\(a.A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\) ( x ≥ 0 ; x # 1 )

\(b.\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{x+2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{2}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}>0\) \(c.\) \(\dfrac{2}{x+\sqrt{x}+1}\) ≤ \(\dfrac{2}{1}=2\left(x\text{≥ }0\right)\)

⇒ \(A_{Max}=2."="\) ⇔ \(x=0\left(TM\right)\)

a)ĐK: \(x\ge0;x\ne1\)

\(A\Leftrightarrow\left(\dfrac{x+2+\sqrt{x}\left(\sqrt{x}+1\right)-1\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\dfrac{\sqrt{x}-1}{2}\right)\)

\(\Leftrightarrow\left(\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}-1}{2}\)

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

Sửa câu a:

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

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

ĐK:\(x\ge0\),\(x\ne1\)

\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{2}\right)=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}-1}{2}\right)=\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right].\dfrac{2}{\sqrt{x}-1}=\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right].\dfrac{2}{\sqrt{x}-1}=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{\left(x-2\sqrt{x}+1\right).2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)

Ta có \(\sqrt{x}\ge0\Leftrightarrow x+\sqrt{x}\ge0\)(vì \(x\ge0\))\(\Leftrightarrow x+\sqrt{x}+1>0\Leftrightarrow\dfrac{2}{x+\sqrt{x}+1}>0\)

Vậy A>0 với mọi x để A có nghĩa

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

(ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\))

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

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

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

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

\(M=\dfrac{1}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le1\)

Dấu "=" xảy ra khi x = 0

Mysterious Person Nguyễn Thanh Hằng Phương An giúp mk với. thanks!!

a: ĐKXĐ: x>=0; x<>1

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

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

b: Vì x+căn x+1>0

nên A>0

https://i.imgur.com/Qx0XV1d.jpg

a) điều kiện xác định : \(x\ge0;x\ne1\)

ta có : \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)

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

a) ta có : \(A=\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{1}{4}}>0\forall x\)

c) ta có : \(A=\dfrac{2}{x+\sqrt{x}+1}\le\dfrac{2}{1}=2\) (vì \(x\ge0\) )

\(\Rightarrow\) \(A_{max}=2\) khi \(x=0\)

Mysterious Person Nguyễn Huy Tú Phương An giup mk voi

a: \(P=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)

\(=\left(x^2+5x+5\right)^2>=0\)

b: \(P=\dfrac{16-5\sqrt{7}}{2}+\dfrac{5\sqrt{7}-25}{2}+5\)

\(=\dfrac{-9}{2}+5=\dfrac{1}{2}\)

a) \(A=\dfrac{1}{\sqrt{x}+1}-\dfrac{3}{x\sqrt{x}+1}+\dfrac{2}{x-\sqrt{x}+1}\)

\(A=\dfrac{1}{\sqrt{x}+1}-\dfrac{3}{\sqrt{x^3}+1}+\dfrac{2}{x-\sqrt{x}+1}\)

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

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

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

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

b) Chứng minh \(A\ge0\)

Ta có \(A=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)

\(A=\dfrac{\sqrt{x}}{\sqrt{x^2}-2\sqrt{x}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}+1}=\dfrac{\sqrt{x}}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)

Mà \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\) và \(\sqrt{x}\ge0\)

\(\Rightarrow A=\dfrac{\sqrt{x}}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\ge0\) (1)

Chứng minh \(A\le1\)

Ta có \(A=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)

\(\Leftrightarrow\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\le1\)

\(\Leftrightarrow\sqrt{x}\le x-\sqrt{x}+1\)

\(\Leftrightarrow2\sqrt{x}\le x+1\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow x+1\ge2\sqrt{x}\) ( luôn đúng với mọi \(x\ge0\) )

Vậy \(A\le1\) (2)

Từ (1) và (2)

\(\Rightarrow0\le A\le1\) ( đpcm )