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1: Ta có: \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}}{2}\)
\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-2}{\sqrt{x}\left(x+\sqrt{x}+1\right)}\)
a)√x−1=2(x≥1)
\(x-1=4
\)
x=5
b)
\(\sqrt{3-x}=4\) (x≤3)
\(\left(\sqrt{3-x}\right)^2=4^2\)
x-3=16
x=19
a: Ta có: \(\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)
hay x=5
b: Ta có: \(\sqrt{3-x}=4\)
\(\Leftrightarrow3-x=16\)
hay x=-13
c: Ta có: \(2\cdot\sqrt{3-2x}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{3-2x}=\dfrac{1}{4}\)
\(\Leftrightarrow-2x+3=\dfrac{1}{16}\)
\(\Leftrightarrow-2x=-\dfrac{47}{16}\)
hay \(x=\dfrac{47}{32}\)
d: Ta có: \(4-\sqrt{x-1}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{7}{2}\)
\(\Leftrightarrow x-1=\dfrac{49}{4}\)
hay \(x=\dfrac{53}{4}\)
e: Ta có: \(\sqrt{x-1}-3=1\)
\(\Leftrightarrow\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=16\)
hay x=17
f:Ta có: \(\dfrac{1}{2}-2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow\sqrt{x+2}=\dfrac{1}{8}\)
\(\Leftrightarrow x+2=\dfrac{1}{64}\)
hay \(x=-\dfrac{127}{64}\)
a) ĐKXĐ có thêm \(x\ne4\)
\(A=\left(\dfrac{x-\sqrt{x}+2}{x-\sqrt{x}-2}-\dfrac{x}{x-2\sqrt{x}}\right):\dfrac{1-\sqrt{x}}{2-\sqrt{x}}\)
\(=\left(\dfrac{x-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\dfrac{2-\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(x-\sqrt{x}+2\right)-x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{2-\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{-2x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{-2\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{-2}{\sqrt{x}+1}\)
\(B=\left(\dfrac{x}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\dfrac{x+1}{\sqrt{x}+3}:\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{x+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+1}{\sqrt{x}+3}:\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{x+1}{\sqrt{x}+3}.\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+1}{\sqrt{x}+1}\)
\(1,\\ a,E=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ b,E>0\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}>0\Leftrightarrow\sqrt{x}-1>0\left(\sqrt{x}>0\right)\\ \Leftrightarrow x>1\\ 2,\\ a,B=\dfrac{x-\sqrt{x}+\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\left(\sqrt{x}+1\right)\\ B=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,B=2\Leftrightarrow\sqrt{x}-1=2\left(\sqrt{x}+1\right)\\ \Leftrightarrow\sqrt{x}-1=2\sqrt{x}+2\\ \Leftrightarrow\sqrt{x}=-3\Leftrightarrow x\in\varnothing\)
a: \(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b: \(=\dfrac{1+\sqrt{a}}{a-\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
\(=\left[\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\cdot\left(\dfrac{1}{2\sqrt{x}}-\dfrac{x}{2\sqrt{x}}\right)^2\)
\(=\left[\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\cdot\dfrac{\left(1-x\right)^2}{\left(2\sqrt{x}\right)^2}\)
\(=\dfrac{-4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{4x}\)
\(=-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
Ta có: \(B=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\cdot\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\left(\dfrac{1}{2\sqrt{x}}-\dfrac{x}{2\sqrt{x}}\right)^2\)
\(=\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)}{4x}\)
\(=\dfrac{-4\sqrt{x}\cdot\left(x-1\right)}{4x}\)
\(=\dfrac{-\left(x-1\right)}{\sqrt{x}}=\dfrac{1-x}{\sqrt{x}}\)
1.
\(Q=\left[\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\sqrt{x}(\sqrt{x}+1)\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}-2)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)(\sqrt{x}-1)-\sqrt{x}(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2x}{x-1}\)
2.
\(A=\left[\frac{\sqrt{x}+2-(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{4\sqrt{x}}{x-4}\right].\frac{x-4}{\sqrt{x}+1}\)
\(=\left(\frac{4}{x-4}-\frac{4\sqrt{x}}{x-1}\right).\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{x-4}.\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{\sqrt{x}+1}\)
trục căn rồi tính bình thường
câu a trục căn nhân với cái như phần tử hả ? câu b thì biết làm rồi.