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\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-....-\frac{1}{\sqrt{24}-\sqrt{25}}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{(\sqrt{1}-\sqrt{2})(\sqrt{1}+\sqrt{2})}-\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{4}}{(\sqrt{3}-\sqrt{4})(\sqrt{3}+\sqrt{4})}-...-\frac{\sqrt{24}+\sqrt{25}}{(\sqrt{24}-\sqrt{25})(\sqrt{24}+\sqrt{25})}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{-1}-\frac{\sqrt{2}+\sqrt{3}}{-1}+\frac{\sqrt{3}+\sqrt{4}}{-1}-...-\frac{\sqrt{24}+\sqrt{25}}{-1}\)
\(=\frac{(1+\sqrt{2})-(\sqrt{2}+\sqrt{3})+(\sqrt{3}+\sqrt{4})-...-(\sqrt{24}+\sqrt{25})}{-1}\)
\(=\frac{1-\sqrt{25}}{-1}=4\)
\(B=\frac{5}{4+\sqrt{11}}+\frac{11-3\sqrt{11}}{\sqrt{11}-3}-\frac{4}{\sqrt{5}-1}+\sqrt{(\sqrt{5}-2)^2}\)
\(=\frac{5(4-\sqrt{11})}{(4+\sqrt{11})(4-\sqrt{11})}+\frac{\sqrt{11}(\sqrt{11}-3)}{\sqrt{11}-3}-\frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}+\sqrt{5}-2\)
\(=\frac{5(4-\sqrt{11})}{5}+\sqrt{11}-\frac{4(\sqrt{5}+1)}{4}+\sqrt{5}-2\)
\(=4-\sqrt{11}+\sqrt{11}-(\sqrt{5}+1)+\sqrt{5}-2\)
\(=1\)
1, vt : \(\left(1-\dfrac{5+\sqrt{2}}{\sqrt{2}+1}\right).\sqrt{3+2\sqrt{2}}\)
=\(\dfrac{\sqrt{2}+1-5-\sqrt{2}}{\sqrt{2}+1}.\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}+1}\)
=\(\dfrac{-4}{\sqrt{2}+1}.\sqrt{\left(\sqrt{2}+1\right)^2}\)
=\(\dfrac{-4\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
=-4
2, A=\(\left(\dfrac{\sqrt{x}}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right)\div\dfrac{2}{x+\sqrt{x}-2}\)
=\(\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{2}\)
=\(\left(\dfrac{x-\sqrt{x}-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2}\)
=\(\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{2}\)
=\(\dfrac{-\sqrt{x}-2}{\sqrt{x}+1}\)
a, Ta có : \(x=9\Rightarrow\sqrt{x}=3\)
Thay vào biểu thức A ta được : \(A=\frac{2}{3-2}=2\)
b, Với \(x\ge0;x\ne4\)
\(B=\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{4\sqrt{x}}{x-4}=\frac{\sqrt{x}\left(\sqrt{x}-2\right)+4\sqrt{x}}{x-4}\)
\(=\frac{x+2\sqrt{x}}{x-4}=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}=\frac{\sqrt{x}}{\sqrt{x}-2}\)( đpcm )
c, Ta có : \(A+B=\frac{3x}{\sqrt{x}-2}\)hay
\(\frac{2}{\sqrt{x}-2}+\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{2+\sqrt{x}}{\sqrt{x}-2}=\frac{3x}{\sqrt{x}-2}\)
\(\Rightarrow2+\sqrt{x}=3x\Leftrightarrow3x-2-\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}=3x-2\Leftrightarrow x=9x^2-12x+4\)
\(\Leftrightarrow\left(9x-4\right)\left(x-1\right)=0\Leftrightarrow x=\frac{4}{9}\left(ktm\right);x=1\)( đk : \(x\ge\frac{2}{3}\))
