Cho \(f\left(x\right)=\dfrac{1+\sqrt{1+x}}{x+1}+\dfrac{1+\sqrt{1-x}}{x-1}\) và \(a=\dfrac{\sqrt{3}}{2}\). Tính f(a)
Cho \(f\left(x\right)=\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\) và \(a=\dfrac{4}{\sqrt{3}+\dfrac{1}{\sqrt{3}}}\). Tính giá trị f(a)
\(a=\dfrac{4}{\sqrt{3}+\dfrac{1}{\sqrt{3}}}\)
\(=4:\dfrac{4\sqrt{3}}{3}\)
\(=\sqrt{3}\)
\(f\left(x\right)=\dfrac{\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1}}{\sqrt{\sqrt{3}+1}-\sqrt{\sqrt{3}-1}}\)
\(=\dfrac{\left(\sqrt{3}+1+\sqrt{3}-1+2\cdot\sqrt{2}\right)}{2}\)
\(=\sqrt{3}+\sqrt{2}\)
Bài 1:Cho \(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-10}{x-1}=5\) ,\(g\left(x\right)=\sqrt{f\left(x\right)+6}-2\sqrt[3]{f\left(x\right)-2}\)
Tính \(\lim\limits_{x\rightarrow1}\dfrac{1}{\left(\sqrt{x}-1\right)g\left(x\right)}\)
Bài 2: Cho \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2ax^2+30}-bx-5}{x^3-3x+2}=c\left(a;b;c\in R\right)\)
Tính giá trị \(P=a^2+b^2+36c\)
Bài 3: Cho a;b là các số nguyên dương. Biết \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{4x^2+ax}+\sqrt[3]{8x^3+2bx^2+3}\right)=\dfrac{7}{3}\)
Tinh P= a+2b
Bài 4:Cho a,b,c thuộc R với a>0 thỏa mãn
\(c^2+a=2\) và \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{ax^2+bx}-cx\right)=-3\)
Tính P= a+b+5c
Bài 5:
Mấy câu này khó nên mong các bạn giúp mình với. Mai mình phải kiểm tra rồi
Mấy câu này bạn cần giải theo kiểu trắc nghiệm hay tự luận nhỉ?
Làm tự luận thì hơi tốn thời gian đấy (đi thi sẽ không bao giờ đủ thời gian đâu)
Câu 1:
Kiểm tra lại đề, \(\lim\limits_{x\rightarrow1}\dfrac{1}{\left(\sqrt[]{x}-1\right)g\left(x\right)}\) hay một trong 2 giới hạn sau: \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[]{x}-1}{g\left(x\right)}\) hoặc \(\lim\limits_{x\rightarrow1}\dfrac{g\left(x\right)}{\sqrt[]{x}-1}\)
Vì đúng như đề của bạn thì \(\lim\limits_{x\rightarrow1}\dfrac{1}{\left(\sqrt[]{x}-1\right)g\left(x\right)}=\dfrac{1}{0}=\infty\), cả \(g\left(x\right)\) lẫn \(\sqrt{x}-1\) đều tiến tới 0 khi x dần tới 1
Cho hàm số :
\(f\left(x\right)=\left\{{}\begin{matrix}\dfrac{\sqrt{x^2-1}+\sqrt[3]{\left(x-1\right)^3}}{\sqrt{x-1}}\forall x>1\\\sqrt{2};.....x=1\\\dfrac{\sqrt[3]{x}-1}{\sqrt{2}-\sqrt{x+1}};....\left|x\right|< 1\end{matrix}\right.\)
