a) Hai đường thẳng y=(3m + 2)x +m-1 và y=(3-m)x-m+2 cắt nhau \(\Leftrightarrow3m+2\ne3-m\)

\(\Leftrightarrow4m\ne1\Leftrightarrow m\ne\frac{1}{4}\)

b) Hai đường thẳng y=(3m + 2)x +m-1 và y=(3-m)x-m+2 song song với nhau \(\Leftrightarrow3m+2=3-m\)

\(\Leftrightarrow4m=1\Leftrightarrow m=\frac{1}{4}\)

P/s: E ms lớp 6, sai thông cảm

\(Q=\frac{2x+2\sqrt{x}+2}{-\sqrt{x}}+\sqrt{x}\)

\(Q=-2\sqrt{x}-2-\frac{2}{\sqrt{x}}+\sqrt{x}\)

\(Q=-\sqrt{x}-\frac{2}{\sqrt{x}}-2\)

\(\sqrt{x}+\frac{2}{\sqrt{x}}\ge2\sqrt{2}\Rightarrow-\left(\sqrt{x}+\frac{2}{\sqrt{x}}\right)\le-2\sqrt{2}\)

\(\Rightarrow Q\le-2\sqrt{2}-2\)

\("="\Leftrightarrow x=\sqrt{2}\)

\(\sqrt{5x+3}=\sqrt{3-\sqrt{2}}\)

\(\Leftrightarrow\sqrt{5x+3}^2=\sqrt{3-\sqrt{2}}^2\)

\(\Leftrightarrow5x+3=3-\sqrt{2}\)

\(\Leftrightarrow5x=-\sqrt{2}\)

\(\Leftrightarrow x=\frac{-\sqrt{2}}{5}\)

\(\sqrt{4\left(1-x\right)^2}-\sqrt{3}=0\)

\(\Leftrightarrow\sqrt{\left(2-2x\right)^2}-\sqrt{3}=0\)

\(\Leftrightarrow\sqrt{\left(2-2x\right)^2}=\sqrt{3}\)

\(\Leftrightarrow\left|2-3x\right|=\sqrt{3}\)

\(\Leftrightarrow\orbr{\begin{cases}2-3x=\sqrt{3}\\2-3x=-\sqrt{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2-\sqrt{3}}{3}\\x=\frac{2+\sqrt{3}}{3}\end{cases}}\)

ĐK \(y^2\ge9\)

\(PT\Leftrightarrow\sqrt{y^2-9}=6-2y\)

Bình phương 2 vế ta được

\(y^2-9=36-24y+4y^2\)

\(\Leftrightarrow3y^2-24y+45=0\)

\(\Leftrightarrow y^2-8y+15=0\)

\(\Leftrightarrow\left(y-3\right)\left(y-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y-3=0\\y-5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}y=3\\y=5\end{cases}}\)

Vậy..................

\(y=5\) không đúng (thử thế y vào là biết)

\(\Leftrightarrow2\left(x-3\right)+\sqrt{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\sqrt{x-3}\left(2\sqrt{x-3}-\sqrt{x+3}\right)=0\)

\(\Rightarrow\sqrt{x-3}\left(2\sqrt{x-3}-\sqrt{8}-\sqrt{x+3}+\sqrt{8}\right)=0\)

\(\Leftrightarrow\sqrt{x-3}\cdot\sqrt{x-5}\left(\frac{1}{2\sqrt{x-3}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{x+3}}\right)=0\)

\(\Rightarrow\left(\frac{1}{2\sqrt{x-3}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{x+3}}\right)>0\left(\forall x\right)\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\)

\(\Leftrightarrow\sqrt{x-3}\left(2\sqrt{x-3}+\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\sqrt{x-3}\left(2\sqrt{x-3}-\sqrt{8}+\sqrt{x+3}+\sqrt{8}\right)=0\)

\(\Leftrightarrow\sqrt{x-3}\cdot\sqrt{x-5}\cdot\left(\frac{1}{2\sqrt{x-3}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{x+3}}\right)=0\)

\(\left(\frac{1}{2\sqrt{x-3}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{x+3}}\right)>0\left(\forall x\right)\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\)

\(\Leftrightarrow\sqrt{x-3}\left(2\sqrt{x-3}+\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\sqrt{x-3}\left(2\sqrt{x-3}-\sqrt{8}+\sqrt{x+3}+\sqrt{8}\right)=0\)

\(\Leftrightarrow\sqrt{x-3}\cdot\sqrt{x-5}\cdot\left(\frac{1}{2\sqrt{x-3}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{x+3}}\right)=0\)

\(\left(\frac{1}{2\sqrt{x-3}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{x+3}}\right)>0\left(\forall x\right)\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\)