bài 1 quy đồng mẫu thức các phân thức ( có thể ddoooir dấu để tìm MTC cho tiên)
a) \(\frac{x-1}{2x+2}\),\(\frac{x+1}{2x-2}\),\(\frac{1}{1-x^2}\)
b)\(\frac{2x+1}{x+a}\),\(\frac{a-x}{-x^2+ax-a^2}\),\(\frac{2x^2-1}{x^3+a^3}\)
c) \(\frac{24}{4x^3-x}\),\(\frac{4x}{x-2x^2}\),\(\frac{18}{2x^2+x}\)
d) \(\frac{x+1}{2x^2-x^4}\),\(\frac{x}{x^4+2x^2+4}\),\(\frac{2x-1}{x^7-8x}\)
a) MTC = 2(x - 1)(x + 1)
\(\frac{x-1}{2x+2}=\frac{\left(x-1\right)\left(x-1\right)}{2\left(x+1\right)\left(x-1\right)}=\frac{\left(x-1\right)^2}{2\left(x+1\right)\left(x-1\right)}\)
\(\frac{x+1}{2x-2}=\frac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}=\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}\)
\(\frac{1}{1-x^2}=-\frac{1}{x^2-1}=-\frac{1.2}{\left(x-1\right)\left(x+1\right).2}=-\frac{2}{2\left(x-1\right)\left(x+1\right)}\)
b) MTC = x3 + a3
\(\frac{2x+1}{x+a}=\frac{\left(2x+1\right)\left(x^2-ax+a^2\right)}{x^3+a^3}\)
\(\frac{a-x}{-x^2+ax-a^2}=\frac{x-a}{x^2-ax+a^2}=\frac{\left(x-a\right)\left(x+a\right)}{x^3+a^3}=\frac{x^2-a^2}{x^3-a^3}\)
\(\frac{2x^2-1}{x^3+a^3}\)
c) MTC = x(2x - 1)(2x + 1)
\(\frac{24}{4x^3-x}=\frac{24}{x\left(2x-1\right)\left(2x+1\right)}\)
\(\frac{4x}{x-2x^2}=\frac{-4x}{2x^2-x}=\frac{-4x}{x\left(2x-1\right)}=\frac{-4x.\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-8x^2-4x}{x\left(2x-1\right)\left(2x+1\right)}\)
\(\frac{18}{2x^2+x}=\frac{18}{x\left(2x+1\right)}=\frac{18\left(2x-1\right)}{x\left(2x-1\right)\left(2x+1\right)}=\frac{36x-18}{x\left(2x-1\right)\left(2x+1\right)}\)
d) MTC = x2(x2 - 2)(x4 + 2x2 + 4)
\(\frac{x+1}{2x^2-x^4}=\frac{-x-1}{x^4-2x^2}=\frac{-x-1}{x^2\left(x^2-2\right)}=\frac{\left(-x-1\right)\left(x^4+2x^2+4\right)}{x^2\left(x^2-2\right)\left(x^4+2x^2+4\right)}\)
\(\frac{x}{x^4+2x^2+4}=\frac{x.x^2\left(x^2-2\right)}{x^2\left(x^2-2\right)\left(x^4+2x^2+4\right)}=\frac{x^3\left(x^2-2\right)}{x^2\left(x^2-2\right)\left(x^4+2x^2+4\right)}\)
\(\frac{2x-1}{x^7-8x}=\frac{2x-1}{x\left(x^6-8\right)}=\frac{2x-1}{x\left(x^2-2\right)\left(x^4+2x^2+4\right)}\)
\(=\frac{\left(2x-1\right).x}{x^2\left(x^2-2\right)\left(x^4+2x^2+4\right)}=\frac{x\left(2x-1\right)}{x^2\left(x^2-2\right)\left(x^4+2x^2+4\right)}\)