Xét tính liên tục của hàm số tại x0=1
\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x^2-1}+\sqrt[3]{\left(x-1\right)^3}}{\sqrt{x-1}}=\lim\limits_{x\rightarrow1^+}\dfrac{\left(x^2-1\right)^{\dfrac{1}{2}}+x-1}{\left(x-1\right)^{\dfrac{1}{2}}}=\lim\limits_{x\rightarrow1^+}\dfrac{\dfrac{1}{2}\left(x^2-1\right)^{-\dfrac{1}{2}}.2+1}{\dfrac{1}{2}\left(x-1\right)^{-\dfrac{1}{2}}}\)
\(=\dfrac{1}{0}=+\infty\)
\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt[3]{x}-1}{\sqrt{2}-\sqrt{x+1}}=\lim\limits_{x\rightarrow1^-}\dfrac{\left(x-1\right)\left(\sqrt{2}+\sqrt{x+1}\right)}{[\left(\sqrt[3]{x}\right)^2+\sqrt[3]{x}+1]\left(1-x\right)}=\lim\limits_{x\rightarrow1^-}\dfrac{-\left(\sqrt{2}+\sqrt{1+1}\right)}{1+1+1}=-\dfrac{2\sqrt{2}}{3}\)
\(f\left(1\right)=\sqrt{2}\)
\(\lim\limits_{x\rightarrow1^-}f\left(x\right)\ne\lim\limits_{x\rightarrow1^+}f\left(x\right)\ne f\left(x\right)\)=> ham gian doan tai x=1
a, tính GT của đa thức \(f\left(x\right)=\left(x^4-3x+1\right)^{2016}\) tại \(x=9-\dfrac{1}{\sqrt{\dfrac{9}{4}-\sqrt{5}}}+\dfrac{1}{\sqrt{\dfrac{9}{4}+\sqrt{5}}}\)
b, so sánh \(\sqrt{2017^2-1}-\sqrt{2016^2-1}và\dfrac{2.2016}{\sqrt{2017^2-1}-\sqrt{2016^2-1}}\)
c, tính GTBT: \(sinx.cosx+\dfrac{sin^2x}{1+cotx}+\dfrac{cos^2x}{1+tanx}\)
d, biết \(\sqrt{5}\) là số hữu tỉ, hãy tìm các số nguyên a,b tm::
\(\dfrac{2}{a+b\sqrt{5}}-\dfrac{3}{a-b\sqrt{5}}=-9-20\sqrt{5}\)
a.
\(x=9-\dfrac{1}{\sqrt{\dfrac{9-4\sqrt{5}}{4}}}+\dfrac{1}{\sqrt{\dfrac{9+4\sqrt{5}}{4}}}\\ x=9-\dfrac{1}{\dfrac{\sqrt{5}-2}{2}}+\dfrac{1}{\dfrac{\sqrt{5}+2}{2}}\\ x=9-\left(\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}\right)=9-8=1\\ \Rightarrow f\left(x\right)=f\left(1\right)=\left(1-1+1\right)^{2016}=1\)
c.
\(=\sin x\cdot\cos x+\dfrac{\sin^2x}{1+\dfrac{\cos x}{\sin x}}+\dfrac{\cos^2x}{1+\dfrac{\sin x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^2x}{\dfrac{\sin x+\cos x}{\sin x}}+\dfrac{\cos^2x}{\dfrac{\sin x+\cos x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^3x}{\sin x+\cos x}+\dfrac{\cos^3x}{\sin x+\cos x}\\ =\sin x\cdot\cos x+\dfrac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x\cdot\cos x+\cos^2x\right)}{\sin x+\cos x}\\ =\sin x\cdot\cos x-\sin x\cdot\cos x+\sin^2x+\cos^2x\\ =1\)
d.
