\(\frac{4x-2x^2}{x^3-4x}\) =\(\frac{1}{2}\)
Thực hiện phép tính:
a) \(\frac{1}{x}-\frac{1}{x+1}\)
b)\(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}\)
c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}\)
d) \(\frac{2x-1}{x}-\frac{2x+5}{3x-4x^2}+\frac{2x^2+x+3}{3x-4x^2}\)
e) \(\frac{x}{2x+1}+\frac{1}{4x^2-1}-\frac{x-2}{2x-1}\)
giúp mình với mình cần gấp ! Help me ===)
\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x^2+x}\)
b, \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{y^2-xy-xy+x^2}{\left(xy-x^2\right)\left(y^2-xy\right)}=\frac{x^2+y^2}{xy^3-xyxy-xyxy+x^3y}\)Tu rut gon tiep
c, tt
d, cx r
a) \(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}\)
\(=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
b) \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)
\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)
c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)
\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)
\(a,\frac{1}{x}-\frac{1}{x+1}\)
\(=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}=\frac{x+1-x}{x\left(x+1\right)}\)
\(=\frac{1}{x\left(x+1\right)}\)
\(b,\frac{1}{xy-x^2}-\frac{1}{y^2-xy}\)
\(=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)
\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{x-y}{xy\left(x-y\right)}=\frac{1}{xy}\)
\(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)
\(\frac{x+4}{x^2-3x+2}-\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
giai phuong trinh
c) \(\frac{4x}{x^2+4x+3}-1=6\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\)
d) \(\frac{3}{2x+1}=\frac{6}{2x+3}+\frac{8}{4x^2+8x+3}\)
\(\frac{4x}{x^2+4x+3}-1=6\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\) \(ĐK:x\ne-1;x\ne-3\)
\(\Leftrightarrow\frac{4x}{x^2+4x+3}-\frac{x^2+4x+3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)}{2\left(x+3\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)\left(x+3\right)}\right]\)
\(\Leftrightarrow\frac{4x-x^2-4x-3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)-x-3}{2\left(x+3\right)\left(x+1\right)}\right]\)
\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=6\left[\frac{2x+2-x-3}{2\left(x^2+4x+3\right)}\right]\)
\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=\frac{6\left(x-1\right)}{2\left(x^2+4x+3\right)}\)
\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=\frac{3\left(x-1\right)}{x^2+4x+3}\)
\(\Leftrightarrow-x^2-3=3x-3\)
\(\Leftrightarrow-x^2-3x=0\)
\(\Leftrightarrow-x\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\left(loại\right)\end{cases}}\)
Vậy x = 0
\(ĐK:x\ne\frac{-1}{2};x\ne\frac{-3}{2}\)
\(\frac{3}{2x+1}=\frac{6}{2x+3}+\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow\frac{3}{2x+1}-\frac{6}{2x+3}=\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow\frac{3\left(2x+3\right)-6\left(2x+1\right)}{\left(2x+1\right)\left(2x+3\right)}=\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow\frac{6x+9-12x-6}{4x^2+8x+3}=\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow-6x+3=8\)
\(\Leftrightarrow x=-\frac{5}{6}\)
Vậy ...
