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A=\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)....\left[1+\frac{1}{n+\left(n+2\right)}\right]< 2\)
Chứng minh rằng với mọi số tự nhiên n\(\ge\)1
Thực hiện phép tính:
1,frac{1-2x}{2x}+frac{2x}{2x-1}+frac{1}{2x-4x^2}
2,frac{x^2+2}{x^3-1}+frac{2}{x^2+x+1}+frac{1}{1-x}
3,frac{x}{x-2y}+frac{x}{x+2y}+frac{4xy}{4y^2-x^2}
4,frac{2x}{x^2+2xy}+frac{y}{xy-2y^2}+frac{4}{x^2-4y^2}
5,left(frac{9}{x^3-9x}+frac{1}{x+3}right):left(frac{x-3}{x^2+3x}-frac{x}{3x+9}right)
Đọc tiếp
Thực hiện phép tính:
1,\(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)
2,\(\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}\)
3,\(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)
4,\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
5,\(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)
1) \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
2)\(\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3x-6}{4-9x^2}\)
3) \(\frac{5}{2x-4}+\frac{7}{x+2}+\frac{-1}{x^2-4}\)
4) \(\frac{x+3}{x^2+x-2}+\frac{4-x}{x^2+5x+6}\)
Giúp mình với
\(a,\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(b,\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
\(c,\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\)
thực hiện phép tính
a) (x3+8y3):(2y+x) b.frac{a-1}{2left(a-4right)}+frac{a}{a-4} c. (x3+3x2y+3xy2+y3):(2x+2y)
d. (x-5)2+(7-x)(x+2) e.frac{3x}{x-2}-frac{2x+1}{2-x} f. left(frac{x+2}{x+1}-frac{2x}{x-1}right)cdotfrac{3x+3}{x}+frac{4x^2+x+7}{x^2-x}
g.left(frac{1}{x+1}-frac{3}{x^{3^{ }}+1}+frac{3}{x^2-x+1}right)cdotleft(frac{3x^2-3x+3}{left(x+1right)left(x+2right)}right) h.frac{1}{3x-2}-frac{1}{3x+2}-frac{3x+6}{4-9x^2}
Nguyễn Nam giúp giùm
Đọc tiếp
thực hiện phép tính
a) (x3+8y3):(2y+x) b.\(\frac{a-1}{2\left(a-4\right)}+\frac{a}{a-4}\) c. (x3+3x2y+3xy2+y3):(2x+2y)
d. (x-5)2+(7-x)(x+2) e.\(\frac{3x}{x-2}-\frac{2x+1}{2-x}\) f. \(\left(\frac{x+2}{x+1}-\frac{2x}{x-1}\right)\cdot\frac{3x+3}{x}+\frac{4x^2+x+7}{x^2-x}\)
g.\(\left(\frac{1}{x+1}-\frac{3}{x^{3^{ }}+1}+\frac{3}{x^2-x+1}\right)\cdot\left(\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}\right)\) h.\(\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3x+6}{4-9x^2}\)
Nguyễn Nam giúp giùm
Bài 1 : Tính
a, \(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{x^2-1}\)
b, \(\frac{x^2}{x+1}+\frac{2x}{x^2-1}-\frac{1}{4-x}+1\)
c, \(\left(\frac{x^2-16}{x^2+8x+16}+\frac{6}{x+4}\right).\frac{2x}{x+2}\)
Bài 2 : Tính
\(\left(x^2-y^2-z^2-2xyz\right):\frac{x+y+z}{x+y-z}\)
\(a,\frac{x-3}{2}+\frac{4x+1}{3}=\frac{2x-7}{6}\)
\(b,\frac{x-4}{5}+\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+5}{6}\)
\(c,\frac{2\left(x-3\right)}{4}+\frac{x-5}{3}=\frac{13x+4}{12}\)
Bài 2: Giải phương trình sau:
\(a,\frac{x+1}{3}+\frac{3\left(2x+1\right)}{4}=\frac{2x+3\left(x+1\right)}{6}+\frac{7+12x}{12}\)
\(b,\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)
CMR nếu \(\frac{1}{x}-\frac{1}{y}-\frac{1}{z}=1\) và x = y + z thì \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=1\)