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7 tháng 10 2023

a) \(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\left(x\ge0;x\ne0\right)\)

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

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

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

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

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

7 tháng 10 2023

b) \(\dfrac{3}{\sqrt{x}-1}-\dfrac{\sqrt{x}+5}{x-1}\left(x\ge0;x\ne1\right)\)

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

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

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

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

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

7 tháng 10 2023

\(a,\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{x-25}\\ =\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{3\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{3\sqrt{x}+15+20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{\sqrt{x}+35}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

7 tháng 10 2023

\(b,\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{x+3\sqrt{x}+2\sqrt{x}-2}{x-9}\\ =\dfrac{x-5\sqrt{x}-2}{x-9}\)

1 tháng 9 2021

1.

\(A=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{x+9\sqrt{x}}{9-x}\)

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

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

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

\(=\dfrac{x-15\sqrt{x}}{x-9}\)

1 tháng 9 2021

2.

\(B=\dfrac{3}{\sqrt{x}-3}+\dfrac{2}{\sqrt{x}+3}+\dfrac{x-5\sqrt{x}-3}{x-9}\)

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

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

\(=\dfrac{x}{x-9}\)

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

\(=2+\sqrt{3}-\sqrt{3}-\sqrt{2}+\sqrt{2}\)

=2

Ta có: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)

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

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

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

\(=\dfrac{3x-9\sqrt{x}+3\sqrt{x}}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

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

18 tháng 10 2021

\(Q=\dfrac{\sqrt{x}-3-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{x-1-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\\ Q=\dfrac{-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\\ Q=\dfrac{-6\sqrt{x}+1+8\sqrt{x}+24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\\ Q=\dfrac{2\sqrt{x}+25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

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

\(=\dfrac{x}{\sqrt{x}-5}\)

31 tháng 8 2023

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

\(=\dfrac{3.\left(\sqrt{x}-3\right)+x+9}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-5-\left(\sqrt{x}-3\right)}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)

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

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

b) \(M< 0\Leftrightarrow\sqrt{x}-2< 0\Leftrightarrow x< 4\)

Kết hợp điều kiện ta được \(0< x< 4\) thì M < 0

c) Từ câu b ta có M < 0 \(\Leftrightarrow0< x< 4\)

nên \(x\inℤ\) để M nguyên âm <=> \(x\in\left\{1;2;3\right\}\)

Thay lần lượt các giá trị vào M được x = 1 thỏa 

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

Vì x > 4 nên \(\sqrt{x}-2>0\)

Áp dụng BĐT Cauchy ta có 

\(M=\left(\sqrt{x}-2+\dfrac{4}{\sqrt{x}-2}\right)+4\ge2\sqrt{\left(\sqrt{x}-2\right).\dfrac{4}{\sqrt{x}-2}}+4=8\)

Dấu "=" xảy ra khi \(\sqrt{x}-2=\dfrac{4}{\sqrt{x}-2}\Leftrightarrow x=16\left(tm\right)\)

31 tháng 8 2023

1) \(M=\left(\dfrac{3}{\sqrt[]{x}+3}+\dfrac{x+9}{x-9}\right):\left(\dfrac{2\sqrt[]{x}-5}{x-3\sqrt[]{x}}-\dfrac{1}{\sqrt[]{x}}\right)\left(x>0;x\ne9\right)\)

\(\Leftrightarrow M=\left(\dfrac{3\left(\sqrt[]{x}-3\right)}{\left(\sqrt[]{x}+3\right)\left(\sqrt[]{x}-3\right)}+\dfrac{x+9}{x-9}\right):\left(\dfrac{2\sqrt[]{x}-5}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}-\dfrac{1}{\sqrt[]{x}}\right)\)

\(\Leftrightarrow M=\left(\dfrac{3\sqrt[]{x}-9+x+9}{x-9}\right):\left(\dfrac{2\sqrt[]{x}-5-\left(\sqrt[]{x}-3\right)}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}\right)\)

\(\Leftrightarrow M=\left(\dfrac{3\sqrt[]{x}+x}{x-9}\right):\left(\dfrac{2\sqrt[]{x}-5-\sqrt[]{x}+3}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}\right)\)

\(\Leftrightarrow M=\left(\dfrac{\sqrt[]{x}\left(\sqrt[]{x}+3\right)}{x-9}\right):\left(\dfrac{\sqrt[]{x}-2}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}\right)\)

\(\Leftrightarrow M=\left(\dfrac{\sqrt[]{x}}{\sqrt[]{x}-3}\right):\left(\dfrac{\sqrt[]{x}-2}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}\right)\)

\(\Leftrightarrow M=\dfrac{\sqrt[]{x}}{\sqrt[]{x}-3}.\dfrac{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}{\sqrt[]{x}-2}\)

\(\Leftrightarrow M=\dfrac{x}{\sqrt[]{x}-2}\)

2) Để \(M< 0\) khi và chỉ chi

\(M=\dfrac{x}{\sqrt[]{x}-2}< 0\left(1\right)\)

Nghiệm của tử là \(x=0\)

Nghiệm của mẫu \(\sqrt[]{x}-2=0\Leftrightarrow\sqrt[]{x}=2\Leftrightarrow x=4\)

Lập bảng xét dấu... ta được

\(\left(1\right)\Leftrightarrow0< x< 4\)

13 tháng 11 2021

\(a,P=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ P=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ P=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\dfrac{-3}{\sqrt{x}+3}\\ b,P=\dfrac{-3}{\sqrt{x}+3}\ge\dfrac{-3}{0+3}=-1\\ P_{min}=-1\Leftrightarrow x=0\)