\(=\lim\limits_{x\rightarrow3}\dfrac{\sqrt{3+2x}-3-\sqrt{7-x}+2}{2x-6}\)

\(=\lim\limits_{x\rightarrow3}\left(\dfrac{2x-6}{\left(2x-6\right)\left(\sqrt{3+2x}+3\right)}-\dfrac{3-x}{\left(2x-6\right)\left(\sqrt{7-x}+2\right)}\right)\)

\(=\dfrac{1}{\sqrt{3+2\cdot3}+3}+\dfrac{1}{2\cdot\left(\sqrt{7-3}+2\right)}=\dfrac{7}{24}\)

141. x > -5

Đặt \(\left\{{}\begin{matrix}2x^2+5x+12=a\\2x^2+3x+2=b\end{matrix}\right.\left(a,b>0\right)\)

pt \(\Leftrightarrow a+b=\dfrac{a^2-b^2}{2}\)

\(\Leftrightarrow\left(a+b\right)\left(1-\dfrac{a-b}{2}\right)=0\)

\(\Leftrightarrow a-b=2\) (vì \(\sqrt{2x^2+5x+12}+\sqrt{2x^2+3x+2}>0\))

\(\Rightarrow\sqrt{2x^2+5x+12}=\sqrt{2x^2+3x+2}+2\)

\(\Rightarrow x+3=2\sqrt{2x^2+3x+2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=-1\end{matrix}\right.\) (thỏa mãn)

Vậy .....

1.A

2.B

3.C

Hình chóp S.ABC có SA = SB = SC = a, \(\widehat{ASB}=90^o,\widehat{BSC}=60^o,\widehat{ASC}=120^o\). Tính góc giữa đt SC và (SAB)

tổng hệ số = 0 mà bạn

\(u_7-u_3=8\Leftrightarrow u_1+6d-\left(u_1+2d\right)=8\Leftrightarrow d=2\)

A

Chặn dưới

\(\Rightarrow1+a+b=0\Leftrightarrow b=-a-1\)

\(\lim\limits_{x\rightarrow1}\dfrac{x^2+ax-a-1}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(x+1+a\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{x+1+a}{x+1}=\dfrac{1+1+a}{1+1}=\dfrac{1}{2}\)

\(\Rightarrow a=-1\Rightarrow b=0\)