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a)\(ĐKXĐ:x\ne m;x\ne2\)
\(\frac{x+1}{m-x}=\frac{x+4}{x-2}\)
\(\Leftrightarrow\left(m-x\right)\left(x+4\right)=\left(x+1\right)\left(x-2\right)\)
\(\Leftrightarrow-x^2+\left(m-4\right)x+4m=x^2-x-2\)
\(\Leftrightarrow-2x^2+\left(m-3\right)x+\left(4m+2\right)=0\)
Để phương trình vô nghiệm thì \(\Delta< 0\)
hay \(\left(m-3\right)^2-4.\left(-2\right).\left(4m+2\right)< 0\)
\(\Leftrightarrow m^2-6m+9+32m+16< 0\)
\(\Leftrightarrow m^2+26m+25< 0\)
\(\Leftrightarrow m^2+26m+169-144< 0\)
\(\Leftrightarrow\left(m+13\right)^2< 144\)
\(\Leftrightarrow\orbr{\begin{cases}m+13< 12\\m+13>-12\end{cases}}\Leftrightarrow\orbr{\begin{cases}m< -1\\m>-25\end{cases}}\)
b) \(ĐKXĐ:x\ne m;x\ne1\)
\(1+\frac{2x+1}{m-x}=\frac{3x-5}{x-1}\)
\(\Leftrightarrow\frac{x+1+m}{m-x}=\frac{3x-5}{x-1}\)
\(\Leftrightarrow\left(x+1+m\right)\left(x-1\right)=\left(3x-5\right)\left(m-x\right)\)
\(\Leftrightarrow x^2+mx-m-1=3xm-5m-3x^2+5x\)
\(\Leftrightarrow4x^2-\left(2m+5\right)x+\left(4m-1\right)=0\)
Để phương trình vô nghiệm thì \(\Delta< 0\)
\(\Rightarrow\left(2m+5\right)^2-4.4.\left(4m-1\right)=4m^2-44m+41< 0\)
\(\Rightarrow4m^2-44m+121-80< 0\)
\(\Rightarrow\left(2m-11\right)^2< 80\)
\(\Rightarrow\orbr{\begin{cases}2m-11< \sqrt{80}\\2m-11>-\sqrt{80}\end{cases}}\)
Vậy \(\orbr{\begin{cases}m< \frac{\sqrt{80}+11}{2}\\m>-\frac{\sqrt{80}+11}{2}\end{cases}}\)
1/a/\(\Leftrightarrow\left(x+5\right)\left(x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-6\end{cases}}}\)
Vậy ...................
b/ ĐKXĐ:\(x\ne2;x\ne5\)
.....\(\Rightarrow3x^2-15x-x^2+2x+3x=0\)
\(\Leftrightarrow2x^2-10x=0\)
\(\Leftrightarrow2x\left(x-5\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}2x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(nhận\right)\\x=5\left(loại\right)\end{cases}}}\)
Vậy ..............
`Answer:`
`1.`
a. \(\left(x+5\right)\left(2x+1\right)-x^2+25=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)-\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)-\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x+1-x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-6\\x=-5\end{cases}}}\)
b. \(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{3x}{\left(x-2\right)\left(x-5\right)}=0\left(ĐKXĐ:x\ne2;x\ne5\right)\)
\(\Leftrightarrow\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}+\frac{3x}{\left(x-2\right)\left(x-5\right)}=0\)
\(\Leftrightarrow\frac{3x\left(x-5\right)-x\left(x-2\right)+3x}{\left(x-2\right)\left(x-5\right)}=0\)
\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)+3x=0\)
\(\Leftrightarrow3x^2-15x-x^2+2x+3x=0\)
\(\Leftrightarrow2x\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\text{(Không thoả mãn)}\end{cases}}}\)
`2.`
\(ĐKXĐ:x\ne-m-2;x\ne m-2\)
Ta có: \(\frac{x+1}{x+2+m}=\frac{x+1}{x+2-m}\left(1\right)\)
a. Khi `m=-3` phương trình `(1)` sẽ trở thành: \(\frac{x+1}{x-1}=\frac{x+1}{x+5}\left(x\ne1;x\ne-5\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\frac{1}{x-1}=\frac{1}{x+5}\end{cases}\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-1=x+5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\-1=5\text{(Vô nghiệm)}\end{cases}}}\)
b. Để phương trình `(1)` nhận `x=3` làm nghiệm thì
\(\Leftrightarrow\hept{\begin{cases}\frac{3+1}{3+2-m}=\frac{3+1}{3+2-m}\\3\ne-m-2\\3\ne m-2\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{4}{5+m}=\frac{4}{5-m}\\m\ne\pm5\end{cases}}\Leftrightarrow\hept{\begin{cases}5+m=5-m\\m\ne\pm5\end{cases}}\Leftrightarrow m=0\)
ĐKXĐ : \(x\ne-5;-m\)
\(\dfrac{x-m}{x+5}+\dfrac{x-5}{x+m}=2\left(1\right)\)
\(\Leftrightarrow\dfrac{\left(x-m\right)\left(x+m\right)+\left(x+5\right)\left(x-5\right)}{\left(x+5\right)\left(x+m\right)}=2\)
\(\Leftrightarrow x^2-m^2+x^2-25=2x^2+2xm+10x+10m\)
\(\Leftrightarrow2xm+10x+m^2+10m+25=0\)
\(\Leftrightarrow2x\left(m+5\right)=-\left(m+5\right)^2\)
\(\Leftrightarrow x=\dfrac{-\left(m+5\right)}{2}\)
PT \(\left(1\right)\) VN \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{-\left(m+5\right)}{2}=-5\\\dfrac{\left(-m+5\right)}{2}=-m\end{matrix}\right.\)
Câu 13 : Phương trình m(x-1) =5-(m-1)x vô nghiệm nếu :
A/ m=1/2 B/ m=1/4 C/ m=3/2 D/ m=1
cho mình lời giải chi tiết với