biết \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{3x^2+2}-\sqrt{2-2x}}{x}=\dfrac{a\sqrt{2}}{b}\). tìm a,b biết a/b tối giản
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NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 1 2021
\(\lim\limits_{x\rightarrow0}\dfrac{x}{\sqrt[7]{x+1}\left(\sqrt[]{x+4}-2\right)+2\left(\sqrt[7]{x+1}-1\right)}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{x}{\dfrac{x\sqrt[7]{x+1}}{\sqrt[]{x+4}+2}+\dfrac{2x}{\sqrt[7]{\left(x+1\right)^6}+\sqrt[7]{\left(x+1\right)^5}+\sqrt[7]{\left(x+1\right)^4}+\sqrt[7]{\left(x+1\right)^3}+\sqrt[7]{\left(x+1\right)^2}+\sqrt[7]{x+1}+1}}\)
\(=\dfrac{1}{\dfrac{1}{2+2}+\dfrac{2}{1+1+1+1+1+1+1}}=\dfrac{28}{15}\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 1 2021
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+9}-3+\sqrt{x+16}-4}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{x}{\sqrt{x+9}+3}+\dfrac{x}{\sqrt{x+16}+4}}{x}\)
\(\lim\limits_{x\rightarrow0}\left(\dfrac{1}{\sqrt{x+9}+3}+\dfrac{1}{\sqrt{x+16}+4}\right)=\dfrac{7}{24}\)
HT
27 tháng 1 2021
Tui ko biết đề bài có sai hay ko, bởi hệ số khác nhau thì đặt x ra là được, kết ủa là dương vô cùng, ko tồn tại a và b.
AH
Akai Haruma
Giáo viên
29 tháng 1 2023
Lời giải:
a.
\(\lim\limits_{x\to 0}\frac{\sqrt{x+1}-\sqrt{x^2+x+1}}{x^2-2x+1}=\lim\limits_{x\to 0}\frac{\sqrt{0+1}-\sqrt{0^2+0+1}}{0^2-2.0+1}=0\)
b.
\(\lim\limits_{x\to 7}\frac{\sqrt{x-3}-2}{49-x^2}=\lim\limits_{x\to 7}\frac{(x-3)-2^2}{(49-x^2)(\sqrt{x-3}+2)}\)
\(=\lim\limits_{x\to 7}\frac{x-7}{-(x-7)(x+7)(\sqrt{x-3}+2)}=\lim\limits_{x\to 7}\frac{1}{-(x+7)(\sqrt{x-3}+2)}=\frac{1}{-(7+7)(\sqrt{7-3}+2)}=\frac{-1}{56}\)
DD
Đoàn Đức Hà
Giáo viên
30 tháng 1 2023
a) \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {x + 1} - \sqrt {{x^2} + x + 1} }}{{{x^2} - 2x + 1}} = \frac{{\sqrt {0 + 1} - \sqrt {{0^2} + 0 + 1} }}{{{0^2} - 2.0 + 1}} = 0\)
b) \(\mathop {\lim }\limits_{x \to 7} \frac{{\sqrt {x - 3} - 2}}{{49 - {x^2}}} = \mathop {\lim }\limits_{x \to 7} \frac{{x - 3 - {2^2}}}{{\left( {7 - x} \right)\left( {7 + x} \right)\left( {\sqrt {x - 3} + 2} \right)}} = \mathop {\lim }\limits_{x \to 7} \frac{{ - 1}}{{\left( {7 + x} \right)\left( {\sqrt {x - 3} + 2} \right)}} = \frac{{ - 1}}{{56}}\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 3 2022
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x+2}+\sqrt{5x+4}-5}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x+2}-2+\sqrt{5x+4}-3}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{2\left(x-1\right)}{\sqrt{2x+2}+2}+\dfrac{5\left(x-1\right)}{\sqrt{5x+4}+3}}{x-1}=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\sqrt{2x+2}+2}+\dfrac{5}{\sqrt{5x+4}+3}\right)=\dfrac{2}{2+2}+\dfrac{5}{3+3}=...\)
Đề câu b là \(...\sqrt{90-6x}\) hay \(\sqrt{9-6x}\) vậy em? Hình như cái sau mới có lý
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 1
\(a+\dfrac{x+1}{\sqrt{x^2-x+1}}-\dfrac{3x+3}{\sqrt{x}}=0\) 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
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 2 2021
\(a=\lim\limits_{x\rightarrow2}\dfrac{\left(x^2-x-2\right)\left(x^2+x\sqrt[3]{3x+2}+\sqrt[3]{\left(3x+2\right)^2}\right)}{\left(x^3-3x-2\right)\left(x+\sqrt[]{x+2}\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+1\right)\left(x^2+x\sqrt[3]{3x+2}+\sqrt[3]{\left(3x+2\right)^2}\right)}{\left(x-2\right)\left(x+1\right)^2\left(x+\sqrt[]{x+2}\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+x\sqrt[3]{3x+2}+\sqrt[3]{\left(3x+2\right)^2}}{\left(x+1\right)\left(x+\sqrt[]{x+2}\right)}=...\)
