(\(\frac{3}{2x+1}+2\))(5x-2)=\(\frac{5x-2}{2x+1}\)
$\frac{4x+3}{5}$ -$\frac{6x-2}{7}$ =$\frac{5x+4}{3}$ +3
b.
$\frac{x+4}{5}$ -x+4=$\frac{x}{3}$ -$\frac{x-2}{2}$
c.$\frac{5x+2}{6}$ -$\frac{8x-1}{3}$ =$\frac{4x+2}{5}$ -5
d.$\frac{2x+3}{3}$ =$\frac{5-4}{2}$
e. $\frac{5x+3}{12}$ =$\frac{1+2x}{9}$
f.$\frac{7x-1}{6}$ =$\frac{16-x}{5}$
g. $\frac{x-3}{5}$ =6-$\frac{1-2x}{3}$
h. $\frac{3x-2}{6}$ -5=$\frac{3-2(x+7)}{4}$
giúp vs ạ, cần gấp
d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
\(\frac{x-3}{3xy}\)+ \(\frac{5x+3}{3xy}\)
\(\frac{5x-7}{2x-3}+\frac{4-3x}{2x-3}\)
\(\frac{3x+5}{7x-1}-\frac{6-4x}{7x-1}\)
\(\frac{11x-7}{3-5x}-\frac{6x+4}{5x-3}\)
\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
\(\frac{1}{2x-10}+\frac{2x}{3x^2-15x}\)
1/ \(\frac{x-3}{3xy}\)+\(\frac{5x+3}{3xy}\)= \(\frac{6x}{3xy}\)=\(\frac{3}{y}\)
2/\(\frac{5x-7}{2x-3}\)+\(\frac{4-3x}{2x-3}\)=\(\frac{2x-3}{2x-3}\)=1
3/\(\frac{11x-7}{3-5x}\)-\(\frac{6x+4}{5x-3}\)=\(\frac{11x-7}{3-5x}\)+\(\frac{6x+4}{3-5x}\)=\(\frac{17x-3}{3-5x}\)
4/\(\frac{3}{2x+6}\)-\(\frac{x-6}{2x^2+6x}\)=\(\frac{3x}{x\left(2x+6\right)}\)-\(\frac{x-6}{x\left(2x+6\right)}\)=\(\frac{2x-6}{x\left(2x+6\right)}\)
5/\(\frac{1}{2x-10}\)+\(\frac{2x}{3x^2-15x}\)=\(\frac{1}{2\left(x-5\right)}\)+\(\frac{2x}{3x\left(x-5\right)}\)=\(\frac{3x}{6x \left(x-5\right)}\)+\(\frac{4x}{6x\left(x-5\right)}\)
=\(\frac{7x}{6x\left(x-5\right)}\)=\(\frac{7}{6\left(x-5\right)}\)
Rút gọn
a)\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
b)\(\left\{\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right\}:\frac{4x}{10x-5}\)
c)\(\frac{x^2+x}{5x^2-10x+5}:\frac{3x+3}{5x-5}\)
\(a,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\) (x khác -3; khác 0)
\(=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x}{2x.\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x-x+6}{2x.\left(x+3\right)}=\frac{2x+6}{x.\left(2x+6\right)}=\frac{1}{x}\)
\(b,\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\) (x khác 0 , khác 1/2 khác -1/2 )
\(=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)
\(=\left(\frac{4x^2+4x+1}{\left(2x-1\right)\left(2x+1\right)}-\frac{4x^2-4x+1}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)
\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{5.\left(2x-1\right)}{4x}=\frac{10}{2x+1}\)
\(c,\frac{x^2+x}{5x^2-10x+5}:\frac{3x+3}{5x-5}\) (x khác 1 ; khác -1)
\(=\frac{x.\left(x+1\right)}{5.\left(x^2-2x+1\right)}.\frac{5x-5}{3x+3}=\frac{x.\left(x+1\right)}{5.\left(x-1\right)^2}.\frac{5\left(x-1\right)}{3.\left(x+1\right)}=\frac{x}{3.\left(x-1\right)}=\frac{x}{3x-3}\)
Thực hiện phép tính
a) \(\left(\frac{2x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{3}\)
b) \(\left(\frac{5x+2}{x^2-10x}+\frac{5x-2}{x^2+10x}\right).\frac{x^2-100}{x^2+4}\)
c) \(\frac{1}{x-1}-\frac{x^3-x}{x^2+1}.\left(\frac{1}{x^2-2x+1}+\frac{1}{1-x^2}\right)\)
Rút gọn
a)\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
b)\(\left\{\hept{\begin{cases}2x+1\\2x-1\end{cases}-\frac{2x-1}{2x+1}}\right\}:\frac{4x}{10x-5}\)
c)\(\frac{x^2+x}{5x^2-10x+5}:\frac{3x+3}{5x-5}\)
1,Thực hiện phép tính
a,\(\frac{3}{2x^2+2x}+\frac{2x-1}{x^2-1}-\frac{2}{x}\)
b,\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
