x và y là 2 nghiệm của pt:\(t^2-10t+9=0\)

ta có:a+b+c=1-10+9=0

⇒Pt có 2 nghiệm phân biệt

t₁=1     : t₂=\(\dfrac{9}{1}\)=9

Vậy (1;9) hoặc (9;1) thì thỏa  \(\left\{{}\begin{matrix}x+y=10\\xy=9\end{matrix}\right.\)

Exercise 1: 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.

Nowadays more and more people are moving to live in the cities because of the facilities that they find there. Many people prefer living in cities (1)______ the opportunities of jobs, market places, big houses and buildings including schools and hospitals. But as the population in the city increases, the (2)______of cars, raw materials and others increase too. All this contributes to the pollution of the city and increases the size of the (3)______ faced by individuals. Our problems in the city contribute to the destruction of the city and the spread of corruption. Also, these problems are (4)______an atmosphere of discomfort and the inability of the individual to adapt and live in peace in his home town. However, there are many solutions (5)______can be used and applied to contribute to reducing the number of problems, such as increasing the number of police members, providing employment opportunities and spreading awareness among members of the society.

Exercise 2: Read the following passage and mark the letter A, B, C or D to indicate the correct word that best fits each of the numbered blanks.

Clinicians and health care staff work tirelessly to care for their patients in an increasingly complex, inefficient, and stressful environment. (1) _______, the structure, incentives, and culture of the system in which they work are often - perhaps usually - poorly aligned to support their efforts to respond to patients' needs as their core priority. Recognizing the imperative to centre on the patient, a learning health care system is one in (2) _______ patients and their families are key drivers of the design and operation of the learning process. When patients, their families, (3) _______ caregivers, and the public are full, active participants in care, health, the experience of care, and economic outcomes can be substantially improved.

Crossing the Quality Chasm underscores patient-centeredness as a core aim of the health care system, yet care often fails to (4) _______ this aim (IOM, 2001). Despite the Quality Chasm's call to action more than a decade ago, patient-centred care still is not the norm, and users continue to find the health care system uncoordinated and stressful to navigate. As the complexity of the system continues to grow with (5) _______ in science (Chapter 2), patient engagement takes on increased importance as a means of ensuring that patients can find the right care for their individual characteristics, needs, preferences, and circumstances.

(Adapted from https://www.ncbi.nlm.nih.gov/books)

1.   A. Consequently         B. Although         C. Therefore          D. However

2.   A. that                         B. whose             C. which                D. what

3.   A. others                     B. one                 C. another             D. other

4.   A. meet                       B. merge             C. focus                 D. fracture

5.   A. increases               B. fame                C. promotion         D. advances

(em ko biết có đúng ko nên mọi ng sửa dùm em nha)

yes ,madam

Câu 1

a, Thay x=25 vào biểu thức B ta có

B=\(\dfrac{\sqrt{25}-3}{\sqrt{25}-1}=\dfrac{5-3}{5-1}=\dfrac{2}{4}=\dfrac{1}{2}\)

b, Ta có M=\(A\cdot B\)

⇒\(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}\right)\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)

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

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

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

=\(\dfrac{3\sqrt{x}}{\sqrt{x}+3}\)

c,  Để M<\(\sqrt{M}\)

Thì\(\text{​​}\text{​​}\text{​​}\text{​​}\dfrac{3\sqrt{x}}{\sqrt{x}+3}< \sqrt{\dfrac{3\sqrt{x}}{\sqrt{x}+3}}\)

⇔\(\text{​​}\text{​​}\text{​​}\text{​​}\dfrac{3\sqrt{x}}{\sqrt{x}+3}< \dfrac{\sqrt{3\sqrt{x}\left(\sqrt{x}+3\right)}}{\sqrt{x}+3}\)

⇔\(\text{​​}\text{​​}\text{​​}\text{​​}3\sqrt{x}< \sqrt{3\sqrt{x}\left(\sqrt{x}+3\right)}\)

⇔\(\text{​​}\text{​​}\text{​​}\text{​​}9x< 3\sqrt{x}\left(\sqrt{x}+3\right)\)

⇔\(\text{​​}\text{​​}\text{​​}\text{​​}3\sqrt{x}< \sqrt{x}+3\)

