19.

\(\left(a+b\right)^2\le2\left(a^2+b^2\right)=4\Rightarrow-2\le a+b\le2\)

\(P=3\left(a+b\right)+ab=3\left(a+b\right)+\dfrac{\left(a+b\right)^2-\left(a^2+b^2\right)}{2}=\dfrac{1}{2}\left(a+b\right)^2+3\left(a+b\right)-1\)

Đặt \(a+b=x\Rightarrow-2\le x\le2\)

\(P=\dfrac{1}{2}x^2+3x-1=\dfrac{1}{2}\left(x+2\right)\left(x+4\right)-5\ge-5\) (đpcm)

Dấu "=" xảy ra khi \(x=-2\) hay \(a=b=-1\)

20.

Đặt \(P=2a+2ab+abc\)

\(P=2a+ab\left(2+c\right)\le2a+\dfrac{a}{4}\left(b+2+c\right)^2=2a+\dfrac{a}{4}\left(7-a\right)^2\)

\(P\le\dfrac{1}{4}\left(a^3-14a^2+57a-72\right)+18=18-\dfrac{1}{4}\left(8-a\right)\left(a-3\right)^2\le18\) (đpcm)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(3;2;0\right)\)

3 câu hỏi là zì vậy

a: \(Q=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-2\sqrt{x}\left(\sqrt{x}-2\right)-5\sqrt{x}-2}{x-4}:\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+2\right)^2}\)

\(=\dfrac{x+3\sqrt{x}+2-2x+4\sqrt{x}-5\sqrt{x}-2}{x-4}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)

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

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

b: Khi x=4-2căn 3 thì \(Q=\dfrac{\sqrt{3}-1+2}{\sqrt{3}-1-3}=\dfrac{\sqrt{3}+1}{\sqrt{3}-4}=\dfrac{-7-5\sqrt{3}}{13}\)

c: Q>1/6

=>Q-1/6>0

=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{1}{6}>0\)

=>\(\dfrac{6\sqrt{x}+12-\sqrt{x}+3}{6\left(\sqrt{x}-3\right)}>0\)

=>\(\dfrac{5\sqrt{x}+9}{6\left(\sqrt{x}-3\right)}>0\)

=>căn x-3>0

=>x>9

em tán thành 

vì Kỉ luật tạo nên sự đoàn kết, thống nhất, đảm bảo dân chủ được thực hiện một cách tốt nhất.

Có tinh thần, trách nhiệm trong học tập và công việc, luôn có ý thức vì tập thể.

1. A

2. D

3. B

4. B

5. C

6. B

7. B

8. C

9. C

10. A

11. B

12. C

1A

2D

3B

4B

5C

6B

7B

8C

9C

10A

11B

12C

\(\dfrac{1}{2}\left(x-2\right)+\dfrac{1}{3}\left(2-x\right)=x\\ \Leftrightarrow\dfrac{1}{2}\left(x-2\right)-\dfrac{1}{3}\left(x-2\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{3-2}{6}\right)=x\\ \Leftrightarrow\left(x-2\right).\dfrac{1}{6}=x\\ \Leftrightarrow\dfrac{1}{6}x-\dfrac{1}{3}-x=0\\ \Leftrightarrow\left(\dfrac{1}{6}-1\right)x=\dfrac{1}{3}\\ \Leftrightarrow\left(\dfrac{1-6}{6}\right)x=\dfrac{1}{3}\\ \Leftrightarrow\dfrac{-5}{6}x=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{3}:\left(-\dfrac{5}{6}\right)\\ \Leftrightarrow x=-\dfrac{2}{5}\)

Vậy \(x=-\dfrac{2}{5}\)

x= \(\dfrac{7\pm\sqrt{37}}{3}\) nha

d: =-76x10=-760

`15x^2-25x+18`

`=5.(3x^2-5x+6)+12`

`=5.2+12`

`=22`