a) ĐKXĐ: x\(\neq\)0, 9, 4

    A=\(\left(\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}+\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{x-4}{x-5\sqrt{x}+6}\right)\):\(\left(\dfrac{2\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)

    A=\(\left(\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}+\dfrac{x-9}{x-5\sqrt{x}+6}-\dfrac{x-4}{x-5\sqrt{x}+6}\right)\):\(\left(\dfrac{2\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)

    A=\(\dfrac{\left(\sqrt{x}+2\right)+\left(x-9\right)-\left(x-4\right)}{x-5\sqrt{x}+6}\bullet\dfrac{\sqrt{x}+1}{\left(2\sqrt{x}+2\right)-\sqrt{x}}\)

    A=\(\dfrac{\sqrt{x}+2+x-9-x+4}{x-5\sqrt{x}+6}\bullet\dfrac{\sqrt{x}+1}{2\sqrt{x}+2-\sqrt{x}}\)

    A=\(\dfrac{\sqrt{x}+\left(2-5\right)+\left(-x+x\right)}{x-5\sqrt{x}+6}\bullet\dfrac{\sqrt{x}+1}{\left(2\sqrt{x}-\sqrt{x}\right)+2}\)

    A=\(\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\bullet\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

    A=\(\dfrac{\sqrt{x}+1}{x-4}\)

b) ĐKXĐ: x\(\neq\)0, 9, 4

    \(\dfrac{1}{A}\le\dfrac{-5}{2}\)

    \(\dfrac{1}{\dfrac{\sqrt{x}+1}{x-4}}\le\dfrac{-5}{2}\)

    \(\dfrac{x-4}{\sqrt{x}+1}\le\dfrac{-5}{2}\)

    2x-8\(\le-5\sqrt{x}\)-5

    2x+5\(\sqrt{x}\le\)3

    0\(\le\)x\(\le\dfrac{1}{4}\)

Vậy 0\(\le\)x\(\le\dfrac{1}{4}\) để \(\dfrac{1}{A}\le\dfrac{-5}{2}\)

a) ĐKXĐ: x\(\neq\)0, 9, 4

    A=\(\left(\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}+\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{x-4}{x-5\sqrt{x}+6}\right)\):\(\left(\dfrac{2\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)

    A=\(\left(\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}+\dfrac{x-9}{x-5\sqrt{x}+6}-\dfrac{x-4}{x-5\sqrt{x}+6}\right)\):\(\left(\dfrac{2\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)

    A=\(\dfrac{\left(\sqrt{x}+2\right)+\left(x-9\right)-\left(x-4\right)}{x-5\sqrt{x}+6}\bullet\dfrac{\sqrt{x}+1}{\left(2\sqrt{x}+2\right)-\sqrt{x}}\)

    A=\(\dfrac{\sqrt{x}+2+x-9-x+4}{x-5\sqrt{x}+6}\bullet\dfrac{\sqrt{x}+1}{2\sqrt{x}+2-\sqrt{x}}\)

    A=\(\dfrac{\sqrt{x}+\left(2-5\right)+\left(-x+x\right)}{x-5\sqrt{x}+6}\bullet\dfrac{\sqrt{x}+1}{\left(2\sqrt{x}-\sqrt{x}\right)+2}\)

    A=\(\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\bullet\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

    A=\(\dfrac{\sqrt{x}+1}{x-4}\)

b) ĐKXĐ: x\(\neq\)0, 9, 4

    \(\dfrac{1}{A}\le\dfrac{-5}{2}\)

    \(\dfrac{1}{\dfrac{\sqrt{x}+1}{x-4}}\le\dfrac{-5}{2}\)

    \(\dfrac{x-4}{\sqrt{x}+1}\le\dfrac{-5}{2}\)

    2x-8\(\le-5\sqrt{x}\)-5

    2x+5\(\sqrt{x}\le\)3

    0\(\le\)x\(\le\dfrac{1}{4}\)

Vậy 0\(\le\)x\(\le\dfrac{1}{4}\) để \(\dfrac{1}{A}\le\dfrac{-5}{2}\)

a) ĐKXĐ: 0<x\(\ne\)1

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

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

    P=\(\dfrac{x\sqrt{x}+2x}{x\sqrt{x}+x-2\sqrt{x}}\)+\(\dfrac{2\sqrt{x}-2+x+2}{x\sqrt{x}+x-2\sqrt{x}}\)

