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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:
\(P=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{\left(2x+\sqrt{x}\right)\sqrt{x}}{x}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}+2\left(\sqrt{x}+1\right)\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x\left(2\sqrt{x}+1\right)}{x}+2\sqrt{x}+2\)
\(=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\sqrt{x}+2\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}+1\\ =\dfrac{x-\sqrt{x}+x+\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{2x+1}{x+\sqrt{x}+1}\)
a: \(A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{2\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}\)
\(=\dfrac{-6}{\sqrt{x}+3}\)
b: Để A<-1/2 thì A+1/2<0
\(\Leftrightarrow-\dfrac{6}{\sqrt{x}+3}+\dfrac{1}{2}< 0\)
\(\Leftrightarrow-12+\sqrt{x}+3< 0\)
=>0<x<81 và x<>9
a/ \(\dfrac{x+3}{x-9}+\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}=\dfrac{x+3+2\left(\sqrt{x}-3\right)-\left(\sqrt{x}+3\right)}{x-9}=\dfrac{x+3+2\sqrt{x}-6-\sqrt{x}-3}{x-9}=\dfrac{x-\sqrt{x}-6}{x-9}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)-3\left(\sqrt{x}+2\right)}{x-9}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
b/ \(\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2\sqrt{x}+6+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{3\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-3\right)^2\left(\sqrt{x}+3\right)}\)
P/s: câu b đề sai phải không bạn, mk nghĩ ngoài dấu ngoặc là phép chia thì đúng hơn
\(A=\left(x-2\right)\cdot\sqrt{\dfrac{9}{\left(x-2\right)^2}}+3=\dfrac{3\left(x-2\right)}{\left|x-2\right|}+3=\dfrac{3\left(x-2\right)}{-\left(x-2\right)}=-3+3=0\)
\(B=\sqrt{\dfrac{a}{6}}+\sqrt{\dfrac{2a}{3}}+\sqrt{\dfrac{3a}{2}}=\dfrac{\sqrt{a}}{\sqrt{6}}+\dfrac{\sqrt{2a}}{\sqrt{3}}+\dfrac{\sqrt{3a}}{\sqrt{2}}=\dfrac{\sqrt{a}+2\sqrt{a}+3\sqrt{a}}{\sqrt{6}}=\dfrac{6\sqrt{a}}{\sqrt{6}}=\sqrt{6a}\)
\(E=\sqrt{9a^2}+\sqrt{4a^2}+\sqrt{\left(1-a\right)^2}+\sqrt{16a^2}=3\left|a\right|+2\left|a\right|+\left|1-a\right|+4\left|a\right|=9\left|a\right|+1-a=-9a+1-a=-10a+1\)
\(F=\left|x-2\right|\cdot\dfrac{\sqrt{x^2}}{x}=\left|x-2\right|\cdot\dfrac{\left|x\right|}{x}=\dfrac{x\left(x-2\right)}{x}=x-2\)
\(H=\dfrac{x^2+2\sqrt{3}\cdot x+3}{x^2-3}=\dfrac{\left(x+\sqrt{3}\right)^2}{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}=\dfrac{x+\sqrt{3}}{x-\sqrt{3}}\)
\(I=\left|x-\sqrt{\left(x-1\right)^2}\right|-2x=\left|x-\left(-\left(x-1\right)\right)\right|-2x=\left|x+x-1\right|-2x=\left|2x-1\right|-2x=1-4x\)
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: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
1, \(P=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)-\left(x-4\sqrt{x}-9\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+\sqrt{x}-6+x-2\sqrt{x}-3-x+4\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
2, Để P = 3 thì \(\dfrac{\sqrt{x}}{\sqrt{x}-3}=3\Rightarrow3\sqrt{x}-9=\sqrt{x}\)
\(\Leftrightarrow2\sqrt{x}-9=0\)
\(\Leftrightarrow\sqrt{x}=\dfrac{9}{2}\Leftrightarrow x=\dfrac{81}{4}\)(thỏa mãn)
3, \(M=\dfrac{\sqrt{x}}{\sqrt{x}-3}:\dfrac{\sqrt{x}+5}{3-\sqrt{x}}=\dfrac{-\sqrt{x}}{\sqrt{x}+5}\)
để \(\left|M\right|< \dfrac{1}{2}\) thì \(\dfrac{\sqrt{x}}{\sqrt{x}+5}< \dfrac{1}{2}\) \(\Leftrightarrow2\sqrt{x}< \sqrt{x}+5\)
\(\Leftrightarrow\sqrt{x}< 5\)
\(\Leftrightarrow0\le x< 25\)
Kết hợp ĐK ta có \(\left\{{}\begin{matrix}0\le x< 25\\x\ne9\end{matrix}\right.\)
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)\(=4\sqrt{6}-3\sqrt{6}+1-\sqrt{6}\)
\(=1\)
b)ĐK: \(x>0,x\ne9\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{x-9}\right):\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-9}{\sqrt{x}+3}.\dfrac{\sqrt{x}}{2\sqrt{x}+4}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\left(\dfrac{x+3\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{\left(x+2\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)