Biết lim(x—>1)\(\frac{\sqrt{x^2+x+2}-\sqrt[3]{7x+1}}{\sqrt{2}\left(x-1\right)}=\frac{a\sqrt{2}}{b}+c\left(a,b,c\in Zvà\frac{a}{b}tốigiản\right)giátrịcủa.a+b+c=?\)
Những câu hỏi liên quan
Biết lim(x->1) \(\frac{\sqrt{x^2+x}+2-\sqrt[3]{7x+1}}{\sqrt{2}\left(x-1\right)}=\frac{a\sqrt{2}}{b}+c\left(a,b,c,\in Z\right)Và\frac{a}{b}tốigiảm\)
Đề đúng là +2 trên tử phải nằm trong căn đầu tiên, nếu ko giới hạn sẽ là dương vô cùng
\(\lim\limits_{x\rightarrow1}\frac{\sqrt{x^2+x+2}-2+2-\sqrt[3]{7x+1}}{\sqrt{2}\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{x^2+x-2}{\sqrt{x^2+x+2}+2}+\frac{8-\left(7x+1\right)}{4+2\sqrt[3]{7x+1}+\sqrt[3]{\left(7x+1\right)^2}}}{\sqrt{2}\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{\frac{\left(x-1\right)\left(x+2\right)}{\sqrt{x^2+x+2}+2}-\frac{x-1}{4+2\sqrt[3]{7x+1}+\sqrt[3]{\left(7x+1\right)^2}}}{\sqrt{2}\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{x+2}{\sqrt{x^2+x+2}+2}-\frac{1}{4+2\sqrt[3]{7x+1}+\sqrt[3]{\left(7x+1\right)^2}}}{\sqrt{2}}\)
\(=\frac{\frac{3}{4}-\frac{1}{4+4+4}}{\sqrt{2}}=\frac{2}{3\sqrt{2}}=\frac{\sqrt{2}}{3}+0\)
\(\Rightarrow a+b+c=1+3+0=4\)
1/ Rút gọn biểu thức:Gleft(frac{x-sqrt{x}}{sqrt{x}-1}-frac{sqrt{x}+1}{x+sqrt{x}}right)divfrac{sqrt{x}+1}{x}2/ Cho biểu thức: Mx-frac{2x-2sqrt{x}}{sqrt{x}-1}+frac{xsqrt{x}+1}{x-sqrt{x}+1}+1a. Tìm ĐKXĐb. Rút gọn Mc. Tìm giá trị nhỏ nhất của M3/ Chứng minh: sqrt{frac{1}{a^2}+frac{1}{b^2}+frac{1}{left(a+bright)^2}}|frac{1}{a}+frac{1}{b}+frac{1}{a+b}|với ane0,bne0,a+bne04/ Biết a,b,c là số dương và ab + bc + ac 1. Hãy tính tổng:Masqrt{frac{left(1+b^2right)left(1+c^2right)}{1+a^2}}+bsqrt{frac{left(1+a...
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1/ Rút gọn biểu thức:\(G=\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}}\right)\div\frac{\sqrt{x}+1}{x}\)
2/ Cho biểu thức: \(M=x-\frac{2x-2\sqrt{x}}{\sqrt{x}-1}+\frac{x\sqrt{x}+1}{x-\sqrt{x}+1}+1\)
a. Tìm ĐKXĐ
b. Rút gọn M
c. Tìm giá trị nhỏ nhất của M
3/ Chứng minh: \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=|\frac{1}{a}+\frac{1}{b}+\frac{1}{a+b}|\)với \(a\ne0,b\ne0,a+b\ne0\)
4/ Biết a,b,c là số dương và ab + bc + ac =1. Hãy tính tổng:
\(M=a\sqrt{\frac{\left(1+b^2\right)\left(1+c^2\right)}{1+a^2}}+b\sqrt{\frac{\left(1+a^2\right)\left(1+c^2\right)}{1+b^2}}+c\sqrt{\frac{\left(1+a^2\right)\left(1+b^2\right)}{1+c^2}}\)
Ai giải giúp mình bài 1 với bài 4 trước đi
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a. limlimits_{xrightarrow a}frac{xsqrt{x}-asqrt{a}}{sqrt{x}-sqrt{a}} e. limlimits_{xrightarrow0}frac{sqrt{1+x}-sqrt[3]{1+x}}{x}
b. limlimits_{xrightarrow1}frac{sqrt[n]{x}-1}{sqrt[m]{x}-1}left(m,nin Z^+right) f. limlimits_{xrightarrow2}frac{sqrt[3]{8x+11}-sqrt{x+7}}{x^2-3x+2}
c. limlimits_{xrightarrow1}frac{left(1-sqrt{x}right)left(1-sqrt[3]{x}right)left(1-sqrt[4]{x}right)left(1-sqrt[5]{x}right)}{left(1-xright)^4}...
