ĐKXĐ: \(x>0,x\ne4\)

B= \(\left(x-\sqrt{x}-2\right).\left(\dfrac{3}{\sqrt{x}-2}-\dfrac{4-\sqrt{x}}{x-2\sqrt{x}}\right)\)

= \(\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right).\dfrac{3\sqrt{x}-4+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

= \(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

= \(\dfrac{4\left(x-1\right)}{\sqrt{x}}\)

ĐKXĐ: \(x\ne\pm3,x\ne\dfrac{9}{2}\)

= \(\left[\dfrac{x}{2\left(x-3\right)}-\dfrac{x^2}{\left(x-3\right)\left(x+3\right)}+\dfrac{x}{2x-9}.\dfrac{3\left(x-3\right)-x}{x\left(x-3\right)}\right]\) : \(\dfrac{x^2-5x-6}{-2\left(x-3\right)\left(x+3\right)}\)

= \(\left[\dfrac{x}{2\left(x-3\right)}-\dfrac{x^2}{\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x-3}\right]:\dfrac{-\left(x^2-5x-6\right)}{2\left(x-3\right)\left(x+3\right)}\)

= \(\dfrac{x\left(x+3\right)-2x^2+2\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}:\dfrac{-\left(x^2-5x-6\right)}{2\left(x-3\right)\left(x+3\right)}\)

= \(\dfrac{-2\left(x^2-5x-6\right)\left(x-3\right)\left(x+3\right)}{-2\left(x^2-5x-6\right)\left(x-3\right)\left(x+3\right)}=1\)

ĐKXĐ: \(x\ge0,x\ne1\)

\(A=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)-1\)

= \(\dfrac{x+\sqrt{x}+1}{x+1}:\left(\dfrac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)-1\)

= \(\dfrac{\left(x+\sqrt{x}+1\right)\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)^2}-1\)

= \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}-1\)

= \(\dfrac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}\)

= \(\dfrac{x+2}{\sqrt{x}-1}\)

\(C=\left(\dfrac{1-\sqrt{3}}{1+\sqrt{3}}-\dfrac{1+\sqrt{3}}{1-\sqrt{3}}\right):\sqrt{108}\)

= \(\dfrac{\left(1-\sqrt{3}\right)^2-\left(1+\sqrt{3}\right)^2}{\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}:6\sqrt{3}\)

= \(\dfrac{1-2\sqrt{3}+3-1-2\sqrt{3}-3}{1-3}.\dfrac{1}{6\sqrt{3}}\)

= \(\dfrac{-4\sqrt{3}}{-2}.\dfrac{1}{6\sqrt{3}}\)

= \(\dfrac{2\sqrt{3}}{6\sqrt{3}}=\dfrac{1}{3}\)

Bài 1:

a/ ĐKXĐ: \(x\ge0,x\ne1\)

\(A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

= \(\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{2-3\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

= \(\dfrac{15\sqrt{x}-11+\left(\sqrt{x}+3\right)\left(2-3\sqrt{x}\right)-\left(\sqrt{x}-1\right)\left(2\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

= \(\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

= \(\dfrac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

= \(\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)

b/ Với \(x\ge0,x\ne1\)

Xét hiệu \(A-\dfrac{2}{3}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}-\dfrac{2}{3}\)

= \(\dfrac{3\left(2-5\sqrt{x}\right)-2\left(\sqrt{x}+3\right)}{3\left(\sqrt{x}+3\right)}\)

= \(\dfrac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}\)

Ta có: \(-17\sqrt{x}\le0\) với mọi \(x\ge0\)

\(3\left(\sqrt{x}+3\right)>0\) với mọi \(x\ge0\)

\(\Rightarrow\dfrac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}\le0\)

\(\Leftrightarrow A-\dfrac{2}{3}\le0\Leftrightarrow A\le\dfrac{2}{3}\) (đccm)

Vậy \(A\le\dfrac{2}{3}\)

a/

\(A=\dfrac{\sqrt{x}}{x+\sqrt{x}}+\dfrac{\sqrt{x}-1}{2\sqrt{x}}\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{x+\sqrt{x}}\right)\)

= \(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}-1}{2\sqrt{x}}\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\)

= \(\dfrac{1}{\sqrt{x}+1}+\dfrac{\sqrt{x}-1}{2\sqrt{x}}.\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

