Câu 3e 

\(\left(2x+1\right)^2=\left(x-1\right)^2\)

\(\left[{}\begin{matrix}2x+1=x-1\\2x+1=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)

bài đâu ạ?

lỗi r bn

đâu?

11.

Do \(\lim\limits_{x\rightarrow2^-}\left(1-x^2\right)=1-2^2=-3< 0\)

\(\lim\limits_{x\rightarrow2^-}\left(x-2\right)=0\)

Và: \(x-2< 0\) khi \(x< 2\)

\(\Rightarrow\lim\limits_{x\rightarrow2^-}\dfrac{1-x^2}{x-2}=+\infty\)

Bài 1:

3: ĐKXĐ: x>=1

\(x-\sqrt{x+3+4\sqrt{x-1}}=1\)

=>\(x-\sqrt{x-1+2\cdot\sqrt{x-1}\cdot2+4}=1\)

=>\(x-\sqrt{\left(\sqrt{x-1}+2\right)^2}=1\)

=>\(x-\left|\sqrt{x-1}+2\right|=1\)

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

=>\(x-\sqrt{x-1}-2-1=0\)

=>\(x-1-\sqrt{x-1}-2=0\)

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

=>\(\left(\sqrt{x-1}-2\right)\left(\sqrt{x-1}+1\right)=0\)

=>\(\sqrt{x-1}-2=0\)

=>\(\sqrt{x-1}=2\)

=>x-1=4

=>x=5(nhận)

61B

62B

6.

SAB cân tại S \(\Rightarrow SH\perp AB\)

Mà \(\left\{{}\begin{matrix}AB=\left(SAB\right)\cap\left(ABCD\right)\\\left(SAB\right)\perp\left(ABCD\right)\end{matrix}\right.\) \(\Rightarrow SH\perp\left(ABCD\right)\)

Hay SH alf đường cao của chóp

\(\lim\limits_{x\rightarrow1}f\left(x\right)=\lim\limits_{x\rightarrow1}\dfrac{mx^2-\left(m+3\right)x+3}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(mx-3\right)}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\left(mx-3\right)=m-3\)

\(f\left(1\right)=m^2-15\)

Hàm liên tục tại \(x=1\) khi:

\(m-3=m^2-15\Rightarrow m^2-m-12=0\Rightarrow\left[{}\begin{matrix}m=4\\m=-3\end{matrix}\right.\)

\(4^2+\left(-3\right)^2=25\)