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a. \(\lim\limits_{x\rightarrow a}\frac{x\sqrt{x}-a\sqrt{a}}{\sqrt{x}-\sqrt{a}}\) e. \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}\)
b. \(\lim\limits_{x\rightarrow1}\frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}\left(m,n\in Z^+\right)\) f. \(\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-\sqrt{x+7}}{x^2-3x+2}\)
c. \(\lim\limits_{x\rightarrow1}\frac{\left(1-\sqrt{x}\right)\left(1-\sqrt[3]{x}\right)\left(1-\sqrt[4]{x}\right)\left(1-\sqrt[5]{x}\right)}{\left(1-x\right)^4}\) g. \(\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{3x-2}-\sqrt{2x-1}}{x^3-1}\)
d. \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)\) h. \(\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+9}+\sqrt[3]{2x-6}}{x^3+1}\)
\(a=\lim\limits_{x\rightarrow a}\frac{\left(\sqrt{x}-\sqrt{a}\right)\left(x+\sqrt{ax}+a\right)}{\sqrt{x}-\sqrt{a}}=\lim\limits_{x\rightarrow a}\left(x+\sqrt{ax}+a\right)=3a\)
\(b=\lim\limits_{x\rightarrow1}\frac{x^{\frac{1}{n}}-1}{x^{\frac{1}{m}}-1}=\lim\limits_{x\rightarrow1}\frac{\frac{1}{n}x^{\frac{1-n}{n}}}{\frac{1}{m}x^{\frac{1-m}{m}}}=\frac{\frac{1}{n}}{\frac{1}{m}}=\frac{m}{n}\)
Ta có:
\(\lim\limits_{x\rightarrow1}\frac{1-\sqrt[n]{x}}{1-x}=\lim\limits_{x\rightarrow1}\frac{1-x^{\frac{1}{n}}}{1-x}=\lim\limits_{x\rightarrow1}\frac{-\frac{1}{n}x^{\frac{1-n}{n}}}{-1}=\frac{1}{n}\)
\(\Rightarrow c=\lim\limits_{x\rightarrow1}\frac{\left(1-\sqrt{x}\right)}{1-x}.\frac{\left(1-\sqrt[3]{x}\right)}{\left(1-x\right)}.\frac{\left(1-\sqrt[4]{x}\right)}{\left(1-x\right)}.\frac{\left(1-\sqrt[5]{x}\right)}{\left(1-x\right)}=\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}=\frac{1}{120}\)
\(d=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x\sqrt{x}}}}+1}=\frac{1}{2}\)
\(e=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+x}-1+1-\sqrt[3]{1+x}}{x}=\lim\limits_{x\rightarrow0}\frac{\frac{x}{\sqrt{1+x}+1}+\frac{x}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\frac{1}{\sqrt{1+x}+1}+\frac{1}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}\right)=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
\(f=\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-3+3-\sqrt{x+7}}{\left(x-1\right)\left(x-2\right)}=\lim\limits_{x\rightarrow2}\frac{\frac{8\left(x-2\right)}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{x-2}{3+\sqrt{x+7}}}{\left(x-1\right)\left(x-2\right)}\)
\(=\lim\limits_{x\rightarrow2}\frac{\frac{8}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{1}{3+\sqrt{x+7}}}{x-1}=\frac{8}{27}-\frac{1}{6}=\frac{7}{54}\)
