Ôn tập: Phân thức đại số
Các câu hỏi tương tự
chứng minh rằng
x2+5x+3\(\ge\)\(-\frac{1}{4}\)
rút gọn biểu thức
1) \(\dfrac{a+b}{3a-b}+\dfrac{b}{a+b}-\dfrac{a^2-b^2}{3a-b}\)
2) \(\left(\dfrac{7}{a+b}+\dfrac{a^2+49}{a^2-49}-\dfrac{7}{a-7}\right)\div\dfrac{a+1}{2}\)
3) \(\left(x^2+\dfrac{4x^2}{x^2-4}\right)\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x^3-4x}\times\dfrac{x^2-4}{x-2}\right)\)
quy đồng các phân thức sau
a,\(\dfrac{x+1}{x-1};\dfrac{x-1}{x+1};\dfrac{4}{1-x^2}\)
b,\(\dfrac{x^3}{x^3-3x^2y+3xy^2-y^3};\dfrac{x}{y^2xy}\)
c,\(\dfrac{4x}{x-2};\dfrac{3x}{x-2};\dfrac{12x}{x^2-4}\)
d,\(\dfrac{7}{x};\dfrac{x}{x+6};\dfrac{36}{x^2+6x}\)
Chứng minh với x thuộc Z, x khác -1 thì \(\dfrac{x^4+3x^3+3x^2+x}{2x^2+4x+2}\) nguyên.
Cho 4x + y = 1. Chứng minh rằng 4x2 + y2 ≥ \(\frac{1}{5}\)
Cho x khác 0 .CM : \(\dfrac{x^2-2x+2004}{x^2}\ge\dfrac{2003}{2004}\)
Chứng minh đẳng thức :
a) left(dfrac{x^2-2x}{2x^2+8}-dfrac{2x^2}{8-4x+2x^2-x^3}right)left(1-dfrac{1}{x}-dfrac{2}{x^2}right)dfrac{x+1}{2x}
b) left[dfrac{2}{3x}-dfrac{2}{x+1}left(dfrac{x+1}{3}-x-1right)right]:dfrac{x-1}{x}dfrac{2x}{x-1}
c) left[dfrac{2}{left(x+1right)^3}.left(dfrac{1}{x}+1right)+dfrac{1}{x^2+2x+1}left(dfrac{1}{x^2}+1right)right]:dfrac{x-1}{x^3}dfrac{x}{x-1}
Đọc tiếp
Chứng minh đẳng thức :
a) \(\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)=\dfrac{x+1}{2x}\)
b) \(\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\left(\dfrac{x+1}{3}-x-1\right)\right]:\dfrac{x-1}{x}=\dfrac{2x}{x-1}\)
c) \(\left[\dfrac{2}{\left(x+1\right)^3}.\left(\dfrac{1}{x}+1\right)+\dfrac{1}{x^2+2x+1}\left(\dfrac{1}{x^2}+1\right)\right]:\dfrac{x-1}{x^3}=\dfrac{x}{x-1}\)
a) \(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{36}{x^2-9}\)
b) \(\dfrac{2x-1}{x+4}-\dfrac{1-3x}{x-4}=5+\dfrac{96}{x^2-16}\)
c) \(\dfrac{x+3}{x+1}-\dfrac{x-1}{x}=\dfrac{3x^2+4x+1}{x\left(x+1\right)}\)
cho a,b,c >0 . Chứng minh
\(\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ac+a^2}\ge\dfrac{\left(a+b+c\right)}{3}\)