Rút gọn : \(\left(\frac{1+x}{x}+\frac{1}{4x^2}\right)\left(\frac{1-2x}{1+2x}-\frac{1}{1-4x^2}\times\frac{1-4x+4x^2}{1+2x}\right)-\frac{1}{2x}\)
\(=\dfrac{4x\left(x+1\right)+1}{4x^2}\cdot\left(\dfrac{-\left(2x-1\right)}{2x+1}+\dfrac{1}{\left(2x+1\right)\left(2x-1\right)}\cdot\dfrac{\left(2x-1\right)^2}{2x+1}\right)-\dfrac{1}{2x}\)
\(=\dfrac{\left(2x+1\right)^2}{4x^2}\cdot\left(\dfrac{-\left(2x-1\right)}{2x+1}+\dfrac{2x-1}{\left(2x+1\right)^2}\right)-\dfrac{1}{2x}\)
\(=\dfrac{\left(2x+1\right)^2}{4x^2}\cdot\dfrac{-\left(2x-1\right)\left(2x+1\right)+2x-1}{\left(2x+1\right)^2}-\dfrac{1}{2x}\)
\(=\dfrac{-4x^2+1+2x-1}{4x^2}-\dfrac{1}{2x}\)
\(=\dfrac{-4x^2+2x}{4x^2}-\dfrac{1}{2x}\)
\(=\dfrac{-2x\left(2x-1\right)}{2x\cdot2x}-\dfrac{1}{2x}\)
\(=\dfrac{-2x+1-1}{2x}=\dfrac{-2x}{2x}=-1\)
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,\(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}\)
1,\(\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x}{x\left(2x+6\right)}+\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x+x-6}{x\left(2x+6\right)}\)=\(\frac{4x-6}{x\left(2x+6\right)}=\frac{2\left(2x-3\right)}{x\left(2x+6\right)}\)
2, \(\frac{1}{1-x}-\frac{2x}{1-x^2}\)=\(\frac{1+x}{\left(1-x\right)\left(1+x\right)}+\frac{2x}{\left(1-x\right)\left(1+x\right)}\)=\(\frac{1+x+2x}{\left(1-x\right)\left(1+x\right)}=\frac{3x+1}{\left(1-x\right)\left(1+x\right)}\)
3,1/x(y-x)-1/y(y-x)=y/xy(y-x)-x/xy(y-x)=(y-x)/xy(x-y)=1/xy
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) \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}\)
a) Ta có: \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}\)
\(=\frac{x+1}{2\left(x+3\right)}+\frac{2x+3}{x\left(x+3\right)}\)
\(=\frac{x\left(x+1\right)}{2x\left(x+3\right)}+\frac{2\cdot\left(2x+3\right)}{2x\left(x+3\right)}\)
\(=\frac{x^2+x+4x+6}{2x\left(x+3\right)}\)
\(=\frac{x^2+5x+6}{2x\left(x+3\right)}\)
\(=\frac{x^2+2x+3x+6}{2x\left(x+3\right)}\)
\(=\frac{x\left(x+2\right)+3\left(x+2\right)}{2x\left(x+3\right)}\)
\(=\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}=\frac{x+2}{2x}\)
b) Ta có: \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
\(=\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}\)
\(=\frac{3x}{x\left(2x+6\right)}-\frac{x-6}{x\left(2x+6\right)}\)
\(=\frac{3x-x+6}{x\left(2x+6\right)}=\frac{2x+6}{x\left(2x+6\right)}=\frac{1}{x}\)
c) Ta có: \(\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}\)
\(=\frac{5\left(x+2\right)\cdot2\cdot\left(2-x\right)}{4\cdot\left(x-2\right)\cdot\left(x+2\right)}\)
\(=\frac{5\cdot2\cdot\left(2-x\right)}{-4\left(2-x\right)}=\frac{5\cdot2}{-4}=\frac{-5}{2}\)
d) Ta có: \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}\)
\(=\frac{\left(1-2x\right)\left(1+2x\right)\cdot3x}{x\left(x+4\right)\cdot2\left(2-x\right)}\)
\(=\frac{\left(1-2x\right)\left(1+2x\right)\cdot3}{2\left(x+4\right)\cdot\left(2-x\right)}=\frac{3\left(1-4x^2\right)}{2\left(-x^2-2x+8\right)}\)
\(=\frac{3-12x^2}{-2x^2-4x+16}\)
a) \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}\)
\(=\frac{x+1}{2\left(x+3\right)}+\frac{2x+3}{x\left(x+3\right)}\) \(\left(ĐKXĐ:x\ne-3;x\ne0\right)\)
