quy đồng mẫu thức các phân thức sau:
a. \(\dfrac{1}{x+2}\), \(\dfrac{8}{2x-x^2}\)
b. \(x^2+1,\dfrac{x^4}{x^2-1}\)
làm chi tiết nha mn tks
QUy đồng
\(\dfrac{3}{x^3-x^2y+xy^2-y^3};\dfrac{1-x}{2xy^2-2x^2y}\)
1.giải pt
a)\(\dfrac{16-x}{4}=\dfrac{2x+1}{3}\)
b)(2x+3)(1-3x)=9x\(^2\)-1
c)\(\dfrac{2x}{x+1}+\dfrac{x-1}{x}=\dfrac{2x^2+3x-1}{x^2+x}\)
Giải các phương trình sau :
a,\(\dfrac{2}{2x+1}-\dfrac{3}{2x-1}=\dfrac{4}{4x^2-1}\)
b,\(\dfrac{2x}{x+1}+\dfrac{18}{x^2+2x-3}=\dfrac{2x-5}{x+3}\)
c,\(\dfrac{1}{x-1}+\dfrac{2x^2-5}{x^3-1}=\dfrac{4}{x^2+x+1}\)
4.Giải phương trình
a) \(\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{20}{x^2-25}\)
b)\(\dfrac{1}{x-1}+\dfrac{2}{x+1}=\dfrac{x}{x^2-1}\)
c)\(5+\dfrac{76}{x^2-16}=\dfrac{2x-1}{x+4}-\dfrac{3x-1}{4-x}\)
d)\(\dfrac{90}{x}-\dfrac{36}{x-6}=2\)
e)\(\dfrac{1}{x}+\dfrac{1}{x+10}=\dfrac{1}{12}\)
f)\(\dfrac{x+3}{x-3}-\dfrac{1}{x}=\dfrac{3}{x\left(x-3\right)}\)
g)\(\dfrac{3}{x+2}-\dfrac{2}{x-2}+\dfrac{8}{x^2-4}=0\)
h)\(\dfrac{3}{x+2}-\dfrac{2}{x-3}=\dfrac{8}{\left(x-3\right)\left(x+2\right)}\)
i)\(\dfrac{x}{2x+6}-\dfrac{x}{2x+2}=\dfrac{3x+2}{\left(x+1\right)\left(x+3\right)}\)
k)\(\dfrac{x}{x+1}-\dfrac{2x-3}{1-x}=\dfrac{3x^2+5}{x^2-1}\)
l)\(\dfrac{5}{x+7}+\dfrac{8}{2x+14}=\dfrac{3}{2}\)
m)\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)
Cần gấp ạ
Giải các phương trình sau:
a) (2x-1)2-(3x+5)(2x-1)=0
b)\(\dfrac{x+5}{4}-\dfrac{2x-3}{3}=\dfrac{2x-1}{12}\)
c)\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{-4}{1-x^2}\)
d)\(\dfrac{1}{x+1}+\dfrac{2x-1}{x^3+1}=\dfrac{2}{x^2-x+1}\)
Tính:
\(a,\dfrac{x+3}{2x-1}-\dfrac{x^2-5}{4x^2-4x+1}-\dfrac{2x^3+5x^2-x-1}{8x^3-12x^2+6x-1}\)
\(b,\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
a. x(x-1)(x+1)(x+2)=24
b.\(\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}+\dfrac{1}{x^2-11x+30}=\dfrac{1}{8}\)
c.\(\dfrac{x-29}{30}+\dfrac{x-30}{29}=\dfrac{29}{x-30}+\dfrac{30}{x-29}\)
thực hiên phép tính
a.\(\dfrac{x^2+y^2}{4\left(x+y\right)}+\dfrac{2xy}{4\left(x+y\right)}\)
b.\(\dfrac{x+5}{2x-2}-\dfrac{4}{x^2-1}:\dfrac{2}{x+1}\)