tích mình đi

Câu này mình giải không biết bao nhiêu lần hết. Bạn lục lại đi nha

http://olm.vn/hoi-dap/question/751345.html

a/ Ta có

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

= \(\frac{-\sqrt{x}}{1+\sqrt{x}+x}\)

mình muốn hỏi câu b cơ bạn ơi

a) Với  \(x\ge0;x\ne1\)

\(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(A=\frac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(A=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(A=\frac{15\sqrt{x}-11-\left(3x-9\sqrt{x}-2\sqrt{x}-6\right)-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(A=\frac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(A=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

Vậy : \(A=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

b) \(A=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}=\frac{-5\left(\sqrt{x}+3\right)+17}{\sqrt{x}+3}=-5+\frac{17}{\sqrt{x}+3}\)

\(A_{max}\Leftrightarrow\left(\frac{17}{\sqrt{x}+3}\right)_{max}\)

Vì \(x\ge0;x\ne1\Rightarrow\hept{\begin{cases}\sqrt{x}\ge0\\\frac{17}{\sqrt{x}+3}>0\end{cases}A_{max}\Leftrightarrow}\left(\sqrt{x}+3\right)_{min}\Leftrightarrow\sqrt{x}_{min}\Leftrightarrow x=0\)

Vậy : \(A_{max}=\frac{17}{3}\Leftrightarrow x=0\)

c,d chưa làm được .-.

c) Để \(A=\frac{1}{2}\)

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

<=> \(-10\sqrt{x}+4=\sqrt{x}+3\)

<=> \(-11\sqrt{x}=-1\)

<=> \(\sqrt{x}=\frac{1}{11}\)

<=> \(x=\frac{1}{121}\left(tm\right)\)

Vậy ...

d) \(A\le\frac{2}{3}\)

<=> \(\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\le\frac{2}{3}\)

<=> \(\frac{-5\sqrt{x}+2}{\sqrt{x}+3}-\frac{2}{3}\le0\)

<=> \(\frac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\sqrt{x}+9}\le0\)

<=> \(\frac{-17\sqrt{x}}{3\sqrt{x}+9}\le0\)

Vì \(\hept{\begin{cases}-17\sqrt{x}\le0\\3\sqrt{x}+9>0\end{cases}}\) \(\Rightarrow\frac{-17\sqrt{x}}{3\sqrt{x}+9}\le0\)(luôn đúng)

=> Ta có ĐPCM

1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4 
--> Pmin=4 khi x=4

2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1

=> M=2x²-8x+\(\sqrt{x^2-4x+5}\)+6=2(t²-5)+t+6

<=> M=2t²+t-4\(\ge\)2.1²+1-4=-1

M_min=-1 khi t=1 hay x=2