\(a)ĐK:x\ge-1\\ \Leftrightarrow x+1=2\sqrt{x+1}\\ \Leftrightarrow x^2+2x+1=4x+4\\ \Leftrightarrow x^2+2x-4x+1-4=0\\ \Leftrightarrow x^2-2x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{3;-1\right\}\)

\(b)ĐK:x\ge2\\ \Leftrightarrow2x-4=\sqrt{x-2}\\ \Leftrightarrow4x^2-16x+16=x-2\\ \Leftrightarrow4x^2-16x-x+16+2=0\\ \Leftrightarrow4x^2-17x+18=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{9}{4};2\right\}\)

\(c)ĐK:x\ge3\\ \Leftrightarrow2\sqrt{9\left(x-3\right)}-\dfrac{1}{5}\sqrt{25\left(x-3\right)}-\dfrac{1}{7}\sqrt{49\left(x-3\right)}=20\\ \Leftrightarrow2.3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=20\\ \Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\\ \Leftrightarrow4\sqrt{x-3}=20\\ \Leftrightarrow\sqrt{x-3}=5\\ \Leftrightarrow x-3=25\\ \Leftrightarrow x=25+3\\ \Leftrightarrow x=28\left(tm\right)\)

Vậy \(S=\left\{28\right\}\)

a) ĐKXĐ: \(x\ge0;x\ne1\)

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

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

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

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

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

\(B=\dfrac{1}{x-1}\)

Đồ thị hàn số y = a\(x\) + b đi qua các điểm A (\(\sqrt{2}\); 4 - \(\sqrt{2}\)) vàB (2; \(\sqrt{2}\))

       Thay tọa độ điểm A, B vào pt đồ thị ta có:

        \(\left\{{}\begin{matrix}\sqrt{2}.a+b=4-\sqrt{2}\\2a+b=2+\sqrt{2}\end{matrix}\right.\) 

      Trừ vế cho vế ta có:  2a + b - (\(\sqrt{2}\)a + b) = 2 + \(\sqrt{2}\) - (4 - \(\sqrt{2}\))

         2a + b - \(\sqrt{2}\)a - b   =  -2 + 2\(\sqrt{2}\)

        2a - \(\sqrt{2}\)a             = - 2 + 2\(\sqrt{2}\)

        a.(2 - \(\sqrt{2}\))         =      -2 + 2\(\sqrt{2}\)

        a                       = (-2 + 2\(\sqrt{2}\)) : (2 - \(\sqrt{2}\))

         a  = \(\sqrt{2}\)

         b = 2 + \(\sqrt{2}\) -  2\(\sqrt{2}\) 

         b = 2 - \(\sqrt{2}\)  

mn giúp e vs ạ , e đg gấp ạ

\(A=4\sqrt{5}\) \(-3\sqrt{5}\)\(+9\sqrt{2}\)\(+12\sqrt{2}\)\(-5\sqrt{2}\)

    \(=\left(4-3\right)\sqrt{5}\)\(+\left(9+12-5\right)\sqrt{2}\)

    \(=\sqrt{5}+16\sqrt{2}\)

   cậu tự tính mấy ý còn lại nhá:333