\(\dfrac{2}{a+b\sqrt{5}}-\dfrac{3}{a-b\sqrt{5}}=-9-20\sqrt{5}\\ \Leftrightarrow\dfrac{-a-5b\sqrt{5}}{\left(a+b\sqrt{5}\right)\left(a-b\sqrt{5}\right)}=-9-20\sqrt{5}\\ \Leftrightarrow\dfrac{a+5b\sqrt{5}}{a^2-5b^2}=9+20\sqrt{5}\\ \Leftrightarrow\left(9+20\sqrt{5}\right)\left(a^2-5b^2\right)=a+5b\sqrt{5}\\ \Leftrightarrow9\left(a^2-5b^2\right)+\sqrt{5}\left(20a^2-100b^2\right)-5b\sqrt{5}=a\\ \Leftrightarrow\sqrt{5}\left(20a^2-100b^2-5b\right)=9a^2-45b^2+a\)
Vì \(\sqrt{5}\) vô tỉ nên để \(\sqrt{5}\left(20a^2-100b^2-5b\right)\) nguyên thì
\(\left\{{}\begin{matrix}20a^2-100b^2-5b=0\\9a^2-45b^2+a=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}180a^2-900b^2-45b=0\\180a^2-900b^2+20a=0\end{matrix}\right.\\ \Leftrightarrow20a+45b=0\\ \Leftrightarrow4a+9b=0\Leftrightarrow a=-\dfrac{9}{4}b\\ \Leftrightarrow9a^2-45b^2+a=\dfrac{729}{16}b^2-45b^2-\dfrac{9}{4}b=0\\ \Leftrightarrow\dfrac{9}{16}b^2-\dfrac{9}{4}b=0\\ \Leftrightarrow b\left(\dfrac{9}{16}b-\dfrac{9}{4}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}b=0\\b=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\\a=9\end{matrix}\right.\)
Với \(\left(a;b\right)=\left(0;0\right)\left(loại\right)\)
Vậy \(\left(a;b\right)=\left(9;4\right)\)
Cho h/s f(x) liên tục và x/đ trên [-1 ; \(+\infty\)] và t/m : \(f\left(x+1\right)+3f\left(3x+2\right)-4f\left(4x+1\right)-f\left(2^x\right)=\dfrac{3}{\sqrt{x+1}+\sqrt{x+2}}\forall x\in\left[-1;+\infty\right]\)
Tính \(\int\limits^2_1\dfrac{f\left(x\right)}{x}dx\) = ?
Từ GT ta lấy tích phân 2 vế cận từ 0 đến 1 ; sẽ được :
\(\int\limits^1_0f\left(x+1\right)dx+\int\limits^1_03f\left(3x+2\right)dx-\int\limits^1_04f\left(4x+1\right)dx-\int\limits^1_0f\left(2^x\right)dx=\int\limits^1_0\dfrac{3dx}{\sqrt{x+1}+\sqrt{x+2}}\left(1\right)\)
\(\int\limits^1_0\dfrac{3dx}{\sqrt{x+1}+\sqrt{x+2}}=\int\limits^1_03\left(\sqrt{x+2}-\sqrt{x+1}\right)dx\) =
\(2\left[\left(x+2\right)\sqrt{x+2}-\left(x+1\right)\sqrt{x+1}\right]\dfrac{1}{0}\) = \(2+6\sqrt{3}-8\sqrt{2}\left(2\right)\)
Dễ thấy : \(\int\limits^1_0f\left(x+1\right)dx=\int\limits^2_1f\left(t\right)dt=\int\limits^2_1f\left(x\right)dx\)
\(\int\limits^1_03f\left(3x+2\right)dx=\int\limits^5_2f\left(t\right)dt=\int\limits^5_2f\left(x\right)dx\) (3)
\(\int\limits^1_04f\left(4x+1\right)=\int\limits^5_1f\left(t\right)dt=\int\limits^5_1f\left(x\right)dx\left(4\right)\)
\(\int\limits^1_0f\left(2^x\right)dx=\int\limits^2_1\dfrac{f\left(t\right)dt}{tln2}=\dfrac{1}{ln2}.\int\limits^2_1\dfrac{f\left(t\right)dt}{t}=\dfrac{1}{ln2}.\int\limits^2_1\dfrac{f\left(x\right)dx}{x}\) (5)
Thay (2) ; (3) ; (4) ; (5) vào (1) ta được :
\(\int\limits^2_1f\left(x\right)dx+\int\limits^5_2f\left(x\right)dx-\int\limits^5_1f\left(x\right)dx-\dfrac{1}{ln2}.\int\limits^2_1\dfrac{f\left(x\right)dx}{x}=2+6\sqrt{3}-8\sqrt{2}\)
\(\Leftrightarrow\int\limits^2_1\dfrac{f\left(x\right)dx}{x}=\left(2+6\sqrt{3}-8\sqrt{2}\right)ln2\)
Rút gọn các biểu thức sau:
\(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)
\(F=\left(\dfrac{3}{\sqrt{1}+x}+\sqrt{1-x}\right):\left(\dfrac{3}{\sqrt{1-x^2}}+1\right)\)
d) Ta có: \(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)
\(=\dfrac{5\sqrt{x}-6-2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{x-9+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{5\sqrt{x}-6-2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{x-3}\)
