a) Rút gọn P

ĐKXĐ: \(x\ge0;x\ne1\)

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

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

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

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

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

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

b) Tìm GTLN

\(P=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}=\dfrac{17-5\left(\sqrt{x}+3\right)}{\sqrt{x}+3}=\dfrac{17}{\sqrt{x}+3}-5\)

Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\Rightarrow P=\dfrac{17}{\sqrt{x}+3}-5\le\dfrac{17}{3}-5=\dfrac{2}{3}\)

Dấu "=" xảy ra khi \(x=0\)

Vậy \(P_{max}=\dfrac{2}{3}\) khi \(x=0\)

a) \(P=\dfrac{3x+3\sqrt{x}-9}{x+\sqrt{x}-2}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)

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

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

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

b) \(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6+2}{\sqrt{x}+2}=3+\dfrac{2}{\sqrt{x}+2}\)

Để \(P\in Z\Rightarrow2⋮\sqrt{x}+2\Rightarrow\sqrt{x}+2=2\left(\sqrt{x}+2\ge2\right)\)

\(\Rightarrow x=0\)

c) Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\Rightarrow3+\dfrac{2}{\sqrt{x}+2}\le4\)

\(\Rightarrow P_{max}=4\) khi \(x=0\)

\(a,A=\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{x-2-x+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

\(a,P=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\\ P=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)

\(b,P=\dfrac{1}{2}\Leftrightarrow4-10\sqrt{x}=\sqrt{x}+3\Leftrightarrow\sqrt{x}=\dfrac{7}{11}\Leftrightarrow x=\dfrac{49}{121}\left(tm\right)\)

\(c,P-\dfrac{2}{3}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}-\dfrac{2}{3}=\dfrac{6-15\sqrt{x}-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}=\dfrac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}\)

Ta có \(3\left(\sqrt{x}+3\right)>0;-17\sqrt{x}\le0,\forall x\)

\(\Rightarrow P-\dfrac{2}{3}\le0\Leftrightarrow P\le\dfrac{2}{3}\left(đpcm\right)\)

Sửa đề: 

1: Ta có: \(P=\dfrac{x-\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)

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

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

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

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

2)

a) Thay \(x=\dfrac{9}{4}\) vào P, ta được:

\(P=\left(\dfrac{3}{2}+2\right):\left(\dfrac{3}{2}+3\right)=\dfrac{7}{2}:\dfrac{11}{2}=\dfrac{7}{11}\)

b) Ta có: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

\(=5+\sqrt{2}-4-\sqrt{2}\)

=1

Thay x=1 vào P, ta được:

\(P=\dfrac{1+2}{1+3}=\dfrac{3}{4}\)

Bạn vui lòng viết đề bằng công thức toán để được hỗ trợ tốt hơn!

giúp mình với ạ

a) Ta có: \(P=\dfrac{3x-2\sqrt{x}-4}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)

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

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

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

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

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

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