\(\sqrt{3}-\sqrt{3}=0\) ở mẫu thì làm sao biểu thức trên xảy ra được ._.

Biết làm mỗi biểu thức A thui, sorry nha

Áp dụng tính chất \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)ta được:

\(A=\left|x-2014\right|+\left|x-2015\right|\\ =\left|x-2014\right|+\left|2015-x\right|\ge\left|x-2014+2015-x\right|=1\)

Vậy min A = 1 khi

\(\left(x-2014\right)\left(2015-x\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2014\ge0\\2015-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2014\le0\\2015-x\le0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2014\\x\le2015\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2014\\x\ge2015\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow2014\le x\le2015\)

(Nhờ mọi người ktr xem mình có sai chỗ nào ko :<)

\(A=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{4+2\sqrt{3}}}}\\ =\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{3+2\cdot\sqrt{3}\cdot1+1}}}\\ =\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\\ =\sqrt{6+2\sqrt{2}\cdot\sqrt{2-\sqrt{3}}}\\ =\sqrt{6+2\sqrt{4-2\sqrt{3}}}\\ =\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\sqrt{6+2\left(\sqrt{3}-1\right)}\\ =\sqrt{4+2\sqrt{3}}\\ =\sqrt{3+2\cdot\sqrt{3}\cdot1+1}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

Chia các động từ sau : (cái này dùng cấu trúc bị động là làm được)

1. That novel is written (write) by a famous artist .

2. The milk is delivered (deliver) every day.

3. Lan's car was repaired (repair) last month

4. Phong won't be gone (go/not) alone tonight by his family

5. My mother let me use (use) her computer.

6. His car needs repairing/ to be repaired .(repair)

7. The date of that meeting hasn't been changed (change/not) yet.

8. Look at this old house ! They are knocked down(knock)

9. That language was spoken (speak) by his grandfather years ago.

10. All the troubles were caused (cause) by his girlfriend yesterday

Task 4: Choose the suitable word in the box to fill each gap of the passage.

Behave, Lessons, Classes, Typing, Mark,
Strict, Teach, Prepare, Homework, Degree

Josephine is a teacher of English in a state secondary school. She is a graduate of Sussex University with a (1) _degree_ in English literature. When she graduated, she first worked in an office but she was bad at (2) _typing_ and soon got bored with the job. She decided to (3) __teach _, so she went to a teacher training college. Josephine teaches six different (4) __classes________ of children between the ages of 12 and 18.
The students enjoy her (5) _lessons_ but she finds it hard work. She gives children a lot of (6) _homework_ to do, and every evening she has to (7) _mark_ it and (8) _prepare_for the next day. One problem is that the children in her school don’t (9) _behave_ very well. They are often impolite. Josephine and the other teachers have to be very (10) _strict_ with them.

11.They have decided that the company will go to the beach together at the weekend.

=> It has been decided that the company will go to the beach together at the weekend.

a) ĐKXĐ \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

b) ĐKXĐ: \(\left\{{}\begin{matrix}a,b\ge0\\a\ne b\end{matrix}\right.\)

\(\frac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}+2\sqrt{ab}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}+2\sqrt{ab}=\sqrt{a}-\sqrt{b}+2\sqrt{ab}\)

a) Với m = 1 ta có

\(x^2-\left(m+5\right)x-m+6=0\\ \Leftrightarrow x^2-\left(1+5\right)x-1+6=0\\ \Leftrightarrow x^2-6x+5=0\)

Vì \(a+b+c=1+\left(-6\right)+5=0\)nên phương trình có 2 nghiệm phân biệt \(\left\{{}\begin{matrix}x_1=1\\x_2=\frac{c}{a}=5\end{matrix}\right.\)

b)

\(\Delta'=b'^2-ac=\left[-\left(m+5\right)\right]^2-\left(-m+6\right)\\=m^2+10m+25+m-6\\ \\=m^2+11m+19\\ =\left(m+\frac{11}{2}\right)^2-\frac{45}{4}\)

Để phương trình có 2 nghiệm thì

\(\Delta'\ge0\Leftrightarrow\left(m+\frac{11}{2}\right)^2-\frac{45}{4}\ge0\\ \Leftrightarrow\left(m+\frac{11}{2}\right)^2\ge\frac{45}{4}\\ \Leftrightarrow\left[{}\begin{matrix}m+\frac{11}{2}\ge\frac{3\sqrt{5}}{2}\\m+\frac{11}{2}\le-\frac{3\sqrt{5}}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}m\ge\frac{3\sqrt{5}-11}{2}\\m\le\frac{-3\sqrt{5}-11}{2}\end{matrix}\right.\)

Vì phương trình có nghiệm là x = -2 nên thay vào pt, ta có:

\(\left(-2\right)^2-\left(m+5\right)\cdot\left(-2\right)-m+6=0\\ \Leftrightarrow4+2m+10-m+6=0\Leftrightarrow m=-20\left(tm\right)\)

Thay -20 vào phương trình, ta được pt: \(x^2+15x+26=0\)

Áp dụng Viet, ta có \(x_1+x_2=-\frac{b}{a}\Leftrightarrow-2+x_2=-15\Rightarrow x_2-13\)

c)

Theo Viet, ta có: \(\left\{{}\begin{matrix}x_1+x_2=m+5\\x_1\cdot x_2=6-m\end{matrix}\right.\)

Ta có:

\(x^2_1\cdot x_2+x_1\cdot x^2_2=24\\ \Leftrightarrow x_1x_2\left(x_1+x_2\right)=24\\ \Leftrightarrow\left(6-m\right)\left(m+5\right)=24\\ \Leftrightarrow-m^2+m+30=24\\ \Leftrightarrow m^2-m-6=0\\ \Leftrightarrow\left(m-3\right)\left(m+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=3\left(tm\right)\\m=-2\left(ktm\right)\end{matrix}\right.\)

Vậy với m = 3 thì.......

P/s: Bạn xem lại chỗ điều kiện của m ở denta hộ mk nha

\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\\ =\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{27-2\cdot6\sqrt{6}+8}\\ =\sqrt{3^2-2\cdot3\cdot\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}\right)^2-2\cdot3\sqrt{3}\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}\\ =\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}\\ =3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)

Mình chỉ rút gọn được đến đó thôi, sorry :<<

Ủa sao chưa thi được vậy ???