\(\dfrac{2x^2-x}{x-1}+\dfrac{x+1}{1-x}+\dfrac{2-x^2}{x-1}.\left(x\ne1\right).\)
\(\dfrac{2x^2-x-x-1+2-x^2}{x-1}=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1.\)
Thực hiện phép tính
a) (x\(^2\)-3x+4)(-3x+1)
b)\(\dfrac{1}{x}\)+\(\dfrac{1}{x+5}\)+\(\dfrac{x-5}{x^2+5x}\)
Mọi người giúp mình nhanh nha mình cần gấp
Thực hiện phép tính:
\(\dfrac{4}{x+2}-\dfrac{3}{2-x}-\dfrac{5x+2}{x^2-4}\)
Tính
a, \(\dfrac{2x}{y+x}\) + \(\dfrac{2y}{x+y}\)
b, \(\dfrac{x}{x+1}\) + \(\dfrac{3x+1}{x^2-1}\)
1/ \(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)
2/ \(\dfrac{5x-2}{5}-2=\dfrac{1-2x}{3}\)
3/ \(\dfrac{x-2}{2}-\dfrac{2}{3}=x-1\)
4/ \(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)
5/ \(\dfrac{x-3}{9}-\dfrac{x+2}{6}=\dfrac{x+4}{18}-\dfrac{1}{2}\)
Tính:
a)\(\dfrac{2x+4}{x^3-1}\)-\(\dfrac{2}{x+1}\)+\(\dfrac{x+2}{x^2+x+1}\)
b) \(\dfrac{x-1}{x^2-5x+6}\)-\(\dfrac{x-3}{x-2}\)+\(\dfrac{x-2}{x-3}\)
Đưa các phân thức sau về cùng mẫu
a) \(\dfrac{x}{2x^2+7x-15}\); \(\dfrac{x+2}{x^2+3x-10}\); \(\dfrac{1}{x+5}\)
b) \(\dfrac{1}{-x^2+3x-2}\); \(\dfrac{1}{x^2+5x-6}\); \(\dfrac{1}{-x^2+4x-3}\)
c)\(\dfrac{3}{x^3-1}\); \(\dfrac{2x}{x^2+x+1}\); \(\dfrac{x}{x-1}\)
d)\(\dfrac{x}{x^2-2xy+y^2-x^2}\); \(\dfrac{y}{x^2+2yz-y^2-z^2}\); \(\dfrac{z}{x^2-2xz-y^2+z^2}\)
1/ \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
2/ \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
3/ \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
4/ \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
5/ \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
a) (X-2)(x+3)-3(4x-2)=(x-4)\(^{^{ }2}\)
b) \(\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\)
c) \(x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)
d) \(\left(2x+5\right)^2=\left(x+2\right)^2\)
e) \(x^2-5+6=0\)
g) \(2x^3+6x^2=x^2+3x\)
h) \(\left(x+\dfrac{1}{2}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\)
mọi người giúp e với ạ
Giải các phương trình sau
d) \(\dfrac{1}{x-2}\)-\(\dfrac{6}{x+3}\)=\(\dfrac{5}{6-x^2-x}\)
e) \(\dfrac{2}{x+2}\)-\(\dfrac{2x^2+16}{x^3+8}\)=\(\dfrac{5}{x^2-2x+4}\)
f) \(\dfrac{x+1}{x^2+x+1}\)-\(\dfrac{x-1}{x^2-x+1}\)=\(\dfrac{2\left(x+2\right)^2}{x^6-1}\)
