Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)
= \(x^2\) - 16 - 5\(x\) - 5 + \(x^2\) + \(x\)
= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)
= 2\(x^2\) - 4\(x\) - 21
b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)
= 3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7
= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)
= - 3\(x^2\) + 3\(xy\) - 3
a)\(\frac{x^2+y^2-1+2xy}{x^2-y^2+1+2x}\)
\(\Leftrightarrow\frac{\left(x+y\right)^2-1}{\left(x+1\right)^2-y^2}\)
\(\Leftrightarrow\frac{\left(x+y+1\right)\left(x+y-1\right)}{\left(x+1-y\right)\left(x+1+y\right)}\)
\(\Leftrightarrow\frac{x+y-1}{x-y+1}\)
b)\(\frac{3x^3-6x^2y+xy^2-2y^3}{9x^5-18x^4y-xy^4+2y^5}\)
\(\Leftrightarrow\frac{3x^2\left(x-2y\right)+y^2\left(x-2y\right)}{9x^4\left(x-2y\right)-y^4\left(x-2y\right)}\)
\(\Leftrightarrow\frac{\left(3x^2+y^2\right)\left(x-2y\right)}{\left(9x^4-y^4\right)\left(x-2y\right)}\)
\(\Leftrightarrow\frac{3x^2+y^2}{\left(3x^2-y^2\right)\left(3x^2+y^2\right)}\)
\(\Leftrightarrow\frac{1}{3x^2-y^2}\)
f: \(=\dfrac{5x-3-x+3}{4x^2y}=\dfrac{4x}{4x^2y}=\dfrac{1}{xy}\)
g: \(=\dfrac{3x+10-x-4}{x+3}=\dfrac{2x+6}{x+3}=2\)
h: \(=\dfrac{4-2+x}{x-1}=\dfrac{x+2}{x-1}\)
n: \(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{x\left(x+3\right)}=\dfrac{2}{x}\)
p: \(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)
k: \(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-6}{x^2-4}\)
m: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
\(\dfrac{2xy-x^2}{3x^3-6x^2y}\\ =\dfrac{x\left(2y-x\right)}{3x^2\left(x-2y\right)}\\ =\dfrac{x\left(2y-x\right)}{-3x^2\left(2y-x\right)}\\ =\dfrac{1}{-3x}\)
\(\dfrac{2xy-x^2}{3x^3-6x^2y}=\dfrac{-\left(x^2-2xy\right)}{3x^3-6x^2y}\)
\(=\dfrac{-x\left(x-2y\right)}{3x^2\left(x-2y\right)}=\dfrac{-1}{3x}\)
a: =18x^3y^2-12x^3y^3+6x^2y^2
b: (-3x+2)(5x^2-1/3x+4)
=-12x^3+x^2-12x+10x^2-2/3x+8
=-12x^3+11x^2-38/3x+8
c: =x^2-x-2+3x-x^2
=2x-2
d: =4x^2+12x+9-4x^2+25-(x-1)(x^2+12)
=12x+34-x^3-12x+x^2+12
=-x^3+x^2+46
a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)
a: \(\left(3x+2\right)\left(9x^2-6x+4\right)\)
\(=27x^3+8\)
b: \(\left(x-2y\right)^3-\left(x^2-2xy+y^2\right)\)
\(=x^3-6x^2y+12xy^2-8y^3-x^2+2xy-y^2\)