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\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
a: \(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
c: Để A=1/2 thì \(\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{1}{2}\)
=>\(-10\sqrt{x}+4=\sqrt{x}+3\)
=>x=1/121
d: \(A-\dfrac{2}{3}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{2}{3}\)
\(=\dfrac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}< =0\)
=>A<=2/3
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 \(x\in\varnothing\)
Bài 2:
a: \(A=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
b: Thay \(x=5-2\sqrt{6}\) vào A, ta được:
\(A=\dfrac{-5\left(\sqrt{3}-\sqrt{2}\right)+2}{\sqrt{3}-\sqrt{2}+3}=\dfrac{-5\sqrt{3}+5\sqrt{2}+2}{\sqrt{3}-\sqrt{2}+3}\simeq0,124\)
d: Để A=1/2 thì \(\sqrt{x}+3=-10\sqrt{x}+4\)
\(\Leftrightarrow11\sqrt{x}=1\)
hay x=1/121
a: ĐKXĐ: x>=0; x<>1
b: \(P=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-\left(5x-7\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(5\sqrt{x}-2\right)}{\sqrt{x}+3}\)
Để P=1/2 thì \(\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{1}{2}\)
\(\Leftrightarrow-10\sqrt{x}+4=\sqrt{x}+3\)
\(\Leftrightarrow-11\sqrt{x}=-1\)
=>x=1/121
c: \(P-\dfrac{2}{3}=\dfrac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}=\dfrac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}< 0\)
=>P<2/3
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: \(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-3x+8\sqrt{x}+5-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}+8}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(a.A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)
\(\left(x\ge0;x\ne1\right)\)
\(b.A=\dfrac{1}{2}\Leftrightarrow\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}-\dfrac{1}{2}=0\)
\(\Leftrightarrow\dfrac{4-10\sqrt{x}-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}=0\)
\(\Leftrightarrow-11\sqrt{x}+1=0\)
\(\Leftrightarrow x=\dfrac{1}{121}\left(TM\right)\)
KL...........
\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-....-\frac{1}{\sqrt{24}-\sqrt{25}}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{(\sqrt{1}-\sqrt{2})(\sqrt{1}+\sqrt{2})}-\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{4}}{(\sqrt{3}-\sqrt{4})(\sqrt{3}+\sqrt{4})}-...-\frac{\sqrt{24}+\sqrt{25}}{(\sqrt{24}-\sqrt{25})(\sqrt{24}+\sqrt{25})}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{-1}-\frac{\sqrt{2}+\sqrt{3}}{-1}+\frac{\sqrt{3}+\sqrt{4}}{-1}-...-\frac{\sqrt{24}+\sqrt{25}}{-1}\)
\(=\frac{(1+\sqrt{2})-(\sqrt{2}+\sqrt{3})+(\sqrt{3}+\sqrt{4})-...-(\sqrt{24}+\sqrt{25})}{-1}\)
\(=\frac{1-\sqrt{25}}{-1}=4\)
\(B=\frac{5}{4+\sqrt{11}}+\frac{11-3\sqrt{11}}{\sqrt{11}-3}-\frac{4}{\sqrt{5}-1}+\sqrt{(\sqrt{5}-2)^2}\)
\(=\frac{5(4-\sqrt{11})}{(4+\sqrt{11})(4-\sqrt{11})}+\frac{\sqrt{11}(\sqrt{11}-3)}{\sqrt{11}-3}-\frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}+\sqrt{5}-2\)
\(=\frac{5(4-\sqrt{11})}{5}+\sqrt{11}-\frac{4(\sqrt{5}+1)}{4}+\sqrt{5}-2\)
\(=4-\sqrt{11}+\sqrt{11}-(\sqrt{5}+1)+\sqrt{5}-2\)
\(=1\)
a: Ta có: \(A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)