Để pt có 2 nghiệm trái dấu

\(\Leftrightarrow x_1x_2< 0\Leftrightarrow m^2-2m< 0\)

\(\Rightarrow0< m< 2\)

nick mik co vip do

\(\hept{\begin{cases}3x-y=2m+3\\x+2y=3m+1\end{cases}}\Leftrightarrow\hept{\begin{cases}6x-2y=4m+6\\x+2y=3m+1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=m+1\\y=m\end{cases}}\)khi đó: \(^{x^2+y^2=5\Leftrightarrow2m^2+2m+1=5\Leftrightarrow2m^2+2m-4=0\Leftrightarrow\orbr{\begin{cases}m=1\\m=-2\end{cases}}}\)

\(4-m=\dfrac{2}{x+1}\)

Đkxđ : x +1 ≠ 0 ⇔x ≠ -1

\(\forall\) x≠-1; \(\dfrac{2}{x+1}\ne0\)

để pt có nghiệm thì 4 - m ≠ 0 ⇔ m ≠ 4

vậy m ≠ 4 thì pt có nghiệm

(a)<=>(b)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\\left(4-m\right)\left(x+1\right)=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m\ne4\\x=\dfrac{2}{4-m}-1=\dfrac{2-\left(4-m\right)}{4-m}=\dfrac{m-2}{4-m}\end{matrix}\right.\)

\(x\ne-1\Leftrightarrow\dfrac{m-2}{4-m}\ne-1\Leftrightarrow m-2\ne m-4\Leftrightarrow-2\ne-4\forall m\)

ket luan : m khac 4

ĐKXĐ: x∉{0;-1}

\(\frac{x+m}{x+1}+\frac{x-2}{x}=2\)

=>\(\frac{x\left(x+m\right)+\left(x-2\right)\left(x+1\right)}{x\left(x+1\right)}=2\)

=>x(x+m)+(x-2)(x+1)=2x(x+1)

=>\(x^2+mx+x^2-x-2=2x^2+2x\)

=>x(m-1)-2=2x

=>x(m-1)-2x=2

=>x(m-3)=2

Để phương trình vô nghiệm thì \(\left[\begin{array}{l}m-3=0\\ \frac{2}{m-3}=-1\\ \frac{2}{m-3}=0\end{array}\right.\Rightarrow\left[\begin{array}{l}m=3\\ m-3=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}m=3\\ m=1\end{array}\right.\)

=>Chọn A

ĐKXĐ: x∉{0;-1}

\(\frac{x+m}{x+1}+\frac{x-2}{x}=2\)

=>\(\frac{x\left(x+m\right)+\left(x-2\right)\left(x+1\right)}{x\left(x+1\right)}=2\)

=>x(x+m)+(x-2)(x+1)=2x(x+1)

=>\(x^2+mx+x^2-x-2=2x^2+2x\)

=>x(m-1)-2=2x

=>x(m-1)-2x=2

=>x(m-3)=2

Để phương trình vô nghiệm thì \(\left[\begin{array}{l}m-3=0\\ \frac{2}{m-3}=-1\\ \frac{2}{m-3}=0\end{array}\right.\Rightarrow\left[\begin{array}{l}m=3\\ m-3=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}m=3\\ m=1\end{array}\right.\)

=>Chọn A