`@Mon`
`@Bài 1:`
`a)x^2-3x=0`
`<=>x(x-3)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\x-3=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=0(t)\\x=-3(tm)\end{array} \right.\)
`Khi đóM=\frac{x-5}{x-4}=\frac{3-5}{3-4}=\frac{-2}{-1}=2`
`b)N=\frac{x+5}{2x}-\frac{x-6}{5-x}-\frac{2x^2-2x-50}{2x^2-10x}`
`=\frac{(x+5)(x=-5)}{2x(x-5)}+\frac{(x-6)-2x}{2x(x-5)}-\frac{2x^2-2x-50}{2x(x-5)}`
`=\frac{x^2-25+2x^2-12x-2x^2+2x+50}{2x(x-5)}`
`=\frac{x^2-10x+25}{2x(x-5)}=\frac{(x-5)^2}{2x(x-5)}=\frac{x-5}{2x}`
`c)M:N=\frac{x-5}{x-4}:\frac{x-5}{2x}=\frac{2x}{x-4}=\frac{2(x-4)+8}{x-4}=2+\frac{8}{x-4}`
`\text{P nguyên}<=>2+\frac{8}{x-4}nguyên`
`<=>(x-4)\text{ là ước của 8}`
\begin{array}{|c|c|c|}\hline \text{x-4}&\text{-8}&\text{-4}&\text{-2}&\text{-1}&\text{1}&\text{2}&\text{4}&\text{8}\\\hline \text{x}&\text{-4}&\text{0}&\text{2}&\text{3}&\text{5}&\text{6}&\text{8}&\text{12}\\\hline\end{array}
`Vậy x in{2;3;6;8;13;-4} thì P in ZZ`
`@Bài 2:`
`a)N=x^2+y^2` `với` `x \ne=-y`
`b) Ta có:`
`N=x^2+y^2=(x^2+2xy+y^2)-2xy=(x+y)^2-2xy`
`Thay` `xy=-\frac{1}{80};x+y=\frac{1}{40}` `\text{ vào N ta có:}`
`N=(\frac{1}{40})^2-2.(-\frac{1}{80})=\frac{41}{1600}`
`Vậy` `N=\frac{41}{1600}`
`@Bài 3:`
`a)ĐKXĐ:{(2m+3 ne 0),(2m+1 ne 0),((2m+3)(2m+1) ne0):}<=>{(2m+b ne0),(2m+1 ne0):}<=>{(m ne-3/2),(m ne-1/2):}`
`b)P=\frac{2}{2m+3}+\frac{3}{2m+1}-\frac{6m+5}{(2m+3)(2m+1)}`
`=\frac{2(2m+1)+3(2m+3)-(6m+5)}{(2m+3)(2n-1)}`
`=\frac{4m+1+6m+9-6m-5}{(2m+3)(2m+1}``=\frac{4m+6}{(2m+5)(2m+1)}=\frac{2(2m+3)}{(2m+3)(2m+1)}`
`=\frac{2}{2m+1}`
`c)P=-1<=>\frac{2}{2m+1}=-1<=>2=-2m+1`
`<=>2m=-1-2=-3`
`<=>m=-\frac{3}{2}(\text{ loại do mâu thuẫn với đkxđ)`
`=>m=1`
