1. This is the fastest way to get to the city center.

There is no faster way than this to get to the city center.

2. If anyone succeeds in solving the problems, it will probably be him.

He is the most likely person to succeed in solving the problems.

3. We arrived too late to see the first film.

We didn't arrive early enough to see the first film.

4. "Would you mind not smoking in here?"

I'd rather you didn't smoke in here.

5. He wrote the letter in two hours.

 It took him two hours to write the letter.

6. "Why don't we go out for a walk?"

 My father suggested going out for a while.

7. In spite of his age, Mr. Benson runs 8 miles before breakfast.

Though Mr.Benson is old, he runs 8 milies before breakfast.

8. I was very interested in our conversation.

 It was interesting to talk to you.

ko bt có đúng ko

đây nhé

\(\Delta'=\left(k-1\right)^2+4k=k^2+2k+1=\left(k+1\right)^2\ge0;\forall k\)

\(\Rightarrow\) Phương trình luôn có nghiệm với mọi k

b. Để phương trình có 2 nghiệm pb \(\Rightarrow k\ne-1\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(k-1\right)\\x_1x_2=-4k\end{matrix}\right.\)

Kết hợp với điều kiện đề bài ta được:

\(\left\{{}\begin{matrix}x_1+x_2=2\left(k-1\right)\\3x_1-x_2=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=2k-2\\4x_1=2k\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{k}{2}\\x_2=\dfrac{3k-4}{2}\end{matrix}\right.\)

Thế vào \(x_1x_2=-4k\)

\(\Rightarrow\dfrac{k}{2}.\left(\dfrac{3k-4}{2}\right)=-4k\)

\(\Leftrightarrow3k^2-4k=-16k\)

\(\Leftrightarrow3k^2+12k=0\Rightarrow\left[{}\begin{matrix}k=0\\k=-4\end{matrix}\right.\)

Gọi vận tốc xe thứ nhất, xe thứ 2 lần lượt là x ; y ( x ; y > 0 ) 

Theo bài ra ta có hpt \(\left\{{}\begin{matrix}x-y=10\\\dfrac{100}{y}-\dfrac{100}{x}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10+y\\\dfrac{100}{y}-\dfrac{100}{10+y}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=40\\x=50\end{matrix}\right.\left(tm\right)\)

Nhiệt lượng m₁ kg nước ở \(t_1=50^oC\) và bình nhôm tỏa ra khi hạ nhiệt độ xuống \(t_3=10^oC\) là:

\(Q_1=m_0c_{Al}\cdot\left(t_0-t_3\right)+m_1\cdot c_1\cdot\left(t_1-t_3\right)\)

      \(=0,26\cdot880\cdot\left(20-10\right)+m_1\cdot4200\cdot\left(50-10\right)\)

      \(=2288+168000m_1\left(J\right)\)

Nhiệt lượng m₂ kg nước ở \(t_2=0^oC\) thu vào khi tăng nhiệt độ lên \(t_3=10^oC\) là:

\(Q_2=m_2\cdot c_1\cdot\left(t_3-t_2\right)=m_2\cdot4200\cdot\left(10-0\right)=42000m_2\left(J\right)\)

Cân bằng nhiệt: \(Q_1=Q_2\)

\(\Rightarrow2288+168000m_1=42000m_2\left(1\right)\)

Mà \(m_1=1,5-m_2\left(2\right)\)

Thay \(\left(2\right)\) vào \(\left(1\right)\) ta được:

\(2288+168000\left(1,5-m_2\right)=42000m_2\)

\(\Rightarrow m_2=1,21kg\)

\(\Rightarrow m_1=1,5-1,21=0,29kg\)

\(T_2=???^oC\)

\(\Delta=\left(2m+3\right)^2-4m=4m^2+12m+9-4m=4m^2+8m+9=4\left(m+1\right)^2+5>0\)

Vậy pt luôn có 2 nghiệm pb 

Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=2m+3\\x_1x_2=m\end{matrix}\right.\)

\(\left(x_1+x_2\right)^2-2x_1x_2=9\Rightarrow\left(2m+3\right)^2-2m=9\)

\(\Leftrightarrow4m^2+12m+9-2m=9\Leftrightarrow4m^2+10m=0\Leftrightarrow2m\left(2m+5\right)=0\Leftrightarrow m=0;m=-\dfrac{5}{2}\)

Để phương trình đã cho có 2 nghiệm phân biệt thì \(\Delta=\left[-\left(2m+3\right)\right]^2-4.1.m=4m^2+12m+9-4m=4m^2+8m+9>0\)

\(\Leftrightarrow\left(4m^2+8m+4\right)+5>0\Leftrightarrow4\left(m^2+2m+1\right)+5>0\Leftrightarrow4\left(m+1\right)^2+5>0\)(luôn đúng)

Vậy pt đã cho luôn có 2 nghiệm phân biệt với mọi m.

