41 on

42 with/and

Tham khảo vế 1 c42

43 for

44 of

45 to

46 on

47 with

48 of

49 from

50 of

41. Roger: " I don't believe it. " - Anita: " ..................... "

→ Đáp án:                C. Neither do I   

42. Jill : " Do you know Lisa's an excellent pianist ? - Maria: " ..................... "

→ Đáp án:                B. Yes, she does 

43. Karen: " ............... this word in English ? " - Betsy: " C-O-Z-Y "

→ Đáp án:                B. How do you spell    

44. Customer: " I think my change is wrong. " - Shop assistant: " ................... "

→ Đáp án:                C. Sorry. I don't think I have any change  

45. Mandy: " I've got a toothache. " - George: " ................. " 

→ Đáp án:                D. Why don't you see the dentist ? 

a/ \(\lim\limits_{x\rightarrow3}\dfrac{\sqrt[3]{x^2-1}-2}{x-3}+\lim\limits_{x\rightarrow3}\dfrac{2-\sqrt[4]{1+5x}}{x-3}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{x^2-1-8}{\left(x-3\right)\left(\sqrt[3]{\left(x^2-1\right)^2}+2.\sqrt[3]{x^2-1}+4\right)}+\lim\limits_{x\rightarrow3}\dfrac{16-1-5x}{\left(x-3\right)\left(\sqrt[4]{\left(1+5x\right)^3}+2\sqrt[3]{\left(1+5x\right)^2}+4.\sqrt[3]{1+5x}+8\right)}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(\sqrt[3]{\left(x^2-1\right)^2}+2.\sqrt[3]{x^2-1}+4\right)}+\lim\limits_{x\rightarrow3}\dfrac{-5\left(x-3\right)}{\left(x-3\right)\left(\sqrt[4]{\left(1+5x\right)^3}+2\sqrt[3]{\left(1+5x\right)^2}+4\sqrt[3]{1+5x}+8\right)}\)

\(=\dfrac{3+3}{\sqrt[3]{\left(3^2-1\right)^2}+2.\sqrt[3]{3^2-1}+4}-\dfrac{5}{\sqrt[4]{\left(1+5.3\right)^3}+2\sqrt[3]{\left(1+5.3\right)^2}+4.\sqrt[3]{1+5.3}+8}=\dfrac{11}{32}\)

\(\Rightarrow a^2+b^2=1145\)

40/ 

\(L=\lim\limits_{x\rightarrow0}\dfrac{af\left(x\right)+b^n-b^n}{f\left(x\right)\left[\sqrt[n]{\left(af\left(x\right)+b^n\right)^{n-1}}+b.\sqrt[n]{\left(af\left(x\right)+b^n\right)^{n-2}}+....+b^{n-1}\right]}\)

\(L=\lim\limits_{x\rightarrow0}\dfrac{a}{\sqrt[n]{\left(af\left(x\right)+b^n\right)^{n-1}}+b.\sqrt[n]{\left(af\left(x\right)+b^n\right)^{n-2}}+...+b^{n-1}}\)

\(L=\lim\limits_{x\rightarrow0}\dfrac{a}{b^{n-1}+b^{n-1}++...+b^{n-1}}=\dfrac{a}{nb^{n-1}}\)

40/ 

\(\sqrt{1+ax}.\sqrt[3]{1+bx}+\sqrt[4]{1+cx}-1=\left(\sqrt{1+ax}-1\right)+\sqrt{1+ax}\left(\sqrt[3]{1+bx}-1\right)+\sqrt{1+ax}.\sqrt[3]{1+bx}.\left(\sqrt[4]{1+cx}-1\right)\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+ax}-1}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+ax}\left(\sqrt[3]{1+bx}-1\right)}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+ax}.\sqrt[3]{1+bx}\left(\sqrt[4]{1+cx}-1\right)}{x}\)

\(I_1=\lim\limits_{x\rightarrow0}\dfrac{1+ax-1}{x\left(\sqrt{1+ax}+1\right)}=\dfrac{a}{\sqrt{1+ax}+1}=\dfrac{a}{2}\)

\(I_2=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+ax}\left(1+bx-1\right)}{x\left(\sqrt[3]{\left(1+bx\right)^2}+\sqrt[3]{1+bx}+1\right)}=\dfrac{b\sqrt{1+ax}}{\sqrt[3]{\left(1+bx\right)^2+\sqrt[3]{1+bx}+1}}=\dfrac{b}{3}\)

\(I_3=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+ax}\sqrt[3]{1+bx}\left(1+cx-1\right)}{x\left(\sqrt[4]{\left(1+cx\right)^3}+\sqrt[3]{\left(1+cx\right)^2}+\sqrt[3]{1+cx}+1\right)}=\dfrac{c}{4}\)

\(\Rightarrow L=\dfrac{a}{2}+\dfrac{b}{3}+\dfrac{c}{4}\)

P/s: Thông cảm mình đang đau đầu nên làm hơi lâu :b

Gọi V là thể tích khi quay phần giới hạn bởi \(y=\dfrac{1}{x}\) ; x=1, y=0; Ox quanh Ox

\(\Rightarrow V=V_1+V_2\)

\(V=\pi\int\limits^5_1\dfrac{1}{x^2}dx=\dfrac{4\pi}{5}\)

\(V_1=\pi\int\limits^k_1\dfrac{1}{x^2}dx=-\dfrac{\pi}{x}|^k_1=\pi-\dfrac{\pi}{k}\)

\(\Rightarrow V_2=V-V_1=\dfrac{4\pi}{5}-\pi+\dfrac{\pi}{k}=\dfrac{\pi}{k}-\dfrac{\pi}{5}\)

\(\Rightarrow\pi-\dfrac{\pi}{k}=2\left(\dfrac{\pi}{k}-\dfrac{\pi}{5}\right)\)

\(\Rightarrow k=\dfrac{15}{7}\)

41. A 42. B 43.C 44.A 45.D

không có hình hả

hình A,B,C,D hay E

lượng mưa cao nhất là hình B

37 C

38 D 

39 B 

40 D 

41 B

\(b,N=\left(2x-1\right)^2-4\ge-4\\ N_{min}=-4\Leftrightarrow x=\dfrac{1}{2}\\ c,P=\left(2x-5\right)^2+6\left(2x-5\right)+9-4\\ P=\left(2x-5+3\right)^2-4=\left(2x-2\right)^2-4\ge-4\\ P_{min}=-4\Leftrightarrow x=1\\ d,Q=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\\ Q=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\\ Q_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

6a.

$M=x^2-x+1=(x^2-x+\frac{1}{4})+\frac{3}{4}$

$=(x-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}$

Vậy $M_{\min}=\frac{3}{4}$ khi $x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}$

ĐỀ:

 \(\dfrac{2}{x}-2+41=\dfrac{7}{2}\) 

ĐÁP ÁN:

 \(x=\dfrac{-4}{71}\)