Tính:
\(\dfrac{x^3+8}{x^2+2x+1}\)X \(\dfrac{x^2+3x+2}{1-x^2}\)
Mn giúp mik với ạ
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x(x + 2) + x(x - 5) - 5(x + 2)
= [x(x + 2) - 5(x + 2)] + x(x - 5)
= (x - 5)(x + 2) + x(x - 5)
= (x - 5)(x + 2 + x)
= (x - 5)(2x + 2)
= 2(x - 5)(x + 1)
a,A = \(\dfrac{3x^2+6xy}{6x^2}\) (đk \(x\) ≠ 0)
A = \(\dfrac{3x.\left(x+2y\right)}{6x^2}\)
A = \(\dfrac{x+2y}{2x}\)
b,B = \(\dfrac{2x^2-x^3}{x^2-4}\) (đk \(x\)2 - 4 ≠ 0 ⇒ \(x\) ≠ \(\pm\) 2)
B = \(\dfrac{x^2\left(2-x\right)}{\left(x+2\right)\left(x-2\right)}\)
B = \(\dfrac{-x^2.\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}\)
B = \(\dfrac{-x^2}{x+2}\)
a) \(\left(x^2-4x\right)+\left(12x-48\right)\)
\(=x\left(x-4\right)+12\left(x-4\right)\)
\(=\left(x-4\right)\left(x+12\right)\)
b) \(x(x+y)-3x-3y\)
\(=x\left(x+y\right)-\left(3x+3y\right)\)
\(=x\left(x+y\right)-3\left(x+y\right)\)
\(=\left(x+y\right)\left(x-3\right)\)
c) \(20xy^2-12x^2+12x^3y\)
\(=4x\left(5y^2-3x+3x^2y\right)\)
d) \(2x\left(x-y\right)+3y\left(y-x\right)\)
\(=2x\left(x-y\right)-3y\left(x-y\right)\)
\(=\left(x-y\right)\left(2x-3y\right)\)
a) \(70a+84b-20ab-24b^2\)
\(=\left(70a+84b\right)-\left(20ab+24b^2\right)\)
\(=14\left(5a+6b\right)-4b\left(5a+6b\right)\)
\(=\left(5a+6b\right)\left(14-4b\right)\)
\(=2\left(5a+6b\right)\left(7-2b\right)\)
b) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)
\(=\left(x^2y+xy^2+xyz\right)+\left(x^2z+xyz+xz^2\right)+\left(xyz+y^2z+yz^2\right)\)
\(=xy\left(x+y+z\right)+xz\left(x+y+z\right)+yz\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(xy+yz+xz\right)\)
c) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)
\(=\left(x^2y+xy^2\right)+\left(xz^2+yz^2\right)+\left(x^2z+2xyz+y^2z\right)\)
\(=xy\left(x+y\right)+z^2\left(x+y\right)+z\left(x^2+2xy+y^2\right)\)
\(=xy\left(x+y\right)+z^2\left(x+y\right)+z\left(x+y\right)^2\)
\(=\left(x+y\right)\left[xy+z^2+z\left(x+y\right)\right]\)
\(=\left(x+y\right)\left(xy+z^2+xz+yz\right)\)
\(=\left(x+y\right)\left[\left(xy+yz\right)+\left(xz+z^2\right)\right]\)
\(=\left(x+y\right)\left[y\left(x+z\right)+z\left(x+z\right)\right]\)
\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
a, 70a + 84b - 20ab - 24b2
= 14.(5a + 6b) - 4b(5a + 6b)
= (5a + 6b).(14 - 4b)
Lời giải:
ĐKXĐ: $x\neq 2; x\neq -3$
\(\frac{x+2}{x+3}-\frac{5}{(x+3)(x-2)}-\frac{1}{x-2}=\frac{(x+2)(x-2)-5-(x+3)}{(x+3)(x-2)}\\ =\frac{x^2-4-5-x-3}{(x+3)(x-2)}=\frac{x^2-x-12}{(x+3)(x-2)}\\ =\frac{(x+3)(x-4)}{(x+3)(x-2)}=\frac{x-4}{x-2}\)
Bài 4:
a. Ta thấy: $x^2-x+2=(x-\frac{1}{2})^2+1,75>0$ với mọi $x$.
Do đó để $B=\frac{x^2-x+2}{x-3}<0$ thì $x-3<0$
$\Leftrightarrow x<3$
b.
$B=\frac{x(x-3)+2(x-3)+8}{x-3}=x+2+\frac{8}{x-3}$
Với $x$ nguyên, để $B$ nguyên thì $x-3$ phải là ước của 8.
$\Rightarrow x-3\in\left\{\pm 1; \pm 2; \pm 4; \pm 8\right\}$
$\Rightarrow x\in \left\{4; 2; 5; 1; -1; 7; 11; -5\right\}$
Bài 5:
\(\frac{\frac{x}{x-y}-\frac{y}{x+y}}{\frac{y}{x-y}+\frac{x}{x+y}}=\frac{\frac{x(x+y)-y(x-y)}{(x-y)(x+y)}}{\frac{y(x+y)+x(x-y)}{(x-y)(x+y)}}\)
\(=\frac{x(x+y)-y(x-y)}{y(x+y)+x(x-y)}=\frac{x^2+y^2}{x^2+y^2}=1\)
\(\dfrac{x^3+8}{x^2+2x+1}.\dfrac{x^2+3x+2}{1-x^2}\left(x\ne\pm1\right)\\ =\dfrac{x^3+2^3}{\left(x+1\right)^2}.\dfrac{\left(x^2+x\right)+\left(2x+2\right)}{1^2-x^2}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{x\left(x+1\right)+2\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{\left(x+2\right)\left(x+1\right)}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{\left(x+2\right)^2\left(x^2-2x+4\right)}{\left(1-x\right)\left(x+1\right)^2}\)