Have you ever asked yourself why it is that we (1) work? If it is, as some people say, just a question of money, would you be prepared to do any job as long as you (2) ______________________ a lot, even if it involved working for long hours in appalling (3) ______________________? Or are you perhaps more interested in the (4) ______________________ you get when you feel that you are good at your job? It's true that there is a great sense of (5) ______________________ behind, for example, the creation of a well-made product, or the clinching of an important deal, though this might not be the (6) ______________________ if you were serving in a shop or delivering letters. Still, as long as the customers are satisfied, then you can (7) ______________________ yourself on a job well done, and feel that you have (8) ______________________ to the success of the company that (9) ______________________ you. On the other hand, would you feel that your work was more worthwhile if you received more praise from your employers? Would you feel happier if they paid you a (10) ______________________ or sent you to a sales conference in Tahiti? Or is praise unnecessary, as long as the job (11) ______________________ you with the company of other people, and the feeling that you belong to the group? Most of us are probably too (12) ______________________ working to wonder too much about this. One day perhaps we'll find that ideal job which gives life meaning, but until then we'll just get up every morning and keep on working.
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ĐK \(x\in\left\{1;\frac{2}{3}\right\}\)Với điều kiện đã cho phương trình đã ch tương đương với
\(2x\left(3x^2+x+2\right)+13x\left(3x^2-5x+2\right)=6\left(3x^2-5x+2\right)\left(3x^2+x+2\right)\)
\(\Leftrightarrow54x^4-117x^3+105x^2-78x+24=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-4\right)\left(9x^2-3x+6\right)=0\)
Phương trình đã cho có 2 nghiệm \(x=\frac{1}{2};x=\frac{3}{4}\)
Áp dụng định lý phân giác:
BDCD=ABAC=23⇒3BD=2CD=2(BC−BD)BDCD=ABAC=23⇒3BD=2CD=2(BC−BD)
⇔5BD=2BC⇒BD=25BC⇒BDBC=25⇔5BD=2BC⇒BD=25BC⇒BDBC=25
AE=35AD=35(AE+DE)⇒2AE=3DE⇒DEAE=23AE=35AD=35(AE+DE)⇒2AE=3DE⇒DEAE=23
Qua D kẻ đường thẳng song song AC cắt AE tại F
Áp dụng định lý Talet:
FDAK=FEKE=DEAE=23FDAK=FEKE=DEAE=23
Talet cho tam giác BCK: FDCK=BDBC=25FDCK=BDBC=25
⇒(FDAK):(FDCK)=(23):(25)⇔CKAK=53⇒(FDAK):(FDCK)=(23):(25)⇔CKAK=53
⇒CKAC−CK=53⇒3CK=5(24−CK)⇒CK=15⇒CKAC−CK=53⇒3CK=5(24−CK)⇒CK=15
AK=AC−CK=9
\(M=\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}\)
Vì xyz=1 nên \(x\ne0;y\ne0;z\ne0\)
Ta có \(\frac{1}{1+x+xy}=\frac{z}{\left(1+y+yz\right)xz}=\frac{xz}{z+xz+1}\)
Tương tự \(\frac{1}{1+y+yz}=\frac{xz}{\left(1+y+yz\right)xz}=\frac{xz}{xz+z+1}\)
Khi đó \(M=\frac{z}{z+xz+1}+\frac{xz}{xz+1+z}+\frac{1}{1+z+xz}=\frac{z+xz+1}{z+zx+1}=1\)
Tổng thời gian ông An đi và về ( không tính thời gian nghỉ ) = 15h50' - 9h30' - 50' = 5h30' = 11/2h
Gọi độ dài quãng đường AB là x ( km ; x > 0 )
Thời gian ông An đi từ A đến B = x/25 (h)
Thời gian ông An đi từ B về A = x/30 (h)
Tổng thời gian đi và về là 11/2h
=> Ta có phương trình : x/25 + x/30 = 11/2
<=> x( 1/25 + 1/30 ) = 11/2
<=> x.11/150 = 11/2
<=> x = 75 ( tm )
Vậy độ dài quãng đường AB là 75km