Rút gọn \(\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\) với x>0.x khác 9 và 25
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Những câu hỏi liên quan
NH
13 tháng 5 2023
\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{2\sqrt{x}+6}{x-9}\right):\dfrac{x-2\sqrt{x}}{\sqrt{x}-3}\left(x>3;x\ne9\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{2\sqrt{x}+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\dfrac{\sqrt{x}-3}{x-2\sqrt{x}}\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left( \sqrt{x}+3\right)}-\dfrac{2\sqrt{x}+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\cdot\dfrac{x-2\sqrt{x}}{\sqrt{x}-3}\)
\(=\dfrac{x+3\sqrt{x}-2\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{x-2\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)-2\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\cdot\dfrac{1}{x-2\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{1}{\sqrt{x}}\)
T
10 tháng 9 2023
\(B=\left(\dfrac{3\sqrt{x}+6}{x-4}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\dfrac{x-9}{\sqrt{x}-3}\left(x\ge0;x\ne4;x\ne9\right)\)
\(=\left[\dfrac{3\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\cdot\dfrac{\sqrt{x}-3}{x-9}\)
\(=\dfrac{3\sqrt{x}+6+x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+5\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{1}{\sqrt{x}+3}\)
\(=\dfrac{x+2\sqrt{x}+3\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{1}{\sqrt{x}-2}\)
#\(Toru\)
18 tháng 10 2021
\(Q=\dfrac{\sqrt{x}-3-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{x-1-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\\ Q=\dfrac{-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\\ Q=\dfrac{-6\sqrt{x}+1+8\sqrt{x}+24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\\ Q=\dfrac{2\sqrt{x}+25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
3 tháng 7 2021
Ta có: \(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{1}{2\sqrt{x}}\right)\)
\(=\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)+8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{2\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)}{2\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{8x-8\sqrt{x}+8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}-2-\sqrt{x}+2}\)
\(=\dfrac{16x-8\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{2\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{2\left(16-8\sqrt{x}\right)}{\sqrt{x}+2}\)
\(=\dfrac{32-16\sqrt{x}}{\sqrt{x}+2}\)
AT
30 tháng 7 2021
a) \(B=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\dfrac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-1\right):\dfrac{9-x+\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-3}{\sqrt{x}+3}:\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=-\dfrac{3}{\sqrt{x}+3}.\dfrac{\sqrt{x}+3}{-\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
b) \(\sqrt{x}=\sqrt{7-4\sqrt{3}}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)
Thế vào B \(\Rightarrow B=\dfrac{3}{2-\sqrt{3}-2}=\dfrac{3}{-\sqrt{3}}=-\sqrt{3}\)
31 tháng 7 2021
a) Ta có: \(B=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\dfrac{x-3\sqrt{x}-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{9-x+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{-x+4\sqrt{x}-4}\)
\(=\dfrac{-3\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)^2}=\dfrac{3}{\sqrt{x}-2}\)
XO
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
\(\left(1\right)\Leftrightarrow0< x< 4\)
3 tháng 3 2021
Ta có: \(P=\left(1-\dfrac{x-3\sqrt{x}}{x-9}\right):\left(\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{3+\sqrt{x}}-\dfrac{9-x}{x+\sqrt{x}-6}\right)\)
\(=\left(1-\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+3}\right):\left(\dfrac{-\left(x-9\right)+x-4\sqrt{x}+4-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}+3}:\dfrac{-x+9+2x-4\sqrt{x}-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)^2}=\dfrac{3}{\sqrt{x}-2}\)
\(=\dfrac{3\sqrt{x}-x+2x}{9-x}:\dfrac{\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{-\sqrt{x}+5}\)
\(=\dfrac{x}{\sqrt{x}-5}\)