Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x^2-2x-2\right)}{\left(x-1\right)\left(x-3\right)}=\lim\limits_{x\rightarrow1}\dfrac{x^2-2x-2}{x-3}=\dfrac{3}{2}\)
Câu b bạn coi lại đề, là \(x\rightarrow-1^-\) hay \(x\rightarrow1^-\) (đúng như đề thì ko phải dạng vô định, cứ thay số rồi bấm máy)
\(c=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)}{\left(x-3\right)\left(x-1\right)\left(\sqrt[3]{\left(x+5\right)^2}+2\sqrt[3]{x+5}+4\right)}\)
\(=\lim\limits_{x\rightarrow3}\dfrac{1}{\left(x-1\right)\left(\sqrt[3]{\left(x+5\right)^2}+2\sqrt[3]{x+5}+4\right)}=\dfrac{1}{2.\left(4+4+4\right)}=...\)
a/ \(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}{\left(x-1\right)\left(x-3\right)}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}{x-3}=....\)
Từ 2 câu kia lát tui làm, ăn cơm đã :D
\(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-2x+1}{x-1}=3\rightarrow\lim\limits_{x\rightarrow1}\left(f\left(x\right)-2x+1\right)=0\\ \rightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=1\)
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{3f\left(x\right)+1}-x-1}{\sqrt{4x+5}-3x-2}=\dfrac{\sqrt{3.1+1}-1-1}{\sqrt{4.1+5}-3.1-2}=0\)
1/ \(=\lim\limits_{x\rightarrow0}\dfrac{3\left(1+3x\right)^2.3+4.4\left(1-4x\right)^3}{1}=...\left(thay-x-vo\right)\)
2/ \(=\lim\limits_{x\rightarrow2}\dfrac{2.2.x-5}{3x^2-3}=\dfrac{4.2-5}{3.4-3}=\dfrac{1}{3}\)
3/ \(=\lim\limits_{x\rightarrow1}\dfrac{4x^3-3}{3x^2+2}=\dfrac{4.1-3}{3.1-2}=1\)
Xai L'Hospital nhe :v
câu 1 bạn lm kiểu j vậy chả hiểu luôn bạn có thể lm lại chi tiết hơn dc ko
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x-1-\sqrt{4x-3}\right)+\left(\sqrt[3]{6x-5}-\left(2x-1\right)\right)}{\left(x-1\right)^2}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{4\left(x-1\right)^2}{2x-1+\sqrt[]{4x-3}}-\dfrac{4\left(x-1\right)^2\left(2x+1\right)}{\sqrt[3]{\left(6x-5\right)^2}+\left(2x-1\right)\sqrt[3]{6x-5}+\left(2x-1\right)^2}}{\left(x-1\right)^2}\)
\(=\lim\limits_{x\rightarrow1}\left(\dfrac{4}{2x-1+\sqrt[]{4x-3}}-\dfrac{4\left(2x+1\right)}{\sqrt[3]{\left(6x-5\right)^2}+\left(2x-1\right)\sqrt[3]{6x-5}+\left(2x-1\right)^2}\right)=-2\)
\(lim\left(x->1\right)\dfrac{-\sqrt{4x-3}+\sqrt[3]{6x-5}}{\left(x-1\right)^2}\)
Đặt \(\sqrt{4x-3}=f\left(x\right);\sqrt[3]{6x-5}=g\left(x\right)\Rightarrow g\left(x\right)^6-f\left(x\right)^6=4\left(x-1\right)^2\left(16x-13\right)\)
\(f\left(1\right)=1;g\left(1\right)=1\)
Ta có
\(lim\left(x->1\right)\dfrac{-f\left(x\right)+g\left(x\right)}{\left(x-1\right)^2}=lim\left(x->1\right)\dfrac{g\left(x\right)^6-f\left(x\right)^6}{\left(x-1\right)^2}\cdot\dfrac{1}{g\left(x\right)^5+g\left(x\right)^4f\left(x\right)+g\left(x\right)^3f\left(x\right)^2+g\left(x\right)^2f\left(x\right)^3+g\left(x\right)f\left(x\right)^4+f\left(x\right)^5}\)