= \(\dfrac{1}{\sqrt{x}+1}+\dfrac{\sqrt{x}-1}{2\sqrt{x}}.\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

= \(\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

= \(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}}\)

b/ Với \(x>0,x\ne1\)

Để A=1 \(\Leftrightarrow\dfrac{1}{\sqrt{x}}=1\)

\(\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\left(ktm\right)\)

Vậy không có giá trị nào của x thỏa mãn A=1

a/ ĐKXĐ: \(x>0,x\ne1,x\ne2\)

b/

\(P=\left[\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right]:\left[\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right]\)

= \(\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

= \(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

= \(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3\sqrt{x}\left(\sqrt{x}-1\right)}\)

= \(\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)

c/ Với \(x>0,x\ne1,x\ne2\)

Để P=\(\dfrac{1}{4}\Leftrightarrow\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)

\(\Leftrightarrow4\left(\sqrt{x}-2\right)=3\sqrt{x}\)

\(\Leftrightarrow\sqrt{x}=8\)

\(\Leftrightarrow x=64\left(tm\right)\)

Vậy để \(P=\dfrac{1}{4}\) thì x=64

Để P=160\(\Leftrightarrow\) \(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=160\)

\(\Leftrightarrow\) \(\left(x+2\right)\left(x+3\right)\left(x-1\right)\left(x+6\right)\) =160

\(\Leftrightarrow\) \(\left(x^2+2x+3x+6\right)\left(x^2+6x-x-6\right)\) =160

\(\Leftrightarrow\) \(\left(x^2+5x+6\right)\left(x^2+5x-6\right)\)=160 (1)

Đặt \(t=x^2+5x+6\left(t\ge0\right)\), (1) \(\Leftrightarrow t\left(t-12\right)=160\)

\(\Leftrightarrow t^2-12t-160=0\)

\(\Leftrightarrow\left(t-20\right)\left(t+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t-20=0\\t+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=20\left(tm\right)\\t=-8\left(ktm\right)\end{matrix}\right.\)

\(\Leftrightarrow x^2+5x+6=20\)

\(\Leftrightarrow x^2+5x-14=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\)

Vậy để P=160 thì \(x=\left\{2;-7\right\}\)

\(\Leftrightarrow5x^3+3x^2+3x-2=\left(\dfrac{x^2}{2}+3x-\dfrac{1}{2}\right)^2\)

\(\Leftrightarrow5x^3+3x^2+3x-2=\dfrac{x^4}{4}+x^2\left(3x-\dfrac{1}{2}\right)+\left(3x-\dfrac{1}{2}\right)^2\)

\(\Leftrightarrow5x^3+3x^2+3x-2=\dfrac{x^4}{4}+3x^3-\dfrac{x^2}{2}+9x^2-3x+\dfrac{1}{4}\)

\(\Leftrightarrow20x^3+12x^2+12x-8=x^4+12x^3-2x^2+36x^2-12x+1\)

\(\Leftrightarrow x^4-8x^3+22x^2-24x+9=0\)

\(\Leftrightarrow\left(x^4-x^3\right)-\left(7x^3-7x^2\right)+\left(15x^2-15x\right)-\left(9x-9\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-7x^2+15x-9\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x^3-x^2\right)-\left(6x^2-6x\right)+\left(9x-9\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-1\right)\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x-3\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy pt có nghiệm \(x=\left\{1;3\right\}\)

Read the following passage and mark the letter A, B, C or D on your answer sheet to indicate the correct word or phrase that best fits each of the numbered blanks.

You will make the interview process easier for the employer if you volunteer relevant information about yourself. Think about how you want to present your (3)_______, experiences, education, work style, skills, and goals. Be prepared to supplement all your answers with examples that support the statements you make. It is also a good idea to review your résumé with a critical eye and identify areas that an employer might see as limitations or want further information. Think about how you can answer difficult questions (4)_______ and positively, while keeping each answer brief.

An interview gives the employer a (5)_______ to get to know you. While you do want to market yourself to the employer, answer each question with an honest response.

Never say anything negative about past experiences, employers, or courses and professors. Always think of something positive about an experience and talk about that. You should also be (6)_______. If you are genuinely interested in the job, let the interviewer know that.

One of the best ways to show you are interested in a job is to demonstrate that you have researched the organization prior to the interview. You can also (7)_______ interest by asking questions about the job, the organization, and its services and products.

Điền ô số 4

A. accurately

B. hardly

C. rightly

D. sharply