a, Ta có : \(x=4\Rightarrow\sqrt{x}=2\)
Thay vào biểu thức A ta được : \(\frac{1}{2-1}=1\)
b, Với \(x\ge0;x\ne1\)
\(Q=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{x-1}-1=\frac{\sqrt{x}\left(\sqrt{x}+1\right)-2-x+1}{x-1}\)
\(=\frac{x+\sqrt{x}-2-x+1}{x-1}=\frac{\sqrt{x}-1}{x-1}=\frac{1}{\sqrt{x}+1}\)
c, Ta có : \(\frac{1}{Q}+P\le4\)hay\(1:\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\le4\)ĐK : \(x\ne1\)
\(\Leftrightarrow\frac{x-1+1}{\sqrt{x}-1}-4\le0\Leftrightarrow\frac{x-4\sqrt{x}+4}{\sqrt{x}-1}\le0\)
\(\Leftrightarrow\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}\le0\Rightarrow\sqrt{x}-1\le0\Leftrightarrow\sqrt{x}\le1\Leftrightarrow x\le1\)do \(\left(\sqrt{x}-2\right)^2\ge0\)
Kết hợp với đk, vậy \(x< 1\)
1, thay x=4 (TMĐKXĐ) vào P ta được:
P=\(\dfrac{1}{\sqrt{4}-1}\)=1
vậy khi x=4 thì P =1
2,với x≥0,x≠1:
Q=\(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)-\(\dfrac{2}{\sqrt{x}-1}-1\)=\(\dfrac{\sqrt{x}-2-\sqrt{x}+1}{\sqrt{x}-1}\)=\(\dfrac{-1}{\sqrt{x}-1}\)
vậy Q=\(\dfrac{-1}{\sqrt{x}-1}\)
3,\(\dfrac{1}{Q}+P\le4\)
⇒1/\(\dfrac{-1}{\sqrt{x}-1}\)+\(\dfrac{1}{\sqrt{x}-1}\)≤4⇔\(\dfrac{-\sqrt{x}-1}{1}+\dfrac{1}{\sqrt{x}-1}\le4\)⇔\(\dfrac{-x+1+1}{\sqrt{x}-1}-4\le0\)⇔\(\dfrac{-x+2-4\sqrt{x}+4}{\sqrt{x}-1}\le0\)⇔\(\dfrac{-x-4\sqrt{x}+6}{\sqrt{x}-1}\le0\)⇔\(\dfrac{x+4\sqrt{x}-6}{\sqrt{x}-1}\le0\)⇔\(\dfrac{x+4\sqrt{x}+4-10}{\sqrt{x}-1}\le0\)
\(\dfrac{ \left(\sqrt{x}+2\right)^2-10}{\sqrt{x}-1}\le0\)⇒\(\sqrt{x}-1\le0\) (vì (\(\sqrt{x}+2\))\(^2\)≥0 ∀ x hay (\(\sqrt{x}+2\))\(^2\)-10>0 ∀ x)
⇔x≤1 (KTM)
vậy không có giá trị nào của x TM để \(\dfrac{1}{Q}+P\le4\)
a) \(\sqrt{ }\)20 + 2\(\sqrt{ }\)45 - 3\(\sqrt{ }\)80 + \(\sqrt{ }\)125
= \(\sqrt{ }\)4.5 +2\(\sqrt{ }\)9.5 - 3\(\sqrt{16.5}\)
= 2\(\sqrt{5}\) + 6\(\sqrt{5}\) - 12\(\sqrt{5}\)
= -4\(\sqrt{5}\)
b) \(\dfrac{2\sqrt{3}+3\sqrt{2}}{\sqrt{3}+\sqrt{2}}\) - \(4\sqrt{\dfrac{3}{2}}\)- \(\dfrac{5}{1-\sqrt{6}}\)
= \(\dfrac{2\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}\)- \(\sqrt{16.\dfrac{3}{2}}\) - \(\dfrac{5\left(1+\sqrt{6}\right)}{\left(1-\sqrt{6}\right)\left(1+\sqrt{6}\right)}\)
= 2 - \(\sqrt{24}\) - \(\dfrac{5\left(1+\sqrt{6}\right)}{1-6}\)
= 2 - \(\sqrt{4.6}\) + 1+\(\sqrt{ }\)6
= 2 - 2\(\sqrt{ }\)6 + 1+\(\sqrt{ }\)6
= 3 - \(\sqrt{ }\)6
c) (đề bài) với x khác 4...
= \(\dfrac{\sqrt{x}}{\sqrt{x}-2}\)- \(\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
= \(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)- ....
= \(x-4\sqrt{x}+4\)/ \(\sqrt{x}\left(\sqrt{x}-2\right)\)
= (căn -2)2/ căn x(căn x -2)
= căn x-2/căn x