a)\(\frac{4x+1}{3x}\)+\(\frac{2x-3}{6x}\)=\(\frac{2\left(4x+1\right)}{2.3x}\)+\(\frac{2x-3}{6x}\)=\(\frac{8x+2}{6x}\)+\(\frac{2x-3}{6x}\)=\(\frac{8x+2+2x-3}{6x}\)=\(\frac{10x-1}{6x}\)
b)\(\frac{x^2-y^2}{6x^2y^2}\):\(\frac{x+y}{3xy}\)=\(\frac{\left(x+y\right)\left(x-y\right)}{6x^2y^2}\) . \(\frac{3xy}{x+y}\)=\(\frac{\left(x+y\right)\left(x-y\right)3xy}{6x^2y^2\left(x+y\right)}\)=\(\frac{x-y}{2xy}\)
5,thực hiện phép tính
1,\(\frac{4y^2}{11x^4}.\left(-\frac{3x^2}{8y}\right)\)
2,\(\frac{4x^2}{5y^2}:\frac{6x}{5y}:\frac{2x}{3y}\)
3,\(\frac{x^2-4}{3x+12}.\frac{x+4}{2x-4}\)
4,\(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}\)
5,\(\frac{x^2-36}{2x+10}.\frac{3}{6-x}\)
6,\(\frac{x^2-9y^2}{x^2y^2}.\frac{3xy}{2x-6y}\)
7,\(\frac{3x^2-3y^2}{5xy}.\frac{15x^2y}{2y-2x}\)
6,Thực hiện phép tính
1,\(\frac{2a^3-2b^3}{3a+3b}.\frac{6a+6b}{a^2-2ab+b^2}\)
2,\(\frac{a^2+ab}{b-a}:\frac{a+b}{2a^2-2b^2}\)
3,\(\frac{x+y}{y-x}:\frac{x^2+xy}{3x^2-3y^2}\)
4,\(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}\)
5,\(\frac{5x-15}{4x+4}:\frac{x^2-9}{x^2+2x+1}\)
6,\(\frac{6x+48}{7x-7}:\frac{x^2-64}{x^2-2x+1}\)
thực hiện các phép tính sau
a) \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}\)
b) \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
c) \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}\)
d) \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}\)
5,Thực hiện phép tính
1,\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
2,\(\frac{1}{1-x}+\frac{2x}{x^2-1}\)
3,\(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}\)
4,\(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}\)
5,\(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}\)
6,\(\frac{12x}{5y^3}.\frac{15y^4}{8x^3}\)
Câu 3: Giải các phương trình sau bằng cách đưa về dạng ax+b=0
1. a, \(\frac{5x-2}{3}=\frac{5-3x}{2}\); b, \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
c, \(2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\); d, \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
e, \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\); f, 4 (0,5-1,5x)=\(\frac{5x-6}{3}\)
g, \(\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}+2x\); h, \(\frac{x+4}{5}.x+4=\frac{x}{3}-\frac{x-2}{2}\)
i, \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\); k, \(\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x+2}{5}-5\)
m, \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\); n, \(\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+1\right).\frac{1}{3}\left(x+2\right)\)
p, \(\frac{x}{3}-\frac{2x+1}{6}=\frac{x}{6}-x\); q, \(\frac{2+x}{5}-0,5x=\frac{1-2x}{4}+0,25\)
r, \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\); s, \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{6}\)
t, \(\frac{2x-8}{6}.\frac{3x+1}{4}=\frac{9x-2}{8}+\frac{3x-1}{12}\); u, \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{3}+\frac{2x-1}{12}\)
v, \(\frac{5x-1}{10}+\frac{2x+3}{6}=\frac{x-8}{15}-\frac{x}{30}\); w, \(\frac{2x-\frac{4-3x}{5}}{15}=\frac{7x\frac{x-3}{2}}{5}-x+1\)
Rút gọn:
a) \(\frac{x^3+2x^2+1}{4x^2-4}.\frac{x+2}{x^2+1}.\frac{2x^2-2}{x^3+2x^2+1}\)
b)\(\frac{x^4-y^4}{x^2+y^2-2xy}.\frac{x-y}{xy+x^2}\)
c)\(\frac{x^2-9}{x+5}.\frac{2x}{x+3}+\frac{x^2-9}{x+5}.\frac{5-x}{x+3}\)
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,\(\dfrac{7}{x};\dfrac{x}{x+6};\dfrac{36}{x^2+6x}\)
Bài 1: Cho \(A=\frac{1}{x-2}+\frac{x^2-x-2}{x^2-7x+10}-\frac{2x-4}{x-5}\)
a, Rút gọn A b,Tìm x nguyên để A nguyên
Bài 2: Cho \(M=\left[\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right]:\left(x-2+\frac{10-x^2}{x+2}\right)\)
a, Tìm điều kiện xác định của M b, Rút gọn M c, Tính giá trị của M khi \(\left|x\right|=\frac{1}{2}\)
Bài 3: Cho biểu thức \(N=\left(\frac{1}{y-1}-\frac{y}{1-y^3}\cdot\frac{y^2+y+1}{y+1}\right):\frac{1}{y^2-1}\)
a, Rút gọn N b,Tính giá trị của N khi \(y=\frac{1}{2}\) c,Tìm giá trị của y để N luôn có giá trị dương
Bài 1: Giải các phương trình sau
a, \(\frac{4x-8}{2x^2+1}\)= 0
b, \(\frac{x^2-x-6}{x-3}\)= 0
c,\(\frac{x+5}{3x-6}\) - \(\frac{1}{2}\)= \(\frac{2x-3}{2x-4}\)
d,\(\frac{12}{1-9x^2}\)= \(\frac{1-3x}{1+3x}\) - \(\frac{1+3x}{1-3x}\)
e, \(\frac{x+5}{x-1}\)= \(\frac{x+1}{x-3}\) - \(\frac{8}{x^2-4x+3}\)
f, \(\frac{x+1}{x-2}\) - \(\frac{5}{x+2}\)= \(\frac{12}{x^2-4}\) + 1