B nằm giữa A và C nên AB+BC=AC

=>AC=12+x

Theo đề, ta có: \(\dfrac{x+12}{x}=\dfrac{5}{2}\)

=>5x=2x+24

hay x=8

Thiếu B nha bn

a) Thay x = 81 vào A ta có:

\(A=\dfrac{4\sqrt{81}}{\sqrt{81}-5}=\dfrac{4\cdot9}{9-5}=\dfrac{4\cdot9}{4}=9\)

b) \(B=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+2}+\dfrac{5-2\sqrt{x}}{x+\sqrt{x}-2}\left(x\ne1;x\ge0\right)\)

\(B-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+2}+\dfrac{5-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(B=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\dfrac{5-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(B=\dfrac{x-4+\sqrt{x}-1+5-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

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

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

\(B=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)

c) \(\dfrac{A}{B}< 4\) khi

\(\dfrac{4\sqrt{x}}{\sqrt{x}-5}:\dfrac{\sqrt{x}}{\sqrt{x}+2}< 4\)

\(\Leftrightarrow\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-5}< 4\)

\(\Leftrightarrow\dfrac{4\sqrt{x}+8-4\left(\sqrt{x}-4\right)}{\sqrt{x}-5}< 0\)

\(\Leftrightarrow\dfrac{24}{\sqrt{x}-5}< 0\)

\(\Leftrightarrow\sqrt{x}-5< 0\)

\(\Leftrightarrow x< 25\)

Kết hợp với đk: 

\(0\le x< 5\)

a, bc^2 = ab^2 +ac^2 

      <=.> (ae+eb)^2   +(af+fc)^2

     <=.>AE^2 +2 AE.EB +EB^2 +AF^2+FC^2+2AF,FC 

<=> EF^2 +EB^2 +CF^2 +2.(EH^2+FH^2)

<=>EB^2 +CF^2 + AH ^2  + 2 AH^2 vì tứ giác EHAF là hcn suy ra AH =EF 

<=>EB^2 +CF^2+3 AH^2  (đpcm)

b, cb =2a là thế nào vậy

đề bài cho vậy