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

\(A=\dfrac{-3\sqrt{x}-x-2-4\sqrt{x}+2x+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)

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

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

CHÚC BẠN HỌC TỐT NHÉ

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

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

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

\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+2}\right):\dfrac{x+4}{\sqrt{x}+2}\left(dkxd:x\ne4\right)\)

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

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

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

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

Vậy \(B=\dfrac{1}{\sqrt{x}-2}\)

ĐKXĐ: \(x\ge0;x\ne1\)

Sửa lại đề chỗ \(\dfrac{\sqrt{x-1}}{\sqrt{x}+2}\) thành \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

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

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

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

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

\(P=\dfrac{2\sqrt{x}+4-\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}=1+\dfrac{3}{\sqrt{x}+2}\)

Để P lớn nhất \(\Rightarrow\dfrac{3}{\sqrt{x}+2}\) lớn nhất

Mà \(\sqrt{x}+2\ge2\Rightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\)

\(\Rightarrow P_{max}=1+\dfrac{3}{2}=\dfrac{5}{2}\) khi \(\sqrt{x}+2=2\Leftrightarrow x=0\)

a) Điều kiện xác định :

x ≠ 3; x ≠ -3; x ≠ 0

M = \(\dfrac{x}{x^2-9}\) - \(\dfrac{1}{x+3}\): ( \(\dfrac{x}{x\left(x-3\right)}\) - \(\dfrac{x-3}{x\left(x-3\right)}\) )

M = \(\dfrac{x}{x^2-9}\) - \(\dfrac{1}{x+3}\) : ( \(\dfrac{x-x+3}{x\left(x-3\right)}\) )

M = \(\dfrac{x}{x^2-9}\) - \(\dfrac{1}{x+3}\) : \(\dfrac{3}{x\left(x-3\right)}\)

M = \(\dfrac{x}{x^2-9}\) - \(\dfrac{x\left(x-3\right)}{3\left(x+3\right)}\) = \(\dfrac{x}{\left(x-3\right)\left(x+3\right)}\) - \(\dfrac{x\left(x-3\right)}{3\left(x+3\right)}\)

M = \(\dfrac{3x}{3\left(x-3\right)\left(x+3\right)}\) - \(\dfrac{x\left(x-3\right)^2}{3\left(x-3\right)\left(x+3\right)}\)

M = \(\dfrac{3x-x\left(x-3\right)^2}{3\left(x-3\right)\left(x+3\right)}\) = \(\dfrac{3x-x\left(x^2-6x+9\right)}{3\left(x-3\right)\left(x+3\right)}\)

M = \(\dfrac{3x-x^3+6x^2-9x}{3\left(x-3\right)\left(x+3\right)}\) = \(\dfrac{-x^3+6x^2-6x}{3\left(x-3\right)\left(x+3\right)}\)

Mk đang mệt sai thì bạn thông cảm cho mk.

a: \(M=\dfrac{x}{\left(x-3\right)\left(x+3\right)}-\dfrac{1}{x+3}:\dfrac{x-x+3}{x\left(x-3\right)}\)

\(=\dfrac{x}{\left(x-3\right)\left(x+3\right)}-\dfrac{1}{x+3}\cdot\dfrac{x\left(x-3\right)}{3}\)

\(=\dfrac{x}{\left(x-3\right)\left(x+3\right)}-\dfrac{x\left(x-3\right)}{3\left(x+3\right)}\)

\(=\dfrac{3x-x\left(x^2-6x+9\right)}{3\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{3x-x^3+6x^2-9x}{3\left(x-3\right)\left(x+3\right)}=\dfrac{-x^3+6x^2-6x}{3\left(x-3\right)\left(x+3\right)}\)

b: Để M>1/2 thì M-1/2>0

=>\(\dfrac{-x^3+6x^2-6x}{3\left(x^2-9\right)}-\dfrac{1}{2}>0\)

=>\(\dfrac{-2x^3+12x^2-12x-3x^2+9}{6\left(x^2-9\right)}>0\)

=>\(\dfrac{-2x^3+9x^2-12x+9}{x^2-9}>0\)

TH1: \(\left\{{}\begin{matrix}-2x^3+9x^2-12x+9>0\\x^2-9>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 3\\\left[{}\begin{matrix}x>3\\x< -3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x< -3\)

TH2: \(\left\{{}\begin{matrix}-2x^3+9x^2-12x+9< 0\\x^2-9< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>3\\-3< x< 3\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Cho \(5\sqrt{x}7\) mk viet nham

Sua lai thanh \(5\sqrt{x}-7\)

a: \(A=\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)

\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)

\(=\dfrac{2\sqrt{x}+3}{\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)

b: Để A là số nguyên thì \(5\sqrt{x}⋮2\sqrt{x}+1\)

=>10 căn x+5-5 chia hết cho 2 căn x+1

=>\(2\sqrt{x}+1\in\left\{1;5\right\}\)

hay \(x\in\varnothing\)

XEM CÓ SAI ĐỀ BÀI KHÔNG, MK RÚT GỌN RA TO LẮM

\(=\dfrac{x+5\sqrt{x}+6-x+5\sqrt{x}-6}{\left(\sqrt{x}+3\right)^2\cdot\left(\sqrt{x}-3\right)}\cdot\dfrac{x-9}{\sqrt{x}}\)

\(=\dfrac{10\sqrt{x}}{\sqrt{x}}\cdot\dfrac{1}{\sqrt{x}+3}=\dfrac{10}{\sqrt{x}+3}\)

a/ ĐKXĐ: \(x\ge0,x\ne1\)

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

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

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

= \(\dfrac{4\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

= \(\dfrac{4\sqrt{x}}{\sqrt{x}+1}\)

b/ Với \(x\ge0,x\ne1\)

Để \(P=\sqrt{x}-1\Leftrightarrow\dfrac{4\sqrt{x}}{\sqrt{x}+1}=\sqrt{x}-1\)

\(\Leftrightarrow4\sqrt{x}=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

\(\Leftrightarrow x-4\sqrt{x}-1=0\)

\(\Leftrightarrow\left(\sqrt{x}-2+\sqrt{5}\right)\left(\sqrt{x}-2-\sqrt{5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2+\sqrt{5}=0\\\sqrt{x}-2-\sqrt{5}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2-\sqrt{5}\left(ktm\right)\\\sqrt{x}=2+\sqrt{5}\left(tm\right)\end{matrix}\right.\)

\(\Leftrightarrow x=9+4\sqrt{5}\)

Vậy để \(P=\sqrt{x}-1\) thì \(x=9+4\sqrt{5}\)

\(=\dfrac{\left(x+2\right)^2}{x}\cdot\dfrac{x+2-x^2}{x+2}-\dfrac{x^2+6x+4}{x}\)

\(=\dfrac{\left(x+2\right)\left(-x^2+x+2\right)-x^2-6x-4}{x}\)

\(=\dfrac{-x^3+x^2+2x-2x^2+2x+4-x^2-6x-4}{x}\)

\(=\dfrac{-x^3-2x^2-2x}{x}=-x^2-2x-2\)