c,\(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}\)
d,\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x+y\right)}\)
\(=\frac{30x\left(x-y\right)-5x\left(x+y\right)}{5\left(x+y\right).10\left(x+y\right)}\)
\(=\frac{5x\left(5x-7y\right)}{50\left(x+y\right)\left(x-y\right)}\)
\(=\frac{x\left(5x-7y\right)}{\left(x+y\right)\left(x-y\right)}\)
chỗ cuối tớ sai
\(=\frac{x\left(5x-7y\right)}{10\left(x+y\right)\left(x-y\right)}\)
đây nha , e xin lỗi
a) \(\frac{3}{2x^2+2x}+\frac{2x-1}{x^2-1}-\frac{2}{x}=\frac{3}{2x\left(x+1\right)}+\frac{2x-1}{\left(x-1\right)\left(x+1\right)}-\frac{2}{x}\)
\(=\frac{3\left(x-1\right)+\left(2x-1\right)-2.2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{3x-2x+4x^2-2x-4x^2+4x-4x+4}{2x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+1}{2x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{1}{2x\left(x-1\right)}\)
b) \(\frac{3x}{5x+5y}-\frac{x}{10x-10y}=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x-y\right)}\)
\(=\frac{3x.10\left(x-y\right)-x.5\left(x+y\right)}{50\left(x-y\right)\left(x+y\right)}\)
\(=\frac{30x\left(x-y\right)+5x\left(x+y\right)}{50\left(x-y\right)\left(x+y\right)}\)
\(=\frac{5x\left[6\left(x-y\right)-\left(x+y\right)\right]}{50\left(x-y\right)\left(x+y\right)}\)
\(=\frac{5x\left(5x-7y\right)}{50\left(x-y\right)\left(x+y\right)}\)
\(=\frac{x\left(5x-7y\right)}{10\left(x-y\right)\left(x+y\right)}\)
c) \(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}=\frac{5x^2-y-x\left(3x-2y\right)}{xy}\)
\(=\frac{5x^2-y-3x^2+2xy}{xy}\)
\(=\frac{2x^2-y+2xy}{xy}\)
d) \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)
\(=\frac{3x-x+6}{2x\left(x+3\right)}\)
\(=\frac{2x+6}{2x\left(x+3\right)}\)
\(=\frac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\frac{2}{2x}=\frac{1}{x}\)
\(\frac{2x-1}{2}-1=\frac{x^2+x-3}{x-1}-\frac{5x-2}{2-2x}\)
\(\frac{2x-1}{2}\)-1=\(\frac{x^2+x-3}{x-1}\)-\(\frac{5x-2}{2-2x}\)
\(\frac{2x-1}{2}\)-1=\(\frac{x^2+x-3}{x-1}\)-\(\frac{5x-2}{2\left(1-x\right)}\)
\(\frac{\left(2x-1\right)\left(x-1\right)}{2\left(x-1\right)}\)-\(\frac{2\left(x-1\right)}{2\left(x-1\right)}\)=\(\frac{2\left(x^2+x-3\right)}{2\left(x-1\right)}\)+\(\frac{5x-2}{2\left(x-1\right)}\)
2x2-x-2x+1-2x+2=2x2+2x-6+5x-2
2x2-x-2x+1-2x+2-2x2-2x+6-5x+2=0
2x2-2x2-x-2x-2x-2x-5x+1+2+6+2=0
11-12x=0
12x=11
x=\(\frac{11}{12}\)
giải phương trình
\(\left(\frac{3}{2x+1}+2\right).\left(5x-2\right)=\frac{5x-2}{2x+1}\)
\(\frac{-2x-1}{2-3x}-\frac{5}{2+3x}=\frac{6x^2}{9x^2-4}\)
\(\frac{x}{x-1}-\frac{2}{x+1}=\frac{x^2+11}{x^2-1}\)
Giai pt : \(\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\left(1\right)\)
Thực hiện các phép đổi tương đương , ta đưa ( 1 ) về dạng :
\(\frac{x+4}{2x^2-5x+2}-\frac{x+4}{2x^2-7x+3}=0\)
\(\Leftrightarrow\left(x+4\right)\left(\frac{1}{2x^2-5x+2}-\frac{1}{2x^2-7x+3}\right)=0\)
\(\Leftrightarrow\frac{\left(x+4\right)\left(1-2x\right)}{\left(2x^2-5x+2\right)\left(2x^2-7x+3\right)}=0\)
\(\Leftrightarrow\left(x+4\right)\left(1-2x\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-4\\x=\frac{1}{2}\end{array}\right.\)
Thữ vào mẫu thức : Với \(x=\frac{1}{2}\) thì \(2x^2-5x+2=0\)
Với \(x=-4\) thì \(\left(2x^2-5x+2\right)\left(2x^2-7x+3\right)\ne0\)
Vậy phương trình ( 1 ) là cho nghiệm duy nhất là \(x=-4\)