a: Để (d) cắt (d') thì \(\frac{1}{m}<>-m\)

=>\(-m^2<>1\)

=>\(m^2<>-1\) (luôn đúng)

=>(d) luôn cắt (d')

ĐKXĐ: \(x\ge1\)

\(\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=0\)

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

\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|=0\)

Do \(\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}-2+3-\sqrt{x-1}\right|=1>0\) với mọi x thuộc TXĐ

\(\Rightarrow\) Phương trình đã cho vô nghiệm

xin loi chi em ko giup dc

VD6: \(\overrightarrow{AM}=3\cdot\overrightarrow{AB}-2\cdot\overrightarrow{AC}\)

=>\(\overrightarrow{AM}=3\left(\overrightarrow{AM}+\overrightarrow{MB}\right)-2\left(\overrightarrow{AM}+\overrightarrow{MC}\right)\)

=>\(\overrightarrow{AM}=3\cdot\overrightarrow{AM}+3\cdot\overrightarrow{MB}-2\cdot\overrightarrow{AM}-2\cdot\overrightarrow{MC}\)

=>\(3\cdot\overrightarrow{MB}-2\cdot\overrightarrow{MC}=\overrightarrow{0}\)

=>\(3\cdot\overrightarrow{MB}=2\cdot\overrightarrow{MC}\)

=>\(\overrightarrow{MB}=\frac23\cdot\overrightarrow{MC}\)

=>\(\overrightarrow{MB};\overrightarrow{MC}\) là hai vecto cùng phương

VD7: \(\frac{\overrightarrow{BC}}{\overrightarrow{AB}}=\frac{3\left(2\cdot\overrightarrow{a}+3\cdot\overrightarrow{b}\right)}{2\cdot\overrightarrow{a}+3\cdot\overrightarrow{b}}=3\)

=>\(\overrightarrow{BC}=3\cdot\overrightarrow{AB}=-3\cdot\overrightarrow{BA}\)

=>B,A,C thẳng hàng