GTLN là \(\frac{1}{2}+\frac{\sqrt{2}}{4}+\frac{\sqrt{3}}{6}\) Sách mình ghi thế nhưng không có lời giải li ke nha

\(ĐKXĐ:0\le x\ne x\)

a) \(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{\left(1-x\right)^2}{2}\)

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

\(P=\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)

\(P=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)

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

b) \(P=-x+\sqrt{x}=-\left(x-2\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}.\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

\(\Rightarrow MAX_P=\frac{1}{4}\text{ khi }x=\frac{1}{4}\)

\(P=\frac{3\left(x+\sqrt{x}-3\right)}{x+\sqrt{x}-2}+\frac{\sqrt{x}+3}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\left(ĐKXĐ:x\ne1;x\ge0\right)\)

\(P=\frac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x+3}}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(P=\frac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{3x-8+5\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{3x-3\sqrt{x}+8\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{\left(3\sqrt{x}+8\right)\left(\sqrt{x-1}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}\)

b)Để \(P< \frac{15}{4}\)thì \(\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}< \frac{15}{4}\)

      Ta có:\(\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}< \frac{15}{4}\)

          \(\Leftrightarrow\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}-\frac{15}{4}< 0\)

           \(\Leftrightarrow\frac{12\sqrt{x}+32-15\sqrt{x}-30}{4\left(\sqrt{x}+2\right)}< 0\)

            \(\Leftrightarrow\frac{-\left(3\sqrt{x}+2\right)}{4\sqrt{x}+8}< 0\)

                 Vì \(x\ge0;x\ne1\)

                              Do đó \(0< 4\sqrt{x}+8\)

   Mà \(-\left(3\sqrt{x}+2\right)< 0\)

          Vậy \(P< \frac{15}{4}\left(đpcm\right)\)

c)Ta có:\(P=\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}\)

             \(\Leftrightarrow P=\frac{3\sqrt{x}+6+2}{\left(\sqrt{x}+2\right)}\)

             \(\Leftrightarrow P=\frac{3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)}+\frac{2}{2\sqrt{x}+2}\)

              \(\Leftrightarrow P=3+\frac{2}{\sqrt{x}+2}\)

Vì \(x\ge0;x\ne1\Rightarrow\frac{2}{\sqrt{x}+2}\le1\)

       Do đó \(P\le4\Leftrightarrow x=1\)

                Vậy Max P=4 khi x=1

P=3x+3√x−9(√x−1)(√x+2) +√x+3√x+2 −√x−2√x−1 

P=3x+3√x−9(√x−1)(√x+2) +(√x+3)(√x−1)(√x+2)(√x−1) −x−4(√x−1)(√x+2) 

P=3x+3√x−9+x+2√x−3−x+4(√x−1)(√x+2) 

P=3x−8+5√x(√x−1)(√x+2) 

P=3x−3√x+8√x−8(√x−1)(√x+2) 

P=(3√x+8)(√x−1)(√x−1)(√x+2) 

P=(3√x+8)(√x+2) 

b)Để P<154 thì (3√x+8)(√x+2) <154 

      Ta có:(3√x+8)(√x+2) <154 

          ⇔(3√x+8)(√x+2) −154 <0

           ⇔12√x+32−15√x−304(√x+2) <0

            ⇔−(3√x+2)4√x+8 <0

                 Vì x≥0;x≠1

                              Do đó 0<4√x+8

   Mà −(3√x+2)<0

          Vậy P<154 (đpcm)

c)Ta có:P=(3√x+8)(√x+2) 

             ⇔P=3√x+6+2(√x+2) 

             ⇔P=3(√x+2)(√x+2) +22√x+2 

              ⇔P=3+2√x+2 

Vì x≥0;x≠1⇒2√x+2 ≤1

       Do đó 

\(A=31-\sqrt{2x+7}\)

Ta có: điều kiện để có căn:\(\sqrt{2x+7}\) thì :\(2x+7\ge0\Rightarrow2x\ge-7\Rightarrow x\ge-3,5\)

Với mọi \(x\ge-3,5\) ta có:

\(\sqrt{2x+7}\ge0\)

\(\Rightarrow A=31-\sqrt{2x+7}\le31\)

Dấu "=" xảy ra khi:

\(\sqrt{2x+7}=0\Rightarrow2x=-7\Rightarrow x=-3,5\)

Vậy \(MAX_A=31\) khi \(x=-3,5\)

\(B=-9+\sqrt{7+x}\)

Ta có: điều kiện để có căn \(\sqrt{7+x}\) thì:

\(x\ge-7\)

Với mọi \(x\ge-7\) ta có:

\(\sqrt{7+x}\ge0\)

\(\Rightarrow-9+\sqrt{7+x}\ge-9\)
Dấu "=" xảy ra khi:

\(\sqrt{7+x}=0\Rightarrow x=-7\)

\(\Rightarrow MIN_B=-9\) khi \(x=-7\)

a, Sửa đề: Tìm GTLN của biểu thức

Vì \(\sqrt{2x+7}\ge0\) \(\Rightarrow-\sqrt{2x+7}\le0\)

\(\Rightarrow31-\sqrt{2x+7}\le31\)

Dấu ''='' xảy ra khi :

\(-\sqrt{2x+7}=0\Rightarrow2x+7=0\Rightarrow x=-3,5\)

Vậy \(A_{Max}=31\) khi và chỉ khi x = -3,5

b, Tìm GTNN của B

Giải: \(B=-9+\sqrt{7+x}=\sqrt{7+x}-9\)

Vì \(\sqrt{7+x}\ge0\Rightarrow\sqrt{7+x}-9\ge-9\)

Dấu ''='' xảy ra khi \(\sqrt{7+x}=0\Rightarrow x=-7\)

Vậy \(B_{Min}=-9\) khi x = -7

p/s: Lần sau gửi đề cẩn thận hơn ||^^

a) \(A=31-\sqrt{2x+7}\)

Ta có: \(-\sqrt{2x+7}\le0\forall x\)

\(\Rightarrow31-\sqrt{2x+7}\le31\forall x\)

Vậy MIN A = 31