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a/ ĐKXĐ: x>= 0 ; x khác 1
b/ \(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{8\sqrt{x}}{x-1}\right):\dfrac{4\sqrt{x}-8}{1-x}\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)^2}{x-1}-\dfrac{\left(\sqrt{x}-1\right)^2}{x-1}-\dfrac{8\sqrt{x}}{x-1}\right):\dfrac{8-4\sqrt{x}}{x-1}\)
\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{2}-1-8\sqrt{x}}{x-1}\cdot\dfrac{x-1}{8-4\sqrt{x}}\)
\(=\dfrac{-4\sqrt{x}}{x-1}\cdot\dfrac{x-1}{4\left(2-\sqrt{x}\right)}=\dfrac{-4\sqrt{x}}{4\left(2-\sqrt{x}\right)}=-\dfrac{\sqrt{x}}{2-\sqrt{x}}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
Làm nốt bài 1 ::v
\(\dfrac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}+\dfrac{3+6\sqrt{3}}{\sqrt{3}}-\dfrac{13}{\sqrt{3}+4}=\dfrac{-\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\dfrac{\sqrt{3}\left(\sqrt{3}+6\right)}{\sqrt{3}}-\dfrac{13}{\sqrt{3}+4}=6-\dfrac{13}{\sqrt{3}+4}=\dfrac{11+6\sqrt{3}}{\sqrt{3}+4}\)
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 \(x\in\varnothing\)
\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-....-\frac{1}{\sqrt{24}-\sqrt{25}}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{(\sqrt{1}-\sqrt{2})(\sqrt{1}+\sqrt{2})}-\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{4}}{(\sqrt{3}-\sqrt{4})(\sqrt{3}+\sqrt{4})}-...-\frac{\sqrt{24}+\sqrt{25}}{(\sqrt{24}-\sqrt{25})(\sqrt{24}+\sqrt{25})}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{-1}-\frac{\sqrt{2}+\sqrt{3}}{-1}+\frac{\sqrt{3}+\sqrt{4}}{-1}-...-\frac{\sqrt{24}+\sqrt{25}}{-1}\)
\(=\frac{(1+\sqrt{2})-(\sqrt{2}+\sqrt{3})+(\sqrt{3}+\sqrt{4})-...-(\sqrt{24}+\sqrt{25})}{-1}\)
\(=\frac{1-\sqrt{25}}{-1}=4\)
\(B=\frac{5}{4+\sqrt{11}}+\frac{11-3\sqrt{11}}{\sqrt{11}-3}-\frac{4}{\sqrt{5}-1}+\sqrt{(\sqrt{5}-2)^2}\)
\(=\frac{5(4-\sqrt{11})}{(4+\sqrt{11})(4-\sqrt{11})}+\frac{\sqrt{11}(\sqrt{11}-3)}{\sqrt{11}-3}-\frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}+\sqrt{5}-2\)
\(=\frac{5(4-\sqrt{11})}{5}+\sqrt{11}-\frac{4(\sqrt{5}+1)}{4}+\sqrt{5}-2\)
\(=4-\sqrt{11}+\sqrt{11}-(\sqrt{5}+1)+\sqrt{5}-2\)
\(=1\)
\(ĐKXĐ:x\ge0,x\ne1\)
\(K=\left[\dfrac{x+3\sqrt{x}+2}{x+\sqrt{x}-2}-\dfrac{x+\sqrt{x}}{x-1}\right]:\left[\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\right]\)
\(K=\left[\dfrac{x+2\sqrt{x}+\sqrt{x}+2}{x+2\sqrt{x}-\sqrt{x}-2}-\dfrac{x+\sqrt{x}}{x-1}\right]:\left[\dfrac{\sqrt{x}-1+\sqrt{x}+1}{x-1}\right]\)
\(K=\left[\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)-\left(\sqrt{x}+2\right)}-\dfrac{x+\sqrt{x}}{x-1}\right]:\dfrac{2\sqrt{x}}{x-1}\)
\(K=\left[\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{x+\sqrt{x}}{x-1}\right]:\dfrac{2\sqrt{x}}{x-1}\)
\(K=\left[\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1}\right]:\dfrac{2\sqrt{x}}{x-1}\)
\(K=\left[\dfrac{\left(\sqrt{x}+1\right)^2}{x-1}-\dfrac{x+\sqrt{x}}{x-1}\right]:\dfrac{2\sqrt{x}}{x-1}\)
\(K=\dfrac{x+2\sqrt{x}+1-x-\sqrt{x}}{x-1}.\dfrac{x-1}{2\sqrt{x}}\)
\(K=\dfrac{\sqrt{x}+1}{x-1}.\dfrac{x-1}{2\sqrt{x}}\)
\(K=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)
b.
