(4-15x+9x^2\right)-\left(3x^2-7x-20\right)\\ =4x-15x^2+9x^3-3x^2+7x+20\\ =9x^3-18x^2+11x+20\)
a) \(=8-x^3-x\left(16-x^2\right)=8-x^3-16x+x^3=-16x+8\)
b) \(=\left[\left(x+3\right)\left(x^2-3x+9\right)\right]:\left(x^2-3x+9\right)-x+7\)
\(=x+3-x+7=10\)
\(1,\\ a,=2x^2+2x\\ b,=x^2+4x+3-4=x^2+4x-1\\ c,=x^2+4x+4+3x-5=x^2+7x-1\\ 2,\\ a,=3\left(x+y\right)\\ b,=\left(x-3\right)^2\\ c,=7\left(x+y\right)\\ 3,\\ \Leftrightarrow\left(x-1\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)
\(=\left(x^2-4\right)\left(x^2+4\right)=x^4-16\)
(x – 2)(x + 2)(x2 + 4)
=(\(x^2\)-4)(\(x^2\)+4)
=\(x^4\)-16
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@ann1234