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21 tháng 1
Bài 1:
a: ĐKXĐ: \(x+4\ne0\)
=>\(x\ne-4\)
b: ĐKXĐ: \(2x-1\ne0\)
=>\(2x\ne1\)
=>\(x\ne\dfrac{1}{2}\)
c: ĐKXĐ: \(x\left(y-3\right)\ne0\)
=>\(\left\{{}\begin{matrix}x\ne0\\y-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\y\ne3\end{matrix}\right.\)
d: ĐKXĐ: \(x^2-4y^2\ne0\)
=>\(\left(x-2y\right)\left(x+2y\right)\ne0\)
=>\(x\ne\pm2y\)
e: ĐKXĐ: \(\left(5-x\right)\left(y+2\right)\ne0\)
=>\(\left\{{}\begin{matrix}x\ne5\\y\ne-2\end{matrix}\right.\)
Bài 2:
a: \(\dfrac{-12x^3y^2}{-20x^2y^2}=\dfrac{12x^3y^2}{20x^2y^2}=\dfrac{12x^3y^2:4x^2y^2}{20x^2y^2:4x^2y^2}=\dfrac{3x}{5}\)
b: \(\dfrac{x^2+xy-x-y}{x^2-xy-x+y}\)
\(=\dfrac{\left(x^2+xy\right)-\left(x+y\right)}{\left(x^2-xy\right)-\left(x-y\right)}\)
\(=\dfrac{x\left(x+y\right)-\left(x+y\right)}{x\left(x-y\right)-\left(x-y\right)}=\dfrac{\left(x+y\right)\left(x-1\right)}{\left(x-y\right)\left(x-1\right)}\)
\(=\dfrac{x+y}{x-y}\)
c: \(\dfrac{7x^2-7xy}{y^2-x^2}\)
\(=\dfrac{7x\left(x-y\right)}{\left(y-x\right)\left(y+x\right)}\)
\(=\dfrac{-7x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{-7x}{x+y}\)
d: \(\dfrac{7x^2+14x+7}{3x^2+3x}\)
\(=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\)
\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)
e: \(\dfrac{3y-2-3xy+2x}{1-3x-x^3+3x^2}\)
\(=\dfrac{3y-2-x\left(3y-2\right)}{1-3x+3x^2-x^3}\)
\(=\dfrac{\left(3y-2\right)\left(1-x\right)}{\left(1-x\right)^3}=\dfrac{3y-2}{\left(1-x\right)^2}\)
g: \(\dfrac{x^2+7x+12}{x^2+5x+6}\)
\(=\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x+3\right)\left(x+2\right)}\)
\(=\dfrac{x+4}{x+2}\)
DM
0
MH
15 tháng 9 2023
a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)
b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)
c) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5}{3}\left(x\ne y\right)\)
d) \(\dfrac{x^2-y^2}{x+y}=x-y\left(đk:x\ne-y\right)\)
e) \(\dfrac{x^3-x^2+x-1}{x^2-1}=\dfrac{x^2+1}{x+1}\left(đk:x\ne\pm1\right)\)
f) \(\dfrac{x^2+4x+4}{2x+4}=\dfrac{x+2}{2}\left(đk:x\ne-2\right)\)
LH
1
9 tháng 11 2021
1) \(\dfrac{5-x}{x^2-3x}=\dfrac{5-x}{x\left(x-3\right)}\left(đk:x\ne0,x\ne3\right)\)
2) \(\dfrac{3x}{2x+3}\left(đk:x\ne-\dfrac{3}{2}\right)\)
25 tháng 2 2021
`a,x^3-8 ne 0`
`=>x^3 ne 8`
`=>x ne 2`
`b,2x^2+5x+3 ne 0`
`=>2x^2+2x+3x+3 ne 0`
`=>2x(x+1)+3(x+1) ne 0`
`=>(x+1)(2x+3) ne 0`
`=>x ne -1,-3/2`
`c,x^2-4 ne 0`
`=>x^2 ne 4`
`=>x ne 2,-2`
25 tháng 2 2021
a) ĐK:
\(x^3-8\ne0\\ \Leftrightarrow x\ne2\)
b) ĐK:
\(2x^2+5x+3\ne0\\ \Leftrightarrow\left[{}\begin{matrix}x\ne-1\\x\ne-\dfrac{3}{2}\end{matrix}\right.\)
c) ĐK:
\(x^2-4\ne0\\ \Leftrightarrow x\ne\pm2\)
16 tháng 5 2022
a: ĐKXĐ: \(x\notin\left\{0;1;2;3;4;5\right\}\)
b: \(P=\dfrac{1}{\left(x-1\right)\cdot x}+\dfrac{1}{\left(x-2\right)\left(x-1\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x-4}-\dfrac{1}{x-3}+\dfrac{1}{x-5}-\dfrac{1}{x-4}\)
\(=\dfrac{1}{x-5}-\dfrac{1}{x}=\dfrac{x-x+5}{x\left(x-5\right)}=\dfrac{5}{x\left(x-5\right)}\)
a) \(\frac{x+5}{x^2+5x-6}\)
ĐKXĐ: \(x^2+5x-6=x^2+6x-x-6\)
\(x\left(x+6\right)-\left(x+6\right)=\left(x-1\right)\left(x+6\right)\)
\(\Rightarrow x\ne1\) và \(x\ne-6\)
b) \(\frac{1-x}{3x^2-7x-6}\)
ĐKXĐ: \(3x^2-7x-6=3x^2-9x+2x-6=\left(x-3\right)\left(3x+2\right)\)
\(\Rightarrow x\ne3\) và \(x\ne\frac{-2}{3}\)