HOC24
Lớp học
Môn học
Chủ đề / Chương
Bài học
1. A
36.
b, \(\lim\limits_{x\rightarrow-\infty}\dfrac{x^2+2x-3}{2x^2-5x}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{1+\dfrac{2}{x}-\dfrac{3}{x^2}}{2-\dfrac{5}{x}}\)
\(=\dfrac{1}{2}\)
a, \(lim\left(\sqrt{n^2+n+1}-n\right)\)
\(=lim\dfrac{n^2+n+1-n^2}{\sqrt{n^2+n+1}+n}\)
\(=lim\dfrac{n+1}{\sqrt{n^2+n+1}+n}\)
\(=lim\dfrac{1+\dfrac{1}{n}}{\sqrt{1+\dfrac{1}{n}+\dfrac{1}{n^2}}+1}\)
Yêu cầu bài toán.
\(f\left(x\right)=\dfrac{\sqrt{1+2x}-\sqrt[3]{1+3x}}{\sqrt{1-x}-x-1}\)
\(=\dfrac{\sqrt{1+2x}-\left(x+1\right)}{\sqrt{1-x}-x-1}+\dfrac{\left(x+1\right)-\sqrt[3]{1+3x}}{\sqrt{1-x}-x-1}\)
\(=\dfrac{\left(1+2x-x^2-2x-1\right)\left(\sqrt{1-x}+x+1\right)}{\left(1-x-x^2-2x-1\right)\left(\sqrt{1+2x}+x+1\right)}+\dfrac{\left(x^3+3x^2+3x+1-1-3x\right)\left(\sqrt{1-x}+x+1\right)}{\left(1-x-x^2-2x-1\right)\left[\left(x+1\right)^2+\sqrt[3]{\left(1+3x\right)^2}+\left(x+1\right)\sqrt[3]{1+3x}\right]}\)
\(=\dfrac{x^2\left(\sqrt{1-x}+x+1\right)}{\left(x^2+3x\right)\left(\sqrt{1+2x}+x+1\right)}-\dfrac{\left(x^3+3x^2\right)\left(\sqrt{1-x}+x+1\right)}{\left(x^2+3x\right)\left[\left(x+1\right)^2+\sqrt[3]{\left(1+3x\right)^2}+\left(x+1\right)\sqrt[3]{1+3x}\right]}\)
\(=\dfrac{x\left(\sqrt{1-x}+x+1\right)}{\left(x+3\right)\left(\sqrt{1+2x}+x+1\right)}-\dfrac{\left(x+3\right)\left(\sqrt{1-x}+x+1\right)}{\left(x+1\right)^2+\sqrt[3]{\left(1+3x\right)^2}+\left(x+1\right)\sqrt[3]{1+3x}}\)
Khi đó:
\(A=\lim\limits_{x\rightarrow0}f\left(x\right)=-\dfrac{3.2}{1+1+1}=-2\)
b, \(\vec{AC}.\vec{BM}=\vec{AC}.\dfrac{1}{2}\left(\vec{BC}+\vec{BD}\right)\)
\(=\vec{AC}.\dfrac{1}{2}\left(\vec{AC}-\vec{AB}+\vec{AD}-\vec{AB}\right)\)
\(=\vec{y}.\dfrac{1}{2}\left(\vec{y}-2\vec{x}+\vec{z}\right)\)
\(=\dfrac{1}{2}\left(a^2-a^2+\dfrac{a^2}{2}\right)\)
\(=\dfrac{a^2}{4}\)
\(\Rightarrow cos\left(\vec{AC};\vec{BM}\right)=\dfrac{\vec{AC}.\vec{BM}}{AC.BM}=\dfrac{\dfrac{a^2}{4}}{\dfrac{a.a\sqrt{3}}{2}}=\dfrac{1}{2\sqrt{3}}\)
\(\Rightarrow cos\left(AC;BM\right)=\dfrac{1}{2\sqrt{3}}\)