Cho A= \(\left(\frac{x}{x-3}-\frac{1}{x+3}+\frac{x^2-1}{9-x^2}\right):\frac{2}{x+3}\)
a) tìm ĐKXĐ ,rút gọn A
b) tìm x thuộc z để A nhận gtri nguyên
giải hộ e vs ạ
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(5x - 2)(x + 1) - 3x(x2 - x - 3) - 2x(x - 5)(x + 4) + 5x(x2 - 7)
= 5x2 + 3x - 2 - 3x3 + 3x2 + 9x - 2x(x2 - x - 20) + 5x3 - 35x
= 2x3 + 8x2 - 23x - 2 - 2x3 + 2x2 + 40x = 10x2 + 17x - 2
\(B=4x\left(2x+y\right)+2y\left(2x+y-y^2+2x\right)\)
\(=4x\left(2x+y\right)+2y\left(4x-y^2+y\right)=8x^2+4xy+8xy-2y^3+2y^2\)
(-3x + 4)(x - 2) - x(x2 - 2)(-4x + 1) + (2x - 5)(x2 - 4x)
= -3x2 + 10x - 8 - x(-4x3 + x2 + 8x - 2) + 2x3 - 13x2 + 20x
= 2x3 - 16x2 + 30x - 8 + 4x4 - x3 - 8x2 + 2x
= 4x4 + x3 - 24x2 + 28x - 8
(8x - 1)(2x - 3)) - (-x + 7)(x - 2) - 6x(x2 - x + 3) + (x2 - 3)(x - 4)
= 16x2 - 26x + 3 + x2 - 9x + 14 - 6x3 + 12x2 - 18x + x3 - 4x2 - 3x + 12
= -5x3 + 25x2 - 56x + 29
(8x - 1) (2x - 3) - (-x + 7)(x - 2) - 6x (x2 - x + 3) + (x2 - 3)(x - 4)
= 16x2-24x-2x-3+x2-2x-7x+14-6x3+6x-18x+x3-4x2-3x+12
=7x3+13x2-36x+23
-7x(2x2- 4x - 5) - (x - 5)(-2x + 3) + (3x - 2)(x+4) - 4x2 (x - 3)
=-14x3+28x2+35x+2x2-3x-10x+15+3x2+12x-2x-8-4x3+12x2
=-18x3+45x2+32x+7
\(\left(3x+5\right)\left(3x-5\right)-\left(4x-3\right)^2\)
\(9x^2-25-\left(16x^2-24x+9\right)\)
\(9x^2-25-16x^2+24x-9\)
\(-7x^2+24x-34\)
(3x - 7)(x + 3) + 5x(x2 - 2x - 4) - (4x - 5)(x-4)-(x + 2)(x + 1)
=3x2+9x-7x-21+5x3-10x2-20x-4x2+16x+5x-20-x2-x-2x-2
=5x3-13x2-43
5x2 (x - 7) - 4x(x - 2)(x + 3) - (x - 5)(x + 6) + (2x-1)(x + 5)
=5x3-35x2-4x(x2+3x-2x-6)-(x2+6x-5x-30)+2x2+10x-x-5
=5x3-35x2-4x3-12x2+8x2+24x-x2-6x+5x+30+2x2+9x-5
=x3+32x2+32x+25
a, \(A=\frac{x^2+3x-x+3-x^2+1}{x^2-9}\)\(.\frac{x+3}{2}\) \(\left(x\ne3;-3\right)\)
\(A=\frac{2x+4}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{2}\)\(=\frac{2\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{2}\)\(=\frac{x+2}{x-3}\)
b, để \(A\in Z\Rightarrow\hept{\begin{cases}x+2⋮x-3\\x-3⋮x-3\end{cases}}\)\(\Rightarrow x+2-x+3=5⋮x-3\)\(\leftrightarrow x+3\in\left(1;5;-1;-5\right)\)
\(\leftrightarrow x\in\left(-2;2;-4;-8\right)\)
Mới 2k9