\(=lim\left(x->1\right)\dfrac{4\left(16x-3\right)}{g\left(x\right)^5+g\left(x\right)^4f\left(x\right)+g\left(x\right)^3f\left(x\right)^2+g\left(x\right)^2f\left(x\right)^3+g\left(x\right)f\left(x\right)^4+f\left(x\right)^5}\)
\(=\dfrac{4\left(16-3\right)}{1^5+1^4\cdot1+1^3\cdot1^2+1^2\cdot1^3+1\cdot1^4+1^5}=\dfrac{26}{3}\)
\(a=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-1+1-\sqrt[3]{2x+1}}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{4x}{\sqrt[]{4x+1}+1}+\dfrac{-2x}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[]{4x+1}+1}+\dfrac{-2}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}\right)=...\)
\(b=\lim\limits_{x\rightarrow1}\dfrac{4\left(x-1\right)\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(x-1\right)\left(\sqrt[]{4x+5}+3\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{4\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(\sqrt[]{4x+5}+3\right)}=...\)
\(c=\lim\limits_{x\rightarrow-1}\dfrac{\left(2x+3\right)^{\dfrac{1}{4}}+\left(2+3x\right)^{\dfrac{1}{3}}}{\left(x+2\right)^{\dfrac{1}{2}}-1}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{1}{2}\left(2x+3\right)^{-\dfrac{3}{4}}+\left(2+3x\right)^{-\dfrac{2}{3}}}{\dfrac{1}{2}\left(x+2\right)^{-\dfrac{1}{2}}}=3\)
1/ \(\lim\limits_{x\rightarrow0^-}\left(\dfrac{x-2}{x^3}\right)=\lim\limits_{x\rightarrow0^-}\dfrac{2-x}{-x^3}=\dfrac{2}{0}=+\infty\)
2/ \(\lim\limits_{x\rightarrow1^+}\dfrac{\left(x^3-x^2\right)^{\dfrac{1}{2}}}{\left(x-1\right)^{\dfrac{1}{2}}+1-x}=\lim\limits_{x\rightarrow1^+}\dfrac{\dfrac{1}{2}\left(x^3-x^2\right)^{-\dfrac{1}{2}}.\left(3x^2-2x\right)}{\dfrac{1}{2}\left(x-1\right)^{-\dfrac{1}{2}}-1}=0\)
3/ \(\lim\limits_{x\rightarrow1^+}\dfrac{1-\left(x^2+x+1\right)}{x^3-1}=\dfrac{1-3}{0}=-\infty\)
4/ \(\lim\limits_{x\rightarrow-\infty}\left(-\infty-\sqrt[3]{1+\infty}\right)=-\left(\infty+\infty\right)=-\infty?\) Cái này ko chắc :v
Mấy câu này bạn cần giải theo kiểu trắc nghiệm hay tự luận nhỉ?
\(a+\dfrac{x+1}{\sqrt{x^2-x+1}}-\dfrac{3x+3}{\sqrt{x}}=0\) có nghiệm \(x=1\)
\(\Rightarrow a+\dfrac{2}{\sqrt{1}}-\dfrac{6}{\sqrt{1}}=0\Rightarrow a=4\)
\(4+\dfrac{x+1}{\sqrt{x^2-x+1}}-\dfrac{3x+3}{\sqrt{x}}=3\left(2-\dfrac{x+1}{\sqrt{x}}\right)+\left(\dfrac{x+1}{\sqrt{x^2-x+1}}-2\right)\)
\(=-3\left(\dfrac{\left(x-1\right)^2}{\sqrt{x}\left(x+1+2\sqrt{x}\right)}\right)+\dfrac{-3\left(x-1\right)^2}{\sqrt{x^2-x+1}\left(x+1-2\sqrt{x^2-x+1}\right)}\)
Rút gọn với \(\left(x-1\right)^2\) bên ngoài rồi thay dố là được
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{4x+5}+x}{x^2+3x+2}=\dfrac{\sqrt{4.1+5}+1}{1^2+3.1+2}=\dfrac{2}{3}\)
(Đề là \(x\rightarrow-1\) thì hợp lý hơn)
dạ em viết lộn đề x-> -1 đấy ạ