a) DK : x > 0; x khác 1

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

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

c )  \(Q=\frac{2\sqrt{x}}{P}=\frac{2\sqrt{x}}{x-\sqrt{x}+1}\)

<=> \(xQ-\left(Q+2\right)\sqrt{x}+Q=0\)(1)

TH1: Q = 0 => x = 0 loại

TH2: Q khác 0

(1) là phương trình bậc 2 với tham số Q ẩn x.

(1) có nghiệm <=> \(\left(Q+2\right)^2-4Q^2\ge0\)

<=> \(-3Q^2+4Q+4\ge0\)

<=> \(-\frac{2}{3}\le Q\le2\)

Vì Q nguyên và khác 0 nên Q =  1 hoặc Q = 2

Với Q = 1 => \(x-3\sqrt{x}+1=0\)

<=> \(\sqrt{x}=\frac{3}{2}\pm\frac{\sqrt{5}}{2}\)----> Tìm được x 

Với Q = 2 => \(2x-4\sqrt{x}+1=0\Leftrightarrow\sqrt{x}=1\pm\frac{1}{\sqrt{2}}\)-----> tìm đc x.

Tự làm tiếp nhé! Kiểm tra lại đề bài câu b.

xin lỗi em mới lớp 8 ko trả lời dc

\(ĐKXĐ:x\ge0\)

Ta có : \(D=\frac{2011x-2\sqrt{x}+1}{\sqrt{x}}=2011\sqrt{x}+\frac{1}{\sqrt{x}}-2\)

Theo BĐT AM - GM ta có :

\(2011\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{2011\sqrt{x}\cdot\frac{1}{\sqrt{x}}}=2\sqrt{2011}\)

\(\Rightarrow2011\sqrt{x}+\frac{1}{\sqrt{x}}-2\ge2\left(\sqrt{2011}-1\right)\)

Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{2011}\)

Vậy \(D_{min}=2\left(\sqrt{2011}-1\right)\) tại \(x=\frac{1}{2011}\)

Bài 1:

\(A=\sqrt{8}-2\sqrt{2}+\sqrt{20}-2\sqrt{5}-2=2\sqrt{2}-2\sqrt{2}+2\sqrt{5}-2\sqrt{5}-2=-2\)\(B=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)

Bài 2:

\(a,P=\left(\frac{1}{x+\sqrt{x}}-\frac{1}{\sqrt{x}+1}\right):\frac{\sqrt{x}}{x+2\sqrt{x}+1}\left(x>0\right)\)

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

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

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

\(=\frac{1-x}{x}\)

\(b,\forall x>0\Leftrightarrow\frac{1-x}{x}>\frac{1}{2}\)

\(\Leftrightarrow2\left(1-x\right)>x\)

\(\Leftrightarrow2-2x>x\)

\(\Leftrightarrow-3x>-2\)

\(\Leftrightarrow x< \frac{2}{3}\)

\(\Rightarrow P>\frac{1}{2}\Leftrightarrow\forall0< x< \frac{2}{3}\)

\(x=\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}-\frac{1}{8}\sqrt{2}\)

\(\Leftrightarrow x+\frac{\sqrt{2}}{8}=\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}\)

\(\Leftrightarrow\left(x+\frac{\sqrt{2}}{8}\right)^2=\frac{1}{4}\left(\sqrt{2}+\frac{1}{8}\right)\)

\(\Leftrightarrow x^2+\frac{x\sqrt{2}}{4}+\frac{1}{32}=\frac{\sqrt{2}}{4}+\frac{1}{32}\)

\(\Leftrightarrow x^2+\frac{x\sqrt{2}}{4}-\frac{\sqrt{2}}{4}=0\)

\(\Leftrightarrow4x^2+x\sqrt{2}-\sqrt{2}=0\)(1)

\(\Leftrightarrow x\sqrt{2}=\sqrt{2}-4x^2\)

\(\Leftrightarrow x=1-2x^2\sqrt{2}\)

Thay vào M ta sẽ được

\(M=x^2+\sqrt{x^4+1-2x^2\sqrt{2}+1}\)

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

     \(=x^2+\left|x^2-\sqrt{2}\right|\)

Từ \(\left(1\right)\Rightarrow\sqrt{2}-x\sqrt{2}=4x^2\ge0\)

           \(\Leftrightarrow\sqrt{2}\left(1-x\right)\ge0\)

           \(\Leftrightarrow x\le1\)

           \(\Leftrightarrow x^2\le1< \sqrt{2}\)

           \(\Rightarrow\left|x^2-\sqrt{2}\right|=\sqrt{2}-x^2\)

Khi đó \(M=x^2+\left|x^2-\sqrt{2}\right|=x^2-\sqrt{2}+x^2=\sqrt{2}\)

|N|

Đề gốc là \(P=\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{z}}+\frac{z}{\sqrt{x}}\)

\(\frac{P}{4}=\frac{x}{2.2\sqrt{y}}+\frac{y}{2.2\sqrt{z}}+\frac{z}{2.2\sqrt{x}}\)

Áp dụng BĐT Côsi:

\(2.2.\sqrt{x}\le x+2^2=x+4\)

\(\Rightarrow\frac{P}{4}\ge\frac{x}{y+4}+\frac{y}{z+4}+\frac{z}{x+4}=\frac{x^2}{xy+4x}+\frac{y^2}{yz+4y}+\frac{z^2}{zx+4z}\)

\(\ge\frac{\left(x+y+z\right)^2}{xy+yz+zx+4\left(x+y+z\right)}\ge\frac{\left(x+y+z\right)^2}{\frac{1}{3}\left(x+y+z\right)^2+4\left(x+y+z\right)}=\frac{3\left(x+y+z\right)}{\left(x+y+z\right)+12}\)

\(=3-\frac{36}{x+y+z+12}\ge3-\frac{36}{12+12}=\frac{3}{2}\)

\(\Rightarrow P\ge6\)

Dấu bằng xảy ra khi \(x=y=z=4\)

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

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

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

\(=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)

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

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

\(x=4\Rightarrow B=\frac{4+2+1}{2}=\frac{7}{2}\)

\(B=\sqrt{x}+\frac{1}{\sqrt{x}}+1\ge2\sqrt{\frac{\sqrt{x}}{\sqrt{x}}}+1=3\)

\(B_{min}=3\) khi \(x=1\)

đk x>0

thay x=4(tm) vào B ta có

\(B=\frac{4+\sqrt{4}+1}{\sqrt{4}}=\frac{7}{2}\)

KL....