\(=\frac{x^2+x}{2x\left(x+3\right)}+\frac{4x+6}{2x\left(x+3\right)}\)
\(=\frac{x^2+2x+3x+6}{2x\left(x+3\right)}=\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}=\frac{x+2}{2x}\)
b) \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\) \(\left(ĐKXĐ:x\ne0;x\ne-3\right)\)
\(=\frac{3x}{2x\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}=\frac{2\left(x+3\right)}{2x\left(x+3\right)}=\frac{1}{x}\)
c) \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}\) \(\left(ĐKXĐ:x\ne\pm2\right)\)
\(=\frac{-5\left(x-2\right)}{2\left(x-2\right)}=\frac{-5}{2}\)
Giải các phương trình sau:
a. \(\frac{4}{2x+3}-\frac{7}{3x-5}=0\)
b. \(\frac{4}{2x-3}+\frac{4x}{4x^2-9}=\frac{1}{2x+3}\)
c. \(\frac{2}{2x+1}+\frac{x}{4x^2-1}=\frac{7}{2x-1}\)
d. \(\frac{x^2+5}{25-x^2}=\frac{3}{x+5}+\frac{x}{x-5}\)
\(\frac{4}{2x+3}-\frac{7}{3x-5}=0\left(đkxđ:x\ne-\frac{3}{2};\frac{5}{3}\right)\)
\(< =>\frac{4\left(3x-5\right)}{\left(2x+3\right)\left(3x-5\right)}-\frac{7\left(2x+3\right)}{\left(2x+3\right)\left(3x-5\right)}=0\)
\(< =>12x-20-14x-21=0\)
\(< =>2x+41=0< =>x=-\frac{41}{2}\left(tm\right)\)
\(\frac{4}{2x-3}+\frac{4x}{4x^2-9}=\frac{1}{2x+3}\left(đk:x\ne-\frac{3}{2};\frac{3}{2}\right)\)
\(< =>\frac{4\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}+\frac{4x}{\left(2x-3\right)\left(2x+3\right)}-\frac{2x-3}{\left(2x+3\right)\left(2x-3\right)}=0\)
\(< =>8x+12+4x-2x+3=0\)
\(< =>10x=15< =>x=\frac{15}{10}=\frac{3}{2}\left(ktm\right)\)
\(\frac{2}{2x+1}+\frac{x}{4x^2-1}=\frac{7}{2x-1}\left(đkxđ:x\ne-\frac{1}{2};\frac{1}{2}\right)\)
\(< =>\frac{2\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\frac{x}{\left(2x+1\right)\left(2x-1\right)}=\frac{7\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\)
\(< =>4x-2+x=14x+7\)
\(< =>14x-5x=-2-7\)
\(< =>9x=-9< =>x=-\frac{9}{9}=-1\left(tm\right)\)
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,\(\frac{12x}{5y^3}.\frac{15y^4}{8x^3}\)
\(1,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2x+6}-\frac{x-6}{x\left(2x-6\right)}=\frac{3x-x+6}{x\left(2x-6\right)}=\frac{2x+6}{x\left(2x-6\right)}\)
\(2,\frac{1}{1-x}+\frac{2x}{x^2-1}=\frac{-1\left(x+1\right)+2x}{x^2-1}=\frac{x-1}{x^2-1}=\frac{1}{x+1}\)
\(3,\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)
\(4,\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}=\frac{-5}{2}\)
\(5,\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}=\frac{3\left(1+2x\right)}{2x\left(x+4\right)}\)
\(6,\frac{12x}{5y^3}.\frac{15y^4}{8x^3}=\frac{9y}{2x^2}\)
Giải các phương trình sau:
a) \(\frac{4}{x-1}-\frac{5}{x-2}=-3\)
b) \(3x-\frac{1}{x-2}=\frac{x-1}{2-x}\)
c) \(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
d) \(\frac{2}{x^2-4}-\frac{1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\)
e) \(\frac{4x}{x^2+4x+3}-1=6\left(\frac{1}{x+3}-\frac{1}{2x-2}\right)\)
f) \(\frac{3}{4x\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{7}{6x+30}\)
g) \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
$\frac{4x+3}{5}$ -$\frac{6x-2}{7}$ =$\frac{5x+4}{3}$ +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