a) \(D=\left(\dfrac{2}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)\)\(=\left(\dfrac{2}{x+2}-\dfrac{4}{\left(x+2\right)^2}\right):\left(\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right)\)

\(=\left(\dfrac{2\left(x+2\right)}{\left(x+2\right)^2}-\dfrac{4}{\left(x+2\right)^2}\right):\left(\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\right)\)

\(=\dfrac{2\left(x+2\right)-4}{\left(x+2\right)^2}:\dfrac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{2x+4-4}{\left(x+2\right)^2}:\dfrac{-x}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{2x}{\left(x+2\right)^2}.\dfrac{\left(x-2\right)\left(x+2\right)}{-x}\)

\(=\dfrac{-2.\left(x-2\right)}{x+2}\)

\(x^2-5x+6=0\\ \Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\\ \Rightarrow\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

\(P=\dfrac{-2.\left(x-2\right)}{x+2}\)

Thay \(x=2\), ta có:

\(P=\dfrac{-2.\left(2-2\right)}{2+2}\)

    \(=0\)

Thay \(x=3\), ta có:

\(P=\dfrac{-2.\left(3-2\right)}{3+2}\)

    \(=-\dfrac{2}{5}\)

D nguyên âm \(\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\left(x-2\right)< 0\\x+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}-2\left(x-2\right)>0\\x+2< 0\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -2\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x>2\\x< -2\end{matrix}\right.\)

a:Ta có: \(D=\left(\dfrac{2}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)\)

\(=\dfrac{2x+4-4}{\left(x+2\right)^2}:\dfrac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)

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

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

b: Ta có: \(x^2-5x+6=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

hay x=3

Thay x=3 vào D, ta được:

\(D=\dfrac{-\left(3-2\right)}{3+2}=-\dfrac{1}{5}\)

3: \(=7-4\sqrt{3}+7+4\sqrt{3}=14\)

4: \(=\left(\dfrac{\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}+3}{3}\right)\cdot\dfrac{1}{3+2\sqrt{2}}=\dfrac{1}{3}\)

\(f'\left(x\right)=2x^2-x\)

\(f'\left(x\right)\ge0\Leftrightarrow2x^2-x\ge0\)

\(\Leftrightarrow x\left(2x-1\right)\ge0\Rightarrow\left[{}\begin{matrix}x\ge\dfrac{1}{2}\\x\le0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}BD\perp AC\left(\text{hai đường chéo hình vuông}\right)\\SA\perp\left(ABCD\right)\Rightarrow SA\perp BD\end{matrix}\right.\)  \(\Rightarrow BD\perp\left(SAC\right)\) 

\(\Rightarrow BD\perp SC\)

Mặt khác \(BD\in\left(SBD\right)\Rightarrow\left(SBD\right)\perp\left(SAC\right)\)

b.

Từ A kẻ \(AH\perp SB\)

Ta có: \(\left\{{}\begin{matrix}AD\perp AB\\SA\perp\left(ABCD\right)\Rightarrow SA\perp AD\end{matrix}\right.\) \(\Rightarrow AD\perp\left(SAB\right)\Rightarrow AD\perp AH\)

\(\Rightarrow AH\) là đường vuông góc chung của AD và SB

\(\Rightarrow AH=d\left(SB;AD\right)\)

\(\dfrac{1}{AH^2}=\dfrac{1}{SA^2}+\dfrac{1}{AB^2}=\dfrac{2}{a^2}\Rightarrow AH=\dfrac{a\sqrt{2}}{2}\)

Gọi O là tâm đáy, từ O kẻ \(OK\perp SC\)

Mà \(BD\perp\left(SAC\right)\) theo câu a \(\Rightarrow BD\perp OK\)

\(\Rightarrow OK\) là đường vuông góc chung của SC và BD hay \(OK=d\left(SC;BD\right)\)

\(AC=AB\sqrt{2}=a\sqrt{2}\) ; \(SC=\sqrt{SA^2+AC^2}=a\sqrt{3}\)

\(OK=OC.sin\widehat{SCA}=\dfrac{1}{2}AC.\dfrac{SA}{SC}=\dfrac{a\sqrt{6}}{6}\)

- Hoc24

Áp dụng t/c dtsbn:

\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}=\dfrac{a+b-c+a-b+c-a+b+c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)

\(\Rightarrow\left\{{}\begin{matrix}a+b-c=c\\a-b+c=b\\-a+b+c=a\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\)

\(M=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{2a.2b.2c}{abc}=8\)

KNJNL

TL:588 tạ nha

588 nhá !

11d

12a

13c

14a

15a

16c

11D

12A

13C

14A

15A

16C