Bài 2: 

a: Xét ΔBHA vuông tại H và ΔBHD vuông tại H có 

BH chung

HA=HD

Do đó: ΔBHA=ΔBHD

b: Ta có: ΔBHA=ΔBHD

nên \(\widehat{ABH}=\widehat{DBH}\)

hay BH là tia phân giác của góc ABD

hi vọng có ai giúp 

\(a,=\dfrac{12\times\left(4+6\right)}{24}=\dfrac{12\times10}{24}=\dfrac{120}{24}=5\\ b,=\dfrac{16\times\left(8-2\right)}{48}=\dfrac{16\times6}{48}=\dfrac{96}{48}=2\)

minh cung dang can gap lam day lam on di ma giup minh di !!! { minh biet on ban day !!! }

 <=> -12x + 60 + 21x = 5 

<=> 9x = -55

<=> x = -55/9

vậy x= -55/9

=> 35 - 15x + 14 +14 x = 0 => -x = - 49 => x = 49 

<=> -12x + 60 + 21x = 5 

<=> 9x = -55

<=> x = -55/ 9 

vậy x = -55/9

\(4,\\ 2.B=\sqrt{x}-1+\dfrac{2-2\sqrt{x}}{\sqrt{x}}\left(x>0\right)\\ B=\dfrac{x-\sqrt{x}+2-2\sqrt{x}}{\sqrt{x}}=\dfrac{x-3\sqrt{x}+2}{\sqrt{x}}\)

\(3.x=\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}=\left(3+\sqrt{2}\right)+\left(3-\sqrt{2}\right)=6\)

Thay vào B, ta được \(B=\dfrac{6-3\sqrt{6}+2}{\sqrt{6}}=\dfrac{6\sqrt{6}-18+2\sqrt{6}}{6}=\dfrac{4\sqrt{6}-9}{3}\)

\(4.B=0\Leftrightarrow\dfrac{x-3\sqrt{x}+2}{\sqrt{x}}=0\Leftrightarrow x-3\sqrt{x}+2=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\)

\(7.B\in Z\Leftrightarrow\dfrac{x-3\sqrt{x}+2}{\sqrt{x}}\in Z\Leftrightarrow\sqrt{x}-3+\dfrac{2}{\sqrt{x}}\in Z\\ \Leftrightarrow\dfrac{2}{\sqrt{x}}\in Z\Leftrightarrow\sqrt{x}\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{1;4\right\}\left(\sqrt{x}>0\right)\)

\(B=\left(\dfrac{1}{9}-\dfrac{5}{7}\right)+\dfrac{3}{6}+\left(\dfrac{-12}{17}+\dfrac{-1}{2}\right)+\dfrac{5}{9}.\)

\(B=\dfrac{1}{9}-\dfrac{5}{7}+\dfrac{1}{2}-\dfrac{12}{17}-\dfrac{1}{2}+\dfrac{5}{9}=\left(\dfrac{1}{9}+\dfrac{5}{9}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\dfrac{5}{7}-\dfrac{12}{17}=\dfrac{2}{3}-\dfrac{5}{7}-\dfrac{12}{17}=\dfrac{238-255-252}{357}=\dfrac{-269}{357}.\)