\(g=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{3x-2}-1+1-\sqrt{2x-1}}{\left(x-1\right)\left(x^2+x+1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{3\left(x-1\right)}{\sqrt[3]{\left(3x-2\right)^2}+\sqrt[3]{3x-2}+1}-\frac{2\left(x-1\right)}{1+\sqrt{2x-1}}}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{\frac{3}{\sqrt[3]{\left(3x-2\right)^2}+\sqrt[3]{3x-2}+1}-\frac{2}{1+\sqrt{2x-1}}}{x^2+x+1}=0\)
\(h=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+9}+\sqrt[3]{2x-6}}{x^3+1}=\frac{\sqrt[3]{10}-\sqrt[3]{4}}{2}\)
Bài 1
a. limlimits_{xrightarrow-infty}left(sqrt[3]{x^3-x^2}-xright)
b. limlimits_{xrightarrow+infty}left(sqrt[3]{x^3+5x^2}-sqrt[3]{x^3+8x}right)
c. limlimits_{xrightarrow+infty}left(sqrt[3]{x^3+1}-xright)
Bài 2
a. limlimits_{xrightarrow1^-}left(frac{2}{x^2-1}-frac{1}{x-1}right)
b. limlimits_{xrightarrow1^+}left(frac{1}{1-x}-frac{3}{1-x^3}right)
c. limlimits_{xrightarrow2^+}left(frac{1}{x^2-3x+2}-frac{1}{x^2-5x+6}right)
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Bài 1
a. \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt[3]{x^3-x^2}-x\right)\)
b. \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt[3]{x^3+5x^2}-\sqrt[3]{x^3+8x}\right)\)
c. \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt[3]{x^3+1}-x\right)\)
Bài 2
a. \(\lim\limits_{x\rightarrow1^-}\left(\frac{2}{x^2-1}-\frac{1}{x-1}\right)\)
b. \(\lim\limits_{x\rightarrow1^+}\left(\frac{1}{1-x}-\frac{3}{1-x^3}\right)\)
c. \(\lim\limits_{x\rightarrow2^+}\left(\frac{1}{x^2-3x+2}-\frac{1}{x^2-5x+6}\right)\)
\(a=\lim\limits_{x\rightarrow-\infty}\left(\frac{-x^2}{\sqrt[3]{\left(x^3-x^2\right)^2}+x\sqrt[3]{x^3-x^2}+x^2}\right)=\lim\limits_{x\rightarrow-\infty}\left(\frac{-1}{\sqrt[3]{\left(1-\frac{1}{x}\right)^3}+\sqrt[3]{1-\frac{1}{x}}+1}\right)=-\frac{1}{3}\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{5x^2-8x}{\sqrt[3]{\left(x^3+5x^2\right)^2}+\sqrt[3]{\left(x^3+5x^2\right)\left(x^3+8x\right)}+\sqrt[3]{\left(x^3+8x\right)^2}}\)
\(=\lim\limits_{x\rightarrow+\infty}\frac{5-\frac{8}{x}}{\sqrt[3]{\left(1+\frac{5}{x}\right)^2}+\sqrt[3]{\left(1+\frac{5}{x}\right)\left(1+\frac{8}{x^2}\right)}+\sqrt[3]{\left(1+\frac{8}{x^2}\right)^2}}=\frac{5}{3}\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{1}{\sqrt[3]{\left(x^3+1\right)^2}+x\sqrt[3]{x^3+1}+x^2}=\frac{1}{+\infty}=0\)
Bài 2:
\(a=\lim\limits_{x\rightarrow1^-}\left(\frac{1-x}{\left(x-1\right)\left(x+1\right)}\right)=\lim\limits_{x\rightarrow1^-}\frac{-1}{x+1}=-\frac{1}{2}\)
