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Read the following passage and mark the letter A, B, C or D to indicate the correct word or phrase that best fits each of the numbered blanks from 31 to 35.

The first question we might ask is: What can you learn in college that will help you in being an employee? The schools teach (31) ______many things of value to the future accountant, doctor or electrician. Do they also teach anything of value to the future employee? Yes, they teach the one thing that it is perhaps most valuable for the future employee to know. But very few students bother to learn it. This basic skill is the ability to organize and express ideas in writing and in speaking. This means that your success as an employee will depend on your ability to communicate with people and to (32) ______ your own thoughts and ideas to them so they will (33) ______understand what you are driving at and be persuaded.

Of course, skill in expression is not enough by itself. You must have something to say in the first place. The effectiveness of your job depends much on your ability to make other people understand your work as they do on the quality of the work itself.

Expressing one's thoughts is one skill that the school can (34) ______ teach. The foundations for skill in expression have to be laidearly: an interest in and an ear (35) ______ language; experience in organizing ideas and data, in brushing aside the irrelevant, and above all the habit of verbal expression. If you do not lay these foundations during your school years, you may never have an opportunity again.

Question 35

A. in

B. for

C. by

D. if

Read the following passage and mark the letter A, B, C, or D on your answer sheet to indicate the correct answer to each of the questions from 1 to 8.

  At 7pm on a dark, cold November evening, thousands of people are making their way across a vast car park. They're not here to see a film, or the ballet, or even the circus. They are all here for what is, bizarrely, a global phenomenon: they are here to see Holiday on Ice. Given that most people don't seem to be acquainted with anyone who's ever been, the show's statistics are extraordinary: nearly 300 million people have seen Holiday on Ice since it began in 1943; it is the most popular live entertainment in the world.

  But what does the production involve? And why are so many people prepared to spend their lives travelling round Europe in caravans in order to appear in it? It can't be glamorous, and it's undoubtedly hard work. The backstage atmosphere is an odd mix of gym class and workplace. A curtained-off section at the back of the arena is laughably referred to as the girls' dressing room, but is more accurately described as a corridor, with beige, cracked walls and cheap temporary tables set up along the length of it. Each girl has a small area littered with pots of orange make-up, tubes of mascara and long false eyelashes.

  As a place to work, it must rank pretty low down the scale: the area round the ice-rink is grey and mucky with rows of dirty blue and brown plastic seating and red carpet tiles. It's an unimpressive picture, but the show itself is an unquestionably vast, polished global enterprise: the lights come from a firm in Texas, the people who make the audio system are in California, but Montreal supplies the smoke effects; former British Olympic skater Robin Cousins is now creative director for the company and conducts a vast master class to make sure they're ready for the show's next performance.

  The next day, as the music blares out from the sound system, the cast start to go through their routines under Cousins' direction. Cousins says, The aim is to make sure they're all still getting to exactly the right place on the ice at the right time - largely because the banks of lights in the ceiling are set to those places, and if the skaters are all half a metre out they'll be illuminating empty ice. Our challenge,' he continues, 'is to produce something they can sell in a number of countries at the same time. My theory is that you take those things that people want to see and you give it to them, but not in the way they expect to see it. You try to twist it. And you have to find music that is challenging to the skaters, because they have to do it every night.

  It may be a job which he took to pay the rent, but you can't doubt his enthusiasm. “The only place you'll see certain skating moves is an ice show,” he says, “because you're not allowed to do them in competition. It's not in the rules. So the ice show world has things to offer which the competitive world just doesn't.” Cousin knows what he's talking about because he skated for the show himself when he stopped competing - he was financially unable to retire. He learnt the hard way that you can't put on an Olympic performance every night. “I'd be thinking, these people have paid their money, now do your stuff, and I suddenly thought”, "I really can't cope. I'm not enjoying it". The solution, he realised, was to give 75 per cent every night, rather than striving for the sort of twice-a-year excellence which won him medals.

          To be honest, for those of us whose only experience of ice-skating is watching top-class Olympic skaters, some of the movements can look a bit amateurish, but then, who are we to judge? Equally, it's impossible not to be swept up in the whole thing; well, you'd have to try pretty hard not to enjoy it.

Which of the following is the writer's conclusion of Holiday on Ice?

A. It is more enjoyable than Holiday on Ice.

B. It is hard to know who really enjoys it.

C. It is difficult to dislike it.

D. It requires more skills than Olympic ice-skating.

Câu trả lời:

Dễ thấy x=0 không là nghiệm của phương trình.

