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`a,ĐKXĐ:x-4 ne 0,2x+2 ne 0`
`<=>x ne 4,x me -1`
`b,ĐKXĐ:4x^2-25 ne 0`
`<=>(2x-5)(2x+5) ne 0`
`<=>x ne +-5/2`
`c,ĐKXĐ:8x^3+27 ne 0`
`<=>8x^3 ne -27`
`<=>2x ne -3`
`<=>x ne -3/2`
`d,2x+2 ne 0,4y^2-9 ne 0`
`<=>2x ne -2,(2y-3)(2y+3) ne 0`
`<=>x ne -1,y ne +-3/2`
b) ĐKXĐ: \(x\notin\left\{\dfrac{5}{2};-\dfrac{5}{2}\right\}\)
c) ĐKXĐ: \(x\ne-\dfrac{3}{2}\)
d) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\notin\left\{\dfrac{3}{2};-\dfrac{3}{2}\right\}\end{matrix}\right.\)
a) Ta có: \(\dfrac{9-3x}{x^2+3x+4}-\dfrac{3x-23}{\left(1-x\right)\left(x+4\right)}\)
\(=\dfrac{9-3x}{x^2+3x+4}+\dfrac{3x-23}{x^2+3x-4}\)
\(=\dfrac{\left(9-3x\right)\left(x^2+3x-4\right)}{\left(x^2+3x+4\right)\left(x^2+3x-4\right)}+\dfrac{\left(3x-23\right)\left(x^2+3x+4\right)}{\left(x^2+3x-4\right)\left(x^2+3x+4\right)}\)
\(=\dfrac{9x^2+27x-36-3x^3-9x^2+12x+3x^3+9x^2+12x-23x^2-69x-92}{\left(x^2+3x-4\right)\left(x^2+3x+4\right)}\)
\(=\dfrac{-14x^2-18x-128}{\left(x^2+3x-4\right)\left(x^2+3x+4\right)}\)
b) Ta có: \(\dfrac{4-x}{x^3+2x}-\dfrac{x+5}{x^3-x^2+2x-2}\)
\(=\dfrac{4-x}{x\left(x^2+2\right)}-\dfrac{x+5}{x^2\left(x-1\right)+2\left(x-1\right)}\)
\(=\dfrac{4-x}{x\left(x^2+2\right)}-\dfrac{x+5}{\left(x-1\right)\left(x^2+2\right)}\)
\(=\dfrac{\left(4-x\right)\left(x-1\right)}{x\left(x-1\right)\left(x^2+2\right)}-\dfrac{x\left(x+5\right)}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\dfrac{4x-4-x^2+x-x^2-5x}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\dfrac{-2x^2-4}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\dfrac{-2\left(x^2+2\right)}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\dfrac{-2}{x\left(x-1\right)}\)
a:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
=>3x-9-10x+2=-4
=>-7x-7=-4
=>-7x=3
=>x=-3/7
b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)
=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)
=>10-2x+7x-14=4x-4+x
=>5x-4=5x-4
=>0x=0(luôn đúng)
Vậy: S=R\{0;2}
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
\(a,\dfrac{3}{2x-1}+1=\dfrac{2x-1}{2x+1};ĐKXĐ:x\ne\pm\dfrac{1}{2}\\ \Leftrightarrow\dfrac{3}{2x-1}-\dfrac{2x-1}{2x+1}+1=0\\ \Leftrightarrow\dfrac{3\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\dfrac{\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}=0\\ \Rightarrow3\left(2x+1\right)-\left(2x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow6x+3-\left(4x^2-4x+1\right)+\left(4x^2-1\right)=0\\ \Leftrightarrow6x+3-4x^2+4x-1+4x^2-1=0\\ \Leftrightarrow10x+1=0\\ \Leftrightarrow10x=-1\\ \Leftrightarrow x=-\dfrac{1}{10}\)
Vậy \(x\in\left\{-\dfrac{1}{10}\right\}\)
\(1,\) thiếu đề
\(2,\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
\(\Leftrightarrow\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)}{30}-\dfrac{150}{30}\)
\(\Leftrightarrow5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)
\(\Leftrightarrow25x+10-80x+10=24x+12-150\)
\(\Leftrightarrow-55x+20=24x-138\)
\(\Leftrightarrow24x-138+55x-20=0\)
\(\Leftrightarrow79x-158=0\)
\(\Leftrightarrow x=2\)
\(3,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne3\end{matrix}\right.\\ \dfrac{x}{2x-6}+\dfrac{x}{2x-2}=\dfrac{-2x}{\left(x+1\right)\left(3-x\right)}\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow x\left(\dfrac{1}{2\left(x-3\right)}+\dfrac{1}{2\left(x-1\right)}-\dfrac{2}{\left(x+1\right)\left(x-3\right)}\right)=0\)
\(\Leftrightarrow x\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4\left(x-1\right)}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)
\(\Leftrightarrow x\left(\dfrac{x^2-1}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{x^2-2x-3}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4x-4}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)
\(\Leftrightarrow x.\dfrac{x^2-1+x^2-2x-3-4x+4}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{2x\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x=0\)
a: \(=\dfrac{x+2y}{xy}\cdot\dfrac{2x^2}{\left(x+2y\right)^2}=\dfrac{2x}{y\left(x+2y\right)}\)
b: \(=\dfrac{x\left(4x^2-y^2\right)}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(2x-y\right)^3}\)
\(=\dfrac{x\left(x-y\right)\left(2x+y\right)\left(2x-y\right)}{\left(2x-y\right)^3}\)
\(=\dfrac{x\left(x-y\right)\left(2x+y\right)}{\left(2x-y\right)^2}\)
c: \(=\dfrac{x+3}{x+2}\cdot\dfrac{2x-1}{3\left(x+3\right)}\cdot\dfrac{2\left(x+2\right)}{2\left(2x-1\right)}\)
=1/3
d: \(=\dfrac{x+1}{x+2}:\left(\dfrac{1}{2x}\cdot\dfrac{3x+3}{2x-3}\right)\)
\(=\dfrac{x+1}{x+2}\cdot\dfrac{2x\left(2x-3\right)}{3\left(x+1\right)}=\dfrac{2x\left(2x-3\right)}{3\left(x+2\right)}\)
\(a,=\dfrac{x^2-20+x^2-7x+10+3x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\\ b,=\dfrac{10x+15-4x+6+2x-9}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{4\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{4}{2x-3}\\ c,=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}\\ =\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{x+4-x}{x\left(x+4\right)}=\dfrac{4}{x\left(x+4\right)}\)
a: \(=\dfrac{2x^3-3x^2+4x^2-6x-2x+3}{2x-3}=x^2+2x-1\)
b: \(=\dfrac{x-5}{\left(x-2\right)^2}\cdot\dfrac{2\left(x-2\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{2}{\left(x-2\right)\left(x+5\right)}\)