\(b=\lim\limits_{x\rightarrow1^+}\left(\frac{x^2+x+1-3}{\left(1-x\right)\left(x^2+x+1\right)}\right)=\lim\limits_{x\rightarrow1^+}\frac{\left(x-1\right)\left(x+2\right)}{\left(1-x\right)\left(x^2+x+1\right)}=\lim\limits_{x\rightarrow1^+}\frac{-x-2}{x^2+x+1}=-1\)
\(c=\lim\limits_{x\rightarrow2^+}\left(\frac{1}{\left(x-1\right)\left(x-2\right)}-\frac{1}{\left(x-2\right)\left(x-3\right)}\right)=\lim\limits_{x\rightarrow2^+}\frac{-2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
Do \(x\rightarrow2^+\Rightarrow x>2\Rightarrow x-2>0\Rightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\rightarrow0^-\)
\(\Rightarrow\lim\limits_{x\rightarrow2^+}\frac{-2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=+\infty\)
Bài 1
a. limlimits_{xrightarrow-infty}frac{sqrt{4x^2}+1}{3x-1}
b. limlimits_{xrightarrow+infty}frac{sqrt{9x^2+x+1}-sqrt{4x^2+2x+1}}{x+1}
c. limlimits_{xrightarrow+infty}frac{sqrt{x+2x+3}+4x+1}{sqrt{4x^2+1}+2-x}
d. limlimits_{xrightarrow+infty}frac{3x-2sqrt{x}+sqrt{x^4-5x}}{2x^2+4x-5}
Bài 2
a. limlimits_{xrightarrow-infty}frac{2x+1}{x-1}
b. limlimits_{xrightarrow-infty}frac{2x^3+3}{x^3-2x^2+1}
c. limlimits_{xrightarrow+infty}frac{left(3x^2+1right)left(5x+3right)}{left(2x^3-1r...
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Bài 1
a. \(\lim\limits_{x\rightarrow-\infty}\frac{\sqrt{4x^2}+1}{3x-1}\)
b. \(\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9x^2+x+1}-\sqrt{4x^2+2x+1}}{x+1}\)
c. \(\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x+2x+3}+4x+1}{\sqrt{4x^2+1}+2-x}\)
d. \(\lim\limits_{x\rightarrow+\infty}\frac{3x-2\sqrt{x}+\sqrt{x^4-5x}}{2x^2+4x-5}\)
Bài 2
a. \(\lim\limits_{x\rightarrow-\infty}\frac{2x+1}{x-1}\)
b. \(\lim\limits_{x\rightarrow-\infty}\frac{2x^3+3}{x^3-2x^2+1}\)
c. \(\lim\limits_{x\rightarrow+\infty}\frac{\left(3x^2+1\right)\left(5x+3\right)}{\left(2x^3-1\right)\left(x+4\right)}\)
Bài 1:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5\)
\(d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}\)
Bài 2:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)
\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)
1. Chứng minh sqrt[3]{3+sqrt[3]{3}}+sqrt[3]{3-sqrt[3]{3}} 2sqrt[3]{3}2. a) Tính Afrac{2b.sqrt{x^2-1}}{x-sqrt{x^2-1}} với xfrac{1}{2}left(sqrt{frac{a}{b}}+sqrt{frac{b}{a}}right)left(a,b0right)
b) Tính Bfrac{xy-sqrt{x^2-1}.sqrt{y^2-1}}{xy+sqrt{x^2-1}.sqrt{y^2-1}} với xfrac{1}{2}left(a+frac{1}{a}right);yfrac{1}{2}left(b+frac{1}{b}right)left(a,bge1right)3. Cho x,y thỏa mãn xyge0. Tính Bleft(left|sqrt{xy}+frac{x}{2}+frac{y}{2}right|-left|xright|right)+left(left|sqrt{xy}-frac{x}{2}-frac{y}{2}right|...