⇔\(\text{​​}\text{​​}\text{​​}\text{​​}2\sqrt{x}< 3\)

⇔\(\text{​​}\text{​​}\text{​​}\text{​​}\sqrt{x}< \dfrac{3}{2}\)

⇒\(\left\{{}\begin{matrix}x\ge0\\x< \dfrac{9}{4}\end{matrix}\right.\)

⇒\(0\le x< \dfrac{9}{4}\)

Gọi vận tốc dự định của hai bố con bạn Dũng là x(km/h)(x>0).Đổi: 10 phút =\(\dfrac{1}{6}\)(h)

thời gian dự định đi về quê là \(\dfrac{60}{x}\)(h)

vận tốc đi trên \(\dfrac{1}{3}\)quãng đường là đường xấu hai bố con bạn Dũng là \(x-10\)(km/h)

Thời gian thực tế đi về quê là \(\dfrac{\dfrac{1}{3}\cdot60}{x-10}+\dfrac{\dfrac{2}{3}\cdot60}{x}\)(h)

Vì hai bố con bạn Dũng đã về tới quê chậm mất 10 phút so với dự kiến

Nên ta có pt sau:

\(\left(\dfrac{\dfrac{1}{3}\cdot60}{x-10}+\dfrac{\dfrac{2}{3}\cdot60}{x}\right)-\dfrac{1}{6}=\dfrac{60}{x}\)

⇔\(\dfrac{20}{x-10}+\dfrac{40}{x}-\dfrac{1}{6}=\dfrac{60}{x}\)

⇔\(20x+40\left(x-10\right)-\dfrac{1}{6}x\left(x-10\right)=60\left(x-10\right)\)

⇔\(-\dfrac{1}{6}x^2+\dfrac{5}{3}x+200=0\)

⇒\(\left[{}\begin{matrix}x=40\left(n\right)\\x=-30\left(l\right)\end{matrix}\right.\)

Vậy ......

Là người con trai, tôi chỉ muốn hỏi các bạn biết bao nhiêu về sở thích của các bạn.Hãy liệt kê ra nha (Nếu là 2k6 cho xin làm quen)

Giải hệ pt:\(\left\{{}\begin{matrix}x^3\left(3+2y\right)=8\\xy\left(y^2+3y+3\right)=4\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}3+2y=\dfrac{8}{x^3}\\xy^3+3xy^2+3xy=4\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}2y=\dfrac{8}{x^3}-3\\xy^3+3xy^2+3xy=4\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}y=\dfrac{\left(\dfrac{8}{x^3}-3\right)}{2}\\x\left(\dfrac{\left(\dfrac{8}{x^3}-3\right)}{2}\right)^3+3x\left(\dfrac{\left(\dfrac{8}{x^3}-3\right)}{2}\right)^2+3x\left(\dfrac{\left(\dfrac{8}{x^3}-3\right)}{2}\right)=4\left(1\right)\end{matrix}\right.\)

Rút (1) ra có: \(x\left(\dfrac{\left(\dfrac{8}{x^3}-3\right)}{2}\right)^3+3x\left(\dfrac{\left(\dfrac{8}{x^3}-3\right)}{2}\right)^2+3x\left(\dfrac{\left(\dfrac{8}{x^3}-3\right)}{2}\right)=4\)

⇔\(\left(\dfrac{8}{x^3}-3\right)^3\cdot x+\left(\dfrac{8}{x^3}-3\right)^2\cdot6x+\left(\dfrac{8}{x^3}-3\right)12x=32\)

⇔\(\left(\dfrac{512}{x^9}-\dfrac{576}{x^6}+\dfrac{216}{x^3}-27\right)\cdot x+\left(\dfrac{64}{x^6}-\dfrac{48}{x^3}+9\right)\cdot6x+\left(\dfrac{8}{x^3}-3\right)12x=32\)

⇔\(\dfrac{512}{x^8}-\dfrac{576}{x^5}+\dfrac{216}{x^2}-27x+\dfrac{384}{x^5}-\dfrac{288}{x^2}+54x+\dfrac{96}{x^2}-36x=32\)

⇔\(\dfrac{512}{x^8}-\dfrac{192}{x^5}+\dfrac{24}{x^2}-9x=32\)(đến đây em ko biết làm j luôn rồi)