có.

prevent sb from smt/Ving

không sai nhé.

who - whose

trước danh từ ( deafness ) không dùng who.

1. The tea was too warm for her to drink

2. I find it interesting to play soccer

3. If the boy eats too many cakes, he will feel sick

4. Mrs Mi got a plumber checked the pipes

5. If he didn't speak too fast, people would understand him

1. I live in a dormitory whose residents come from many countries

2. In spite of the fact that the test was very difficult, we could do it

3. They suggested that we should leave the place at once

28 song

29 musician

30 education

31 decision

32 scientists

\(\Delta=\left(2m+1\right)^2-4.2m\\ =4m^2+4m+1-8m\\ =4m^2-4m+1\\ =\left(2m-1\right)^2\ge0\forall m\)

⇒ pt có 2 nghiệm phân biệt \(\forall m\)

Áp dụng định lý Viet ta được:

\(x_1+x_2=2m+1\left(1\right)\\ x_1x_2=2m\left(2\right)\)

Biến đổi điều kiện đề bài:

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

Thay \(\left(1\right),\left(2\right)\) vào \(\left(3\right)\) ta được

\(A=\left(2m+1\right)^2-3.2m\\ =4m^2+4m+1-6m\\ =4m^2-2m+1\)

\(=\left(2m-\dfrac{1}{4}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall m\\ \Rightarrow A_{min}=\dfrac{3}{4}\Leftrightarrow m=\dfrac{1}{8}\)

Điều kiện xác định: \(\left\{{}\begin{matrix}5x^2+4x\ge0\\x^2-3x-18\ge0\\x\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\left(5x+4\right)\ge0\\\left(x-6\right)\left(x+3\right)\ge0\\x\ge0\end{matrix}\right.\)  \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge0\\x\le\dfrac{-4}{5}\end{matrix}\right.\\\left[{}\begin{matrix}x\ge6\\x\le-3\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow x\ge6\) (*)

Khi đó phương trình \(\Leftrightarrow\) \(\sqrt{5x^2+4x}=\sqrt{x^2-3x-18}+5\sqrt{x}\)

         \(\Leftrightarrow5x^2+4x=x^2+22x-18+10\sqrt{x\left(x^2-3x-18\right)}\\ \Leftrightarrow4x^2-18x+18=10\sqrt{x\left(x^2-3x-18\right)}\\ \Leftrightarrow5\sqrt{x\left(x-6\right)\left(x+3\right)}=2x^2-9x+9\\ \Leftrightarrow5\sqrt{\left(x^2-6x\right)\left(x+3\right)}=2\left(x^2-6x\right)+3\left(x+3\right)\left(1\right)\)

Đặt \(\left\{{}\begin{matrix}a=\sqrt{x^2-6x}\ge0\\b=\sqrt{x+3}\ge0\end{matrix}\right.\)

Khi đó pt \(\left(1\right)\) trở thành: \(2a^2+3b^2-5ab=0\\ \Leftrightarrow\left(a-b\right)\left(2a-3b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=b\\2a=3b\end{matrix}\right.\)

- TH1: \(a=b\Rightarrow x^2-6x=x+3\Leftrightarrow x^2-7x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{7+\sqrt{61}}{2}\left(tm\right)\\\dfrac{7-\sqrt{61}}{2}\left(ktm\right)\end{matrix}\right.\)

-TH2: \(2a=3b\Leftrightarrow4a^2=9b^2\\ \Leftrightarrow4\left(x^2-6x\right)=9\left(x+3\right)\\ \Leftrightarrow4x^2-33x-27=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9\left(tm\right)\\x=\dfrac{-3}{4}\left(ktm\right)\end{matrix}\right.\)

Vậy pt có 2 nghiệm \(x=\dfrac{7+\sqrt{61}}{2};x=9\)