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

    P=\(\dfrac{x\sqrt{x}+2x+2\sqrt{x}+x}{x\sqrt{x}+x-2\sqrt{x}}\)

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

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

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

    P=\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

b) ĐKXĐ: 0<x\(\ne\)1

     x=3+2\(\sqrt{2}\)

     x=(\(\sqrt{2}\))²+2\(\bullet\sqrt{2}\bullet\)1+1

     x=(\(\sqrt{2}\)+1)²

     Thay x=3+2\(\sqrt{2}\)(TM)vào P:

       P=\(\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}+1}{\sqrt{\left(\sqrt{2}+1\right)^2}-1}\)

       P=\(\dfrac{\left|\sqrt{2}+1\right|+1}{\left|\sqrt{2}+1\right|-1}\)

       P=\(\dfrac{\sqrt{2}+1+1}{\sqrt{2}+1-1}\)

       P=1+\(\sqrt{2}\)

Vậy P=1+\(\sqrt{2}\) khi x=3+2\(\sqrt{2}\)

a) ĐKXĐ: 0<x\(\ne\)1

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

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

    P=\(\dfrac{x\sqrt{x}+2x}{x\sqrt{x}+x-2\sqrt{x}}\)+\(\dfrac{2\sqrt{x}-2+x+2}{x\sqrt{x}+x-2\sqrt{x}}\)

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

    P=\(\dfrac{x\sqrt{x}+2x+2\sqrt{x}+x}{x\sqrt{x}+x-2\sqrt{x}}\)

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

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

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

    P=\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

b) ĐKXĐ: 0<x\(\ne\)1

     x=3+2\(\sqrt{2}\)

     x=(\(\sqrt{2}\))²+2\(\bullet\sqrt{2}\bullet\)1+1

     x=(\(\sqrt{2}\)+1)²

     Thay x=3+2\(\sqrt{2}\)(TM)vào P:

       P=\(\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}+1}{\sqrt{\left(\sqrt{2}+1\right)^2}-1}\)

       P=\(\dfrac{\left|\sqrt{2}+1\right|+1}{\left|\sqrt{2}+1\right|-1}\)

       P=\(\dfrac{\sqrt{2}+1+1}{\sqrt{2}+1-1}\)

       P=1+\(\sqrt{2}\)

Vậy P=1+\(\sqrt{2}\) khi x=3+2\(\sqrt{2}\)

A=2(4\(\sqrt{5}\)-7\(\sqrt{5}\)+6\(\sqrt{5}\))

A=             6\(\sqrt{5}\)

A=2(4\(\sqrt{5}\)-7\(\sqrt{5}\)+6\(\sqrt{5}\))

A=             6\(\sqrt{5}\)

Có ACD vuông tại A(ACBA)

  Nên tan =(TSLG của góc nhọn)

          tan ==

          33

Có CDphương BC(GT)

  Nên =90

+=

           =-

             90 -  33

                 57

Có BCA vuông tại A(ACBA)

  Nên tan =(TSLG của góc nhọn)

          tan 57  

          tan 5730AB

                    46    AB(m)

Có ACD vuông tại A(ACBA)

  Nên tan =(TSLG của góc nhọn)

          tan ==

          33

Có CDphương BC(GT)

  Nên =90

+=

           =-

             90 -  33

                 57

Có BCA vuông tại A(ACBA)

  Nên tan =(TSLG của góc nhọn)

          tan 57  

          tan 5730AB

                    46    AB(m)

Có ACD vuông tại A(ACBA)

  Nên tan =(TSLG của góc nhọn)

          tan ==

          33

Có CDphương BC(GT)

  Nên =90

+=

           =-

             90 -  33

                 57

Có BCA vuông tại A(ACBA)

  Nên tan =(TSLG của góc nhọn)

          tan 57  

          tan 5720AB

                    31    AB(m)

Có ACD vuông tại A(ACBA)

  Nên tan =(TSLG của góc nhọn)

          tan ==

          33

Có CDphương BC(GT)

  Nên =90

+=

           =-

             90 -  33

                 57

Có BCA vuông tại A(ACBA)

  Nên tan =(TSLG của góc nhọn)

          tan 57  

          tan 5720AB

                    31    AB(m)