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1. Chứng minh \(\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}< 2\sqrt[3]{3}\)
2. a) Tính \(A=\frac{2b.\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\) với \(x=\frac{1}{2}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)\left(a,b>0\right) \)
b) Tính \(B=\frac{xy-\sqrt{x^2-1}.\sqrt{y^2-1}}{xy+\sqrt{x^2-1}.\sqrt{y^2-1}}\) với \(x=\frac{1}{2}\left(a+\frac{1}{a}\right);y=\frac{1}{2}\left(b+\frac{1}{b}\right)\left(a,b\ge1\right)\)
3. Cho x,y thỏa mãn \(xy\ge0\). Tính \(B=\left(\left|\sqrt{xy}+\frac{x}{2}+\frac{y}{2}\right|-\left|x\right|\right)+\left(\left|\sqrt{xy}-\frac{x}{2}-\frac{y}{2}\right|-\left|y\right|\right)\)
4. Cho \(\frac{2x+2\sqrt{x}+13}{\left(\sqrt{x}-2\right)\left(x+1\right)^2}=\frac{A}{\sqrt{x}-2}+\frac{B\sqrt{x}+C}{x+1}+\frac{D\sqrt{x}+E}{\left(x+1\right)^2}\). Tìm các số A,B,C,D,E để đẳng thức trên là đúng với mọi x
a/ ^{lim}_{x-0}frac{sqrt{1+x}-sqrt[3]{1+x}}{x}
b/^{lim}_{x-1}left(frac{1}{1-x}-frac{1}{1-x^3}right)
c/ ^{lim}_{x-+infty}left(sqrt[3]{2x-1}-sqrt[3]{2x+1}right)
d/ ^{lim}_{x--infty}left(sqrt[3]{3x^3-1}+sqrt{x^2+2}right)
e/^{lim}_{x-2}left(frac{1}{x^2-3x+2}+frac{1}{x^2-5x+6}right)
f/ ^{lim}_{x-0^{+-}}left(frac{2x}{sqrt{4x^2+x^3}}right)
Đọc tiếp
a/ \(^{lim}_{x->0}\frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}\)
b/\(^{lim}_{x->1}\left(\frac{1}{1-x}-\frac{1}{1-x^3}\right)\)
c/ \(^{lim}_{x->+\infty}\left(\sqrt[3]{2x-1}-\sqrt[3]{2x+1}\right)\)
d/ \(^{lim}_{x->-\infty}\left(\sqrt[3]{3x^3-1}+\sqrt{x^2+2}\right)\)
e/\(^{lim}_{x->2}\left(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}\right)\)
f/ \(^{lim}_{x->0^{+-}}\left(\frac{2x}{\sqrt{4x^2+x^3}}\right)\)
Bạn tự hiểu là giới hạn tiến đến đâu nhé, làm biếng gõ đủ công thức
a. \(\frac{\sqrt{1+x}-1+1-\sqrt[3]{1+x}}{x}=\frac{\frac{x}{\sqrt{1+x}+1}-\frac{x}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}}{x}=\frac{1}{\sqrt{1+x}+1}-\frac{1}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)
b.
\(\frac{1-x^3-1+x}{\left(1-x\right)^2\left(1+x+x^2\right)}=\frac{x\left(1-x\right)\left(1+x\right)}{\left(1-x\right)^2\left(1+x+x^2\right)}=\frac{x\left(1+x\right)}{\left(1-x\right)\left(1+x+x^2\right)}=\frac{2}{0}=\infty\)
c.
\(=\frac{-2}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{\left(2x+1\right)^2}+\sqrt[3]{\left(2x-1\right)\left(2x+1\right)}}=\frac{-2}{\infty}=0\)
d.
\(=x\sqrt[3]{3-\frac{1}{x^3}}-x\sqrt{1+\frac{2}{x^2}}=x\left(\sqrt[3]{3-\frac{1}{x^3}}-\sqrt{1+\frac{2}{x^2}}\right)=-\infty\)
e.
\(=\frac{2x^2-8x+8}{\left(x-1\right)\left(x-2\right)\left(x-2\right)\left(x-3\right)}=\frac{2\left(x-2\right)^2}{\left(x-1\right)\left(x-3\right)\left(x-2\right)^2}=\frac{2}{\left(x-1\right)\left(x-3\right)}=\frac{2}{-1}=-2\)
f.