\(b=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt[]{1+2x}-x-1\right)+\left(x+1-\sqrt[3]{1+3x}\right)}{x^2}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{x^2}{\sqrt[]{1+2x}+x+1}+\dfrac{x^3+3x^2}{\left(x+1\right)^2+\left(x+1\right)\sqrt[3]{1+3x}+\sqrt[3]{\left(1+3x\right)^2}}}{x^2}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{1}{\sqrt[]{1+2x}+x+1}+\dfrac{x+3}{\left(x+1\right)^2+\left(x+1\right)\sqrt[3]{1+3x}+\sqrt[3]{\left(1+3x\right)^2}}\right)\)
\(=...\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 2 2021
\(c=\lim\limits_{x\rightarrow-1}\dfrac{\left(\sqrt[]{5+4x}-2x-3\right)+\left(2x+3-\sqrt[3]{7+6x}\right)}{x^3+x^2-x-1}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{5+4x-\left(2x+3\right)^2}{2x+3+\sqrt[]{5+4x}}+\dfrac{\left(2x+3\right)^3-\left(7+6x\right)}{\left(2x+3\right)^2+\left(2x+3\right)\sqrt[3]{7+6x}+\sqrt[3]{\left(7+6x\right)^2}}}{\left(x-1\right)\left(x+1\right)^2}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{-4\left(x+1\right)^2}{2x+3+\sqrt[]{5+4x}}+\dfrac{\left(x+1\right)^2\left(8x+20\right)}{\left(2x+3\right)^2+\left(2x+3\right)\sqrt[3]{7+6x}+\sqrt[3]{\left(7+6x\right)^2}}}{\left(x-1\right)\left(x+1\right)^2}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{-4}{2x+3+\sqrt[]{5+4x}}+\dfrac{8x+20}{\left(2x+3\right)^2+\left(2x+3\right)\sqrt[3]{7+6x}+\sqrt[3]{\left(7+6x\right)^2}}}{x-1}\)
\(=...\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 2 2021
\(a=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-1+1-\sqrt[3]{2x+1}}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{4x}{\sqrt[]{4x+1}+1}+\dfrac{-2x}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[]{4x+1}+1}+\dfrac{-2}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}\right)=...\)
\(b=\lim\limits_{x\rightarrow1}\dfrac{4\left(x-1\right)\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(x-1\right)\left(\sqrt[]{4x+5}+3\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{4\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(\sqrt[]{4x+5}+3\right)}=...\)
\(c=\lim\limits_{x\rightarrow-1}\dfrac{\left(2x+3\right)^{\dfrac{1}{4}}+\left(2+3x\right)^{\dfrac{1}{3}}}{\left(x+2\right)^{\dfrac{1}{2}}-1}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{1}{2}\left(2x+3\right)^{-\dfrac{3}{4}}+\left(2+3x\right)^{-\dfrac{2}{3}}}{\dfrac{1}{2}\left(x+2\right)^{-\dfrac{1}{2}}}=3\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 2 2021
\(a=\lim\limits_{x\rightarrow3}\dfrac{2x+3-x^2}{\left(x^2-4x+3\right)\left(\sqrt[]{2x+3}+x\right)}=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)\left(-x-1\right)}{\left(x-3\right)\left(x-1\right)\left(\sqrt[]{2x+3}+x\right)}\)
\(=\lim\limits_{x\rightarrow3}\dfrac{-x-1}{\left(x-1\right)\left(\sqrt[]{2x+3}+x\right)}=...\)
\(b=\lim\limits_{x\rightarrow0}\dfrac{\left(x+1\right)^{\dfrac{1}{3}}-1}{\left(2x+1\right)^{\dfrac{1}{4}}-1}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{3}\left(x+1\right)^{-\dfrac{2}{3}}}{\dfrac{1}{2}\left(2x+1\right)^{-\dfrac{3}{4}}}=\dfrac{2}{3}\)
\(c=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt[]{1+4x}-2x-1\right)+\left(2x+1-\sqrt[3]{1+6x}\right)}{x^2}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{-4x^2}{2x+1+\sqrt[]{4x+1}}+\dfrac{x^2\left(8x+12\right)}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}}{x^2}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{-4}{2x+1+\sqrt[]{4x+1}}+\dfrac{8x+12}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}\right)=...\)
\(\lim\limits_{x\rightarrow0}\dfrac{3x^2+2-\left(2-2x\right)}{x\left(\sqrt{3x^2+2}+\sqrt{2-2x}\right)}=\lim\limits_{x\rightarrow0}\dfrac{x\left(3x+2\right)}{x\left(\sqrt{3x^2+2}+\sqrt{2-2x}\right)}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{3x+2}{\sqrt{3x^2+2}+\sqrt{2-2x}}=\dfrac{2}{2\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)