\(\Rightarrow\hept{\begin{cases}x_1+x_2=-\frac{-\left(2m+3\right)}{1}=2m+3\\x_1x_2=\frac{m}{1}=m\end{cases}}\)

Lại có \(x_1^2+x_2^2=9\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=9\Leftrightarrow\left(2m+3\right)^2-2m=9\Leftrightarrow4m^2+12m+9-2m=9\)

\(\Leftrightarrow4m^2+10m=0\)\(\Leftrightarrow2m^2+5m=0\)\(\Leftrightarrow m\left(2m+5\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}m=0\\m=-\frac{5}{2}\end{cases}}\)

ủa má, đây đâu phải toán 9, toán 6 hoặc toán 7 mà

Trả lời:

a, \(\Delta=b^2-4ac=\left(-5\right)^2-4.2.1=17>0\)

=> pt có 2 nghiệm phân biệt 

Áp dụng hệ thức Vi-ét, ta có: \(\hept{\begin{cases}x_1+x_2=\frac{-b}{a}=\frac{-\left(-5\right)}{2}=\frac{5}{2}\\x_1x_2=\frac{c}{a}=\frac{1}{2}\end{cases}}\)  (*)

b, \(A=3x_1^2+3x_2^2-5x_1x_2+7=3\left(x_1^2+x_2^2\right)-5x_1x_2+7\)

\(=3\left(x_1^2+2x_1x_2+x_2^2-2x_1x_2\right)-5x_1x_2+7\)

\(=3\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-5x_1x_2+7\) (1)

Thay (*) vào (1), ta được:

\(A=3\left[\left(\frac{5}{2}\right)^2-2\cdot\frac{1}{2}\right]-5\cdot\frac{1}{2}+7=\frac{81}{4}\)

c, \(B=4x_1+4x_2-8x_1^2-8x_2^2-5=4\left(x_1+x_2\right)-8\left(x_1^2+x_2^2\right)-5\) 

\(=4\left(x_1+x_2\right)-8\left(x_1^2+2x_1x_2+x_2^2-2x_1x_2\right)-5\)

\(=4\left(x_1+x_2\right)-8\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-5\) (2)

Thay (*) vào (2), ta được:

\(B=4\cdot\frac{5}{2}-8\left[\left(\frac{5}{2}\right)^2-2\cdot\frac{1}{2}\right]-5=-37\)

d, \(C=2x_1^3+2x_2^3+5=2\left(x_1^3+x_2^3\right)+5\)

\(=2\left(x_1^3+3x_1^2x_2+3x_1x_2^2+x_2^3-3x_1^2x_2-3x_1x_2^2\right)+5\)

\(=2\left[\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\right]+5\) (3)

Thay (*) vào (3), ta được:

\(C=2\left[\left(\frac{5}{2}\right)^3-3\cdot\frac{1}{2}\cdot\frac{5}{2}\right]+5=\frac{115}{4}\)

Xét: \(\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\ge0\) nên ta có thể chứng minh được:

\(\left(a+b-c\right)\ge0;\left(b+c-a\right)\ge0;\left(c+a-b\right)\ge0\)

Đặt: \(x=a+b-c;y=b+c-a;z=c+a-b\)

\(\Rightarrow a=\frac{x+z}{2};b=\frac{x+y}{2};c=\frac{y+z}{2}\)

\(\Rightarrow64xyz\left(x+y+z\right)^3\le27\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2\)

Ta có:

\(3xyz\left(x+y+z\right)\le\left(xy+yz+zx\right)^2\)

\(\Rightarrow64\cdot3xyz\left(x+y+z\right)^3\le64\left(x+y+z\right)^2\left(xy+yz+zx\right)^2\)

Vậy ta cần chứng minh:

\(64\left(x+y+z\right)^2\left(xy+yz+zx\right)^2\le3\cdot27\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)

Lấy căn bậc 2 của 2 vế ta được:

\(9\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge8\left(x+y+z\right)\left(xy+yz+zx\right)\)

Đến đây bài toán được chứng minh.