\(=\frac{2x}{x\sqrt{4+x}}=\frac{2}{\sqrt{4+x}}=1\)
Câu 1:
Xác đinh k để hàm: f(x)left{{}begin{matrix}frac{x^{2016}+x-2}{sqrt{2018x+1}-sqrt{x+2018}}kend{matrix}right.liên tục tại 1
Câu 2: Cho lim(x--1) frac{x^2+ax+b}{x^2-1}frac{1}{2}. Tổng S a^2+b^2 bằng bao nhiêu
Câu 3: lim(x-1) frac{sqrt{x^2+x+2}-sqrt[3]{7x+1}}{sqrt{2}left(x-1right)}frac{asqrt{2}}{b}+c với a/b là phân số tối giản. Tính a+b+c
Đọc tiếp
Câu 1:
Xác đinh k để hàm: f(x)=\(\left\{{}\begin{matrix}\frac{x^{2016}+x-2}{\sqrt{2018x+1}-\sqrt{x+2018}}\\k\end{matrix}\right.\)liên tục tại 1
Câu 2: Cho \(lim\)(x-->1) \(\frac{x^2+ax+b}{x^2-1}=\frac{1}{2}\). Tổng S= \(a^2+b^2\) bằng bao nhiêu
Câu 3: lim(x->1) \(\frac{\sqrt{x^2+x+2}-\sqrt[3]{7x+1}}{\sqrt{2}\left(x-1\right)}=\frac{a\sqrt{2}}{b}+c\) với a/b là phân số tối giản. Tính a+b+c
\(\lim\limits_{x\rightarrow1}\frac{x^{2016}+x-2}{\sqrt{2018x+1}-\sqrt{x+2018}}=\lim\limits_{x\rightarrow1}\frac{2016x^{2015}+1}{\frac{1009}{\sqrt{2018x+1}}-\frac{1}{2\sqrt{x+2018}}}=\frac{2017}{\frac{1009}{\sqrt{2019}}-\frac{1}{2\sqrt{2019}}}=2\sqrt{2019}\)
Để hàm liên tục tại \(x=1\)
\(\Rightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=f\left(1\right)\Rightarrow k=2\sqrt{2019}\)
2.
\(\lim\limits_{x\rightarrow1}\frac{x^2+ax+b}{x^2-1}=\frac{1}{2}\Leftrightarrow\left\{{}\begin{matrix}a+b+1=0\\\lim\limits_{x\rightarrow1}\frac{2x+a}{2x}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-1\\\frac{a+2}{2}=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-1\\b=0\end{matrix}\right.\) \(\Rightarrow S=1\)
3.
\(\lim\limits_{x\rightarrow1}\frac{\sqrt{x^2+x+2}-2+2-\sqrt[3]{7x+1}}{\sqrt{2}\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{\left(x-1\right)\left(x+2\right)}{\sqrt{x^2+x+2}+2}-\frac{7\left(x-1\right)}{\sqrt[3]{\left(7x+1\right)^2}+2\sqrt[3]{7x+1}+4}}{\sqrt{2}\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{1}{\sqrt{2}}\left(\frac{x+2}{\sqrt{x^2+x+2}+2}-\frac{7}{\sqrt[3]{\left(7x+1\right)^2}+2\sqrt[3]{7x+1}+4}\right)\)
\(=\frac{1}{\sqrt{2}}\left(\frac{3}{4}-\frac{7}{12}\right)=\frac{\sqrt{2}}{12}\)
\(\Rightarrow a+b+c=1+12+0=13\)
Rút gọn:
a) \(A=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{1-\sqrt{x}}{1-x}\right)^2\left(x\ge0,x\ne1\right)\)
b) \(B=\left(\frac{2-a\sqrt{a}}{2-\sqrt{a}}+\sqrt{a}\right)\left(\frac{2-\sqrt{a}}{2-a}\right)\left(a\ge0,a\ne2,a\ne4\right)\)
c) \(C=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\left(x>0,x\ne1\right)\)
a) Ta có: \(A=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(=\left(\frac{1-x\sqrt{x}+\sqrt{x}\left(1-\sqrt{x}\right)}{1-\sqrt{x}}\right)\cdot\left(\frac{1}{1+\sqrt{x}}\right)^2\)
\(=\frac{1-x\sqrt{x}+\sqrt{x}-x}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-\left(x-1\right)\left(-1-\sqrt{x}\right)}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{\left(1+\sqrt{x}\right)\cdot\left(-1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-1\cdot\left(1+\sqrt{x}\right)^2}{\left(1+\sqrt{x}\right)^2}=-1\)
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