\(=\dfrac{3\sqrt{x}}{x-3}\)
f) Ta có: \(\left(\dfrac{3}{\sqrt{1+x}}+\sqrt{1-x}\right):\left(\dfrac{3}{\sqrt{1-x^2}}+1\right)\)
\(=\dfrac{3+\sqrt{1-x^2}}{\sqrt{1+x}}:\dfrac{3+\sqrt{1-x^2}}{\sqrt{1-x^2}}\)
\(=\dfrac{\sqrt{1-x^2}}{\sqrt{1+x}}=\sqrt{1-x}\)
Tìm x
a)\(\sqrt{x-1}=2\left(x\ge1\right)\)
b)\(\sqrt{3-x}=4\left(x\le3\right)\)
c)\(2.\sqrt{3-2x}=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\)
d)\(4-\sqrt{x-1}=\dfrac{1}{2}\left(x\ge1\right)\)
e)\(\sqrt{x-1}-3=1\)
f)\(\dfrac{1}{2}-2.\sqrt{x+2}=\dfrac{1}{4}\)
a)√x−1=2(x≥1)
\(x-1=4
\)
x=5
b)
\(\sqrt{3-x}=4\) (x≤3)
\(\left(\sqrt{3-x}\right)^2=4^2\)
x-3=16
x=19
a: Ta có: \(\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)
hay x=5
b: Ta có: \(\sqrt{3-x}=4\)
\(\Leftrightarrow3-x=16\)
hay x=-13
c: Ta có: \(2\cdot\sqrt{3-2x}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{3-2x}=\dfrac{1}{4}\)
\(\Leftrightarrow-2x+3=\dfrac{1}{16}\)
\(\Leftrightarrow-2x=-\dfrac{47}{16}\)
hay \(x=\dfrac{47}{32}\)
d: Ta có: \(4-\sqrt{x-1}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{7}{2}\)
\(\Leftrightarrow x-1=\dfrac{49}{4}\)
hay \(x=\dfrac{53}{4}\)
e: Ta có: \(\sqrt{x-1}-3=1\)
\(\Leftrightarrow\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=16\)
hay x=17
f:Ta có: \(\dfrac{1}{2}-2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow\sqrt{x+2}=\dfrac{1}{8}\)
\(\Leftrightarrow x+2=\dfrac{1}{64}\)
hay \(x=-\dfrac{127}{64}\)
F = \(\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\left(\dfrac{1}{x}+\dfrac{1}{y}\right).\dfrac{1}{x+y+2\sqrt{xy}}+\dfrac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\right]\)
\(F=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\dfrac{x+y}{xy}\cdot\dfrac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\dfrac{2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)
\(=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\dfrac{x+y+2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)
\(=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}\cdot xy=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
Cho biểu thức:
A=\(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
a) Tìm ĐKXĐ và rút gọn A
b) Tính giá trị của A khi x=\(3-2\sqrt{2}\)
a,\(ĐK:x>0,x\ne1,x\ne4\)
\(A=\left[\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\left[\dfrac{x-1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b,\(x=3-2\sqrt{2}=2-2\sqrt{2}+1=\left(\sqrt{2}-1\right)^2\)
\(=>A=\dfrac{\sqrt{2}-3}{3\sqrt{2}-3}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}-1>0\\\sqrt{x}-2>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>1\\x>4\end{matrix}\right.\) \(\Leftrightarrow x>4\)
\(A=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
\(=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b) Ta có \(x=3-2\sqrt{2}=2-2\sqrt{2}+1=\left(2-1\right)^2=1\)
Thay \(x=1\) vào \(A\), ta được:
\(A=\dfrac{\sqrt{1}-2}{3\sqrt{1}}=\dfrac{1-2}{3}=-\dfrac{1}{3}\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\notin\left\{1;4\right\}\end{matrix}\right.\)
Ta có: \(A=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
\(=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)