\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sq...">
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Bài 2:

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

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

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

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

b: Để A=1/2 thì \(\dfrac{-5\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{1}{2}\)

\(\Leftrightarrow-10\sqrt{x}+2=\sqrt{x}+3\)

hay \(x\in\varnothing\)

18 tháng 11 2022

\(Q=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

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

\(=\dfrac{-\sqrt{x}}{x-4}\)

17 tháng 7 2018

\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)

\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)

\(\Leftrightarrow\sqrt{x}-2< 0\)

\(\Leftrightarrow x< 4\)

Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)

KL............

\(2.\) Tương tự bài 1.

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

\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)

13 tháng 6 2017

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

E = \(\sqrt{x}+1+\sqrt{x}\) = \(2\sqrt{x}+1\)

F = \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}+1}{3-\sqrt{x}}-\dfrac{3-11\sqrt{x}}{x-9}\)

F = \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

F = \(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-\left(3-11\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

F = \(\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}+\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

F = \(\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\) = \(\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\) = \(\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)

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

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

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

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

G = \(\dfrac{x+2\sqrt{x}+3\sqrt{x}+6-\left(x-2\sqrt{x}-\sqrt{x}+2\right)-\left(4\sqrt{x}-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

G = \(\dfrac{x+5\sqrt{x}+6-x+2\sqrt{x}+\sqrt{x}-2-4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

G = \(\dfrac{4\sqrt{x}+8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\) = \(\dfrac{4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\) = \(\dfrac{4}{\sqrt{x}-2}\)

a: \(A=\dfrac{5\sqrt{x}-11-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)-\left(2x-5\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{5\sqrt{x}-11-3x-7\sqrt{x}+6-2x+5\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

b: Q-2/3

\(=\dfrac{-15x+9\sqrt{x}-24-2\left(x+2\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-17x+5\sqrt{x}-18}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}< 0\)

=>Q<2/3

21 tháng 6 2017

1/

a) \(\left(\dfrac{2\sqrt{2}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

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

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

\(=\dfrac{2\sqrt{2x}-6\sqrt{2}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

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

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

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

21 tháng 6 2017

bài 2 : đk : \(x\ge0;x\ne1\)

a) P = \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

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

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

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

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

b) P = \(\dfrac{1}{2}\) \(\Leftrightarrow\) \(\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}=\dfrac{1}{2}\) \(\Leftrightarrow\) \(\sqrt{x}+3=4-10\sqrt{x}\)

\(\Leftrightarrow\) \(11\sqrt{x}-1=0\) \(\Leftrightarrow\) \(11\sqrt{x}=1\) \(\Leftrightarrow\) \(\sqrt{x}=\dfrac{1}{11}\) \(x=\left(\dfrac{1}{11}\right)^2=\dfrac{1}{121}\)

6 tháng 12 2018

a ,rút gọn P (dkxd x\(\ge0,x\ne0\)

P=\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

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

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

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

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

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

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

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

..............=\(\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)

a: ĐKXĐ: x>=0; x<>1

b: \(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

c: Để K=1/2 thì \(\dfrac{-5x+5\sqrt{x}-8}{x+2\sqrt{x}-3}=\dfrac{1}{2}\)

=>\(-10x+10\sqrt{x}-16-x-2\sqrt{x}+3=0\)

=>\(-11x+8\sqrt{x}-13=0\)

hay \(x\in\varnothing\)

2 tháng 9 2018

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

ta có : \(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(\Leftrightarrow P=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(\Leftrightarrow P=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(\Leftrightarrow P=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(\Leftrightarrow P=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(2-5\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\) b) ta có : \(P=\dfrac{1}{2}\Leftrightarrow\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}=\dfrac{1}{2}\) \(\Leftrightarrow4-10\sqrt{x}=\sqrt{x}+3\)

\(\Leftrightarrow11\sqrt{x}=1\Leftrightarrow\sqrt{x}=\dfrac{1}{11}\Leftrightarrow x=\dfrac{1}{121}\)

c) ta có : \(P-\dfrac{2}{3}\Leftrightarrow\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}-\dfrac{2}{3}=\dfrac{2-5\sqrt{x}-\dfrac{2}{3}\sqrt{x}-2}{\sqrt{x}+3}\)

\(=\dfrac{\dfrac{-17}{3}\sqrt{x}}{\sqrt{x}+3}\le0\) \(\Rightarrow P-\dfrac{2}{3}\le0\Leftrightarrow P\le\dfrac{2}{3}\left(đpcm\right)\)

2 tháng 9 2018

Sai rồi bạn ạ

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

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

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

\(=2x\sqrt{x}+x-2x-\sqrt{x}+2\sqrt{x}+2\)

\(=2x\sqrt{x}-x+\sqrt{x}+2\)

b: \(=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{x-4-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

c: \(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

\(=\dfrac{-5x+7\sqrt{x}+8}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)