\(1.\sqrt{10,6^2-5,6^2}=\sqrt{\left(10,6-5,6\right)\left(10,6+5,6\right)}=\sqrt{5\times16,2}=\sqrt{81}=9\)

\(2.\sqrt{20+9+2.3.2\sqrt{5}}+\sqrt{20+9-2.3.2\sqrt{5}}=\sqrt{\left(2\sqrt{5}+3\right)^2}+\sqrt{\left(2\sqrt{5}-3\right)^2}\)

\(=2\sqrt{5}+3+2\sqrt{5}-3=4\sqrt{5}\)

\(3.\frac{\sqrt{10}+\sqrt{26}}{2\sqrt{5}+\sqrt{52}}=\frac{\sqrt{2}\left(\sqrt{5}+\sqrt{13}\right)}{2\left(\sqrt{5}+\sqrt{13}\right)}=\frac{1}{\sqrt{2}}\)

\(4.\left(1-\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)=1-\left(\sqrt{2}-\sqrt{3}\right)^2=1-\left(5-2\sqrt{6}\right)=2\sqrt{6}-4\)

\(5.\left(\sqrt{5+\sqrt{21}}+\sqrt{5-\sqrt{21}}\right)^2=10+2\sqrt{5+\sqrt{21}}.\sqrt{5-\sqrt{21}}\)

\(=10+2\sqrt{25-21}=14\)

dk : \(\frac{3}{2}\le x\le\frac{5}{2}\)

\(\sqrt{2x-3}=2-\sqrt{5-2x}\)

\(\Leftrightarrow\sqrt{2x-3}+\sqrt{5-2x}-2=0\)

\(\Leftrightarrow\frac{\left(\sqrt{2x-3}-1\right)\left(\sqrt{2x-3}+1\right)}{\sqrt{2x-3}+1}+\frac{\left(\sqrt{5-2x}-1\right)\left(\sqrt{5-2x}+1\right)}{\sqrt{5-2x}+1}=0\)

\(\Leftrightarrow\frac{2x-3-1}{\sqrt{2x-3}+1}+\frac{5-2x-1}{\sqrt{5-2x}+1}=0\)

\(\Leftrightarrow\frac{2x-4}{\sqrt{2x-3}+1}-\frac{2x-4}{\sqrt{5-2x}+1}=0\)

\(\Leftrightarrow\left(2x-4\right)\left(\frac{1}{\sqrt{2x-3}+1}-\frac{1}{\sqrt{5-2x}+1}\right)=0\)

th1 : 2x - 4 = 0<=> x = 2 (thỏa mãn)

th2 : \(\frac{1}{\sqrt{2x-3}+1}-\frac{1}{\sqrt{5-2x}+1}=0\)

\(\Leftrightarrow\frac{1}{\sqrt{2x-3}+1}=\frac{1}{\sqrt{5-2x}+1}\)

\(\Leftrightarrow\sqrt{2x-3}+1=\sqrt{5-2x}+1\)

\(\Leftrightarrow\sqrt{2x-3}=\sqrt{5-2x}\)

\(\Leftrightarrow x=2\left(tm\right)\)

vậy x = 2

mấy dạng này nếu nhẩm đc nghiệm thì liên hợp nka b

Tham khảo ạ!

Thấy ảnh ko ạ?

( Không thấy ib )

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\left(\sqrt{n+1}\right)^2\sqrt{n}+\left(\sqrt{n}\right)^2\sqrt{n+1}}\)

\(=\frac{1}{\sqrt{n}\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n}\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)(đpcm)

đk : \(\hept{\begin{cases}x\ge5\\x\le5\end{cases}}\Rightarrow x=5\)

thay x = 5 vào pt ta đc : 0 + 0 = 1

=> 0 = 1 (vô lí)

vậy phương trình vô nghiệm

ĐK : \(\hept{\begin{cases}\sqrt{x-5}\ge0\\\sqrt{5-x}\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-5\ge0\\5-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge5\\x\le5\end{cases}}\Leftrightarrow x=5\)

Dễ thấy với x = 5 thì không xảy ra đẳng thức

Vậy pt đã cho vô nghiệm

P₁ = (\(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\)) : \(\frac{\sqrt{x}}{x+\sqrt{x}}\)= \(\frac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\):\(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)=\(\frac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\).

(\(\sqrt{x}+1\)) =\(\frac{x+\sqrt{x}+1}{\sqrt{x}}\)(ĐKXĐ : x > 0 )

P₂ =\(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)=\(\frac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)= \(\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)= \(\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)=\(\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)=\(\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

(ĐKXĐ: x\(\ge\)0,  x\(\ne\)1)

dạng này dễ mà bạn 

bạn tìm ĐK, đối chiếu giá trị với ĐK thấy thỏa mãn rồi thay vô 

toàn SCP nên tính cũng đơn giản:)

1) Thay x = 64 (TMĐK ) vào A, có :

           A = \(\frac{\sqrt{64}}{\sqrt{64}-2}\)=\(\frac{4}{3}\)

     Vậy A = \(\frac{4}{3}\)khi x = 64

2)  Thay x = 36 ( TMĐK ) vào A, có

        A =\(\frac{\sqrt{36}+4}{\sqrt{36}+2}\)=\(\frac{5}{4}\)

     Vậy A =\(\frac{5}{4}\)khi x = 36

3)   Thay x=9 (TMĐK  ) vào A, có :

         A= \(\frac{\sqrt{9}-5}{\sqrt{9}+5}\)=  \(\frac{-1}{4}\)

     Vậy A=\(\frac{-1}{4}\)khi x = 9

4)   Thay x = 25( TMĐK ) vào A có:

         A =\(\frac{2+\sqrt{25}}{\sqrt{25}}\)=\(\frac{7}{5}\)

      Vậy A=\(\frac{7}{5}\) khi x = 25