Xét x khác 0, chia cả 2 vế của phương trình cho \(x^2\ne0\) ta có:

\(x^2+\text{ax}+b+\dfrac{a}{x}+\dfrac{1}{x^2}=0\)

<=> \(\left(x^2+\dfrac{1}{x^2}\right)+a\left(x+\dfrac{1}{x}\right)+b=0\)

<=>\(\left(x+\dfrac{1}{x}\right)^2-2+a\left(a+\dfrac{1}{x}\right)+b=0\)(*)

Đặt \(y=x+\dfrac{1}{x}\)

Ta có: \(y^2-4=\left(x+\dfrac{1}{x}\right)^2-4=x^2+2.x.\dfrac{1}{x}+\dfrac{1}{x^2}-4.x.\dfrac{1}{x}\)

=\(x^2-2.x.\dfrac{1}{x}+\dfrac{1}{x^2}=\left(x-\dfrac{1}{x}\right)^2\ge0\) với mọi x khác 0

=>\(y^2\ge4\)

=>\(\left|y\right|\ge2\)

(*) trở thành: y2-2+ay+b=0

<=>\(2-y^2=ay+b\)

=>\(\left|2-y^2\right|=\left|ay+b\right|\)(1)

Ta có: \(0\le\left(a-by\right)^2\) (với mọi \(a\ne0\) , b, \(\left|y\right|\ge2\))

<=>\(0\le a^2-2aby+b^2y^2\)

<=>\(a^2y^2+2aby+b^2\le a^2y^2+a^2+b^2y^2+b^2\)

<=>\(\left(ay+b\right)^2\le\left(a^2+b^2\right)\left(y^2+1\right)\)

<=>\(\left|ay+b\right|\le\sqrt{a^2+b^2}\sqrt{y^2+1}\)(2)

Từ (1) và (2) => \(\left|2-y^2\right|\le\sqrt{a^2+b^2}\sqrt{y^2+1}\)

<=>\(\left(2-y^2\right)^2\le\left(a^2+b^2\right)\left(y^2+1\right)\)

<=>\(\left(a^2+b^2\right)^2\ge\dfrac{\left(2-y^2\right)^2}{y^2+1}\)(3) (vì y2+1>0 với mọi \(\left|y\right|\ge2\))

\(y^2\ge4\)

=> \(y^2-\dfrac{12}{5}\ge4-\dfrac{12}{5}=\dfrac{8}{5}\) > 0

=> \(\left(y^2-\dfrac{12}{5}\right)^2\ge\left(\dfrac{8}{5}\right)^2\)

<=>\(y^4-\dfrac{24}{5}y^2+\dfrac{144}{25}\ge\dfrac{64}{25}\)

<=>\(y^4-\dfrac{24}{5}y^2+\dfrac{16}{5}\ge0\)

<=>\(5y^4-24y^2+16\ge0\)

<=>\(20-20y^2+5y^4\ge4y^2+4\)

<=>\(5\left(4-4y^2+y^4\right)\ge4\left(y^2+1\right)\)

<=>\(5\left(2-y^2\right)^2\ge4\left(y^2+1\right)\)

<=>\(\dfrac{\left(2-y^2\right)^2}{y^2+1}\ge\dfrac{4}{5}\) (4) (vì y2+1>0 với mọi \(\left|y\right|\ge2\))

Từ (3) và (4)=> \(a^2+b^2\ge\dfrac{4}{5}\)

Vậy giá trị nhỏ nhất của a2+b2\(\dfrac{4}{5}\) khi và chỉ khi:

\(\left\{{}\begin{matrix}\left|y\right|=2\\a=by\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\\a=by\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}\left\{{}\begin{matrix}y=2\\a=2b\end{matrix}\right.\\\left\{{}\begin{matrix}y=-2\\a=-2b\end{matrix}\right.\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\a=-\dfrac{4}{5}\\b=\dfrac{-2}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\a=\dfrac{4}{5}\\b=\dfrac{-2}{5}\end{matrix}\right.\end{matrix}\right.\)(I)

Vì a > 0 nên trường hợp thứ nhất loại.

Do đó:\(\left(I\right)\)<=>\(\left\{{}\begin{matrix}x=-1\\a=\dfrac{4}{5}\\b=\dfrac{-2}{5}\end{matrix}\right.\)

Khi đó giá trị của a cần tìm là \(\dfrac{4}{5}.\)