Ta có: \(24+\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=24+\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2.2\sqrt{5}.3+9}}}\)
\(=24+\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=24+\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}=24+\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}=24+\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=24+\sqrt{\sqrt{5}-\sqrt{5}+1}=24+1=25\)
Thay \(x=25\) vào \(K\) ta được:
\(K=\dfrac{\sqrt{x}+1}{2\sqrt{x}}=\dfrac{\sqrt{25}+1}{2.\sqrt{25}}=\dfrac{6}{10}=\dfrac{3}{5}\)
c.
Ta có: \(\dfrac{1}{K}-\dfrac{\sqrt{x}+1}{8}\ge1\)
\(\Rightarrow\dfrac{1}{K}-\dfrac{\sqrt{x}+1}{8}-1\ge0\)
\(\Rightarrow\dfrac{2\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{8}-1\ge0\)
\(\Rightarrow\dfrac{16\sqrt{x}}{8\sqrt{x}+8}-\dfrac{x+2\sqrt{x}+1}{8\sqrt{x}+8}-\dfrac{8\sqrt{x}+8}{8\sqrt{x}+8}\ge0\)
\(\Rightarrow\dfrac{16\sqrt{x}-x-2\sqrt{x}-1-8\sqrt{x}-8}{8\sqrt{x}+8}\ge0\)
\(\Rightarrow\dfrac{6\sqrt{x}-x-9}{8\sqrt{x}+8}\ge0\)
\(\Rightarrow\dfrac{-\left(\sqrt{x}-3\right)^2}{8\sqrt{x}+8}\ge0\)
Ta có: \(\left\{{}\begin{matrix}-\left(\sqrt{x}-3\right)^2\le0\\8\sqrt{x}+8\ge0\end{matrix}\right.\)
⇒ Không có \(x\) thỏa mãn
Bài 1:
a: ĐKXĐ: 2x+3>=0 và x-3>0
=>x>3
b: ĐKXĐ:(2x+3)/(x-3)>=0
=>x>3 hoặc x<-3/2
c: ĐKXĐ: x+2<0
hay x<-2
d: ĐKXĐ: -x>=0 và x+3<>0
=>x<=0 và x<>-3
a: \(A=\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(=-\sqrt{x}+3-\sqrt{x}+3-6=-2\sqrt{x}\)
b: \(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{1}{x+1}\)
g: \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{x-1}\cdot\left(\sqrt{x}-1-2\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-1}\)
Mình làm mấy bài rút gọn thôi nhé :v (mấy cái kia mình làm sợ không đúng)
\(P=\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{1}{\sqrt{x}-1}-\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}+1-\left(x+2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{x+\sqrt{x}+1-x-2-\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+1-2-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+0-x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left[-\left(\sqrt{x}-1\right)\right]}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(-1\right)}{x+\sqrt{x}+1}\\ =-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
Bài 3:
\(P=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{\left(2x+\sqrt{x}\right)\sqrt{x}}{x}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}+2\left(\sqrt{x}+1\right)\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x\left(2\sqrt{x}+1\right)}{x}+2\sqrt{x}+2\)
\(=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\sqrt{x}+2\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}+1\\ =\dfrac{x-\sqrt{x}+x+\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{2x+1}{x+\sqrt{x}+1}\)
1. \(\dfrac{2\sqrt{3}-6}{\sqrt{8}-2}=\dfrac{2\left(\sqrt{3}-3\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{3}-3}{\sqrt{2}-1}=\dfrac{\left(\sqrt{3}-3\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\dfrac{\sqrt{6}+\sqrt{3}-3\sqrt{2}-3}{2-1}=\sqrt{6}+\sqrt{3}-3\sqrt{2}-3\)
2. \(\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}=\dfrac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\left(2\sqrt{x}+1\right)=\sqrt{x}\left(\sqrt{x}+1\right)-2\sqrt{x}-1=x+\sqrt{x}-2\sqrt{x}-1=x-\sqrt{x}-1\)
3. \(\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}=\dfrac{x\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{x\sqrt{x}-2x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}\left(x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\sqrt{x}-1\)
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)