\(e,\dfrac{\sqrt{4x-1}}{\sqrt{7-2x}-2}\) có nghĩa \(\Leftrightarrow\left[{}\begin{matrix}4x-1\ge0\\7-2x\ne4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{1}{4}\\x\ne-\dfrac{3}{2}\end{matrix}\right.\) \(\Leftrightarrow x\ge\dfrac{1}{4}\)

\(d,\dfrac{\sqrt{2x-1}}{\sqrt{2x+17}+1}\) có nghĩa \(\Leftrightarrow\left[{}\begin{matrix}2x-1\ge0\\2x+17\ge0\end{matrix}\right.\)  \(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{1}{2}\\x\ge-\dfrac{17}{2}\end{matrix}\right.\) \(\Leftrightarrow x\ge\dfrac{1}{2}\)

 \(b,c,\dfrac{3}{\sqrt{2x-17}}\) có nghĩa \(\Leftrightarrow2x-17>0\Leftrightarrow x>\dfrac{17}{2}\)

\(a,\sqrt{2-5x}\) có nghĩa \(\Leftrightarrow2-5x\ge0\Leftrightarrow x\le\dfrac{2}{5}\)

Chiều dài hình chữ nhật là:

\(\sqrt{5^2-1^2}=5\left(m\right)\)

Diện tích hình chữ nhật là:

\(5\cdot1=5\left(m^2\right)\)

Đáp số: \(5m^2\)

Hạ đường cao AH của tam giác ABD => AH=14,4cm

Pytago => AD^2-AH^2=DH^2

            => DH^2=116,64

            => DH=10,8cm

HT lượng => HA^2=HB.HC

                => HB=HA^2/HB=14,4^2/10,8=19,2cm

=> BD=HD+HB=10,8+19,2=30m

Pytago => AB^2=AH^2+HB^2=576

            => AB=24cm

=> chu vi HCN ABCD là: 2(AB+AD)=2(18+24)=84(cm^2)

a. \(\dfrac{\sqrt{2}.\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{7}.\left(\sqrt{3}+\sqrt{5}\right)}=\dfrac{\sqrt{2}}{\sqrt{7}}=\sqrt{\dfrac{2}{7}}\)

d. \(\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}=\dfrac{\sqrt{5-2\sqrt{5}+1}}{\sqrt{5}-1}=\dfrac{\left(\sqrt{5}-1\right)^2}{\sqrt{5}-1}=\sqrt{5}-1\)

\(\sqrt{3-2\sqrt{2}}\)

\(\dfrac{1}{x}\)+  2\(\sqrt{x-8}\)

ĐK: \(x\) ≠ 0; \(x\) - 8 ≥ 0; ⇒ \(x\) ≥ 8 vậy \(x\) ≥ 8

Đk: 2-x ≥ 0 hay x ≤ 2

Đặt \(\sqrt{2-x}=t\) với t ≥ 0

PT tương đương

t -3t+ 4t = 16

\(\Leftrightarrow\)2t = 16

\(\Rightarrow\) t = 8 (TMĐK)

Vậy \(\sqrt{2-x}=8\)

2 - x = 64

vậy x = -62

\(\sqrt{2x-1}\) - \(\sqrt{8x-4}\) + \(\sqrt{50x-25}\) = 24 đk \(x\ge\dfrac{1}{2}\)

\(\sqrt{2x-1}\) - \(\sqrt{4.\left(2x-1\right)}\) + \(\sqrt{25.\left(2x-1\right)}\) = 24

\(\sqrt{2x-1}\) - 2\(\sqrt{2x-1}\) + 5\(\sqrt{2x-1}\) = 24

\(\sqrt{2x-1}\) (1 - 2 + 5) = 24

 4\(\sqrt{2x-1}\) = 24

   \(\sqrt{2x-1}\) = 24: 4

   \(\sqrt{2x-1}\) = 6

    \(2x-1=36\)

    2\(x\) = 37

      \(x=\dfrac{37}{2}\) (thỏa mãn)

Vậy \(x=\dfrac{37}{2}\)

\(\sqrt{x-1}\) - \(\sqrt{9x-9}\) + \(\sqrt{16x-16}\) = 4 (đk \(x\ge\)1)

\(\sqrt{x-1}-\) \(\sqrt{9\left(x-1\right)}\) + \(\sqrt{16\left(x-1\right)}\) = 4

\(\sqrt{x-1}\) - 3\(\sqrt{x-1}\) + \(4\sqrt{x-1}\) = 4  

 \(\sqrt{x-1}\)( 1 - 3 + 4 ) = 4

  \(\sqrt{x-1}\) . 2 = 4

  \(\sqrt{x-1}\) = 4 : 2

  \(\sqrt{x-1}\) = 2

   \(x-1\) =4

  \(x=4+1\)

 \(x=5\) (thỏa mãn)

Vậy \(x\) = 5 

\(\sqrt{x+3}\) + \(\sqrt{9x+27}\) - \(\sqrt{4x-12}\) = 10  đk  \(x+3\) ≥ 0 ⇒ \(x\) ≥ -3

\(\sqrt{x+3}\) + \(\sqrt{9\left(x+3\right)}\) - \(\sqrt{4\left(x+3\right)}\) = 10

\(\sqrt{x+3}\) + 3\(\sqrt{x+3}\) - 2\(\sqrt{x+3}\) = 10

(1 + 3 - 2)\(\sqrt{x+3}\) = 10

2\(\sqrt{x+3}\) = 10

   \(\sqrt{x+3}\) = 10: 2

   \(\sqrt{x+3}\) = 5

    \(x+3\) = 10

    \(x\) = 10 - 3

    \(x\) = 7 ( thỏa mãn) 

Vậy \(x\) = 7

\(\sqrt{x+2}\) + \(\sqrt{16x+32}\) - \(\sqrt{4x+8}\) = 16 (đk \(x\ge\) -2)

\(\sqrt{x+2}\) + \(\sqrt{16\left(x+2\right)}\) - \(\sqrt{4\left(x+2\right)}\) = 16

\(\sqrt{x+2}\) + 4\(\sqrt{x+2}\) - 2\(\sqrt{x+2}\) = 16

( 1 + 4 - 2)\(\sqrt{x+2}\) = 16

         3\(\sqrt{x+2}\) = 16

           \(\sqrt{x+2}\) = \(\dfrac{16}{3}\)

             \(x+2\) = \(\dfrac{256}{9}\)

             \(x\) = \(\dfrac{256}{9}\) - 2

            \(x\) = \(\dfrac{238}{9}\) (thỏa mãn)

Vậy \(x=\dfrac{238}{9}\)