Rút gọn BT
B=\(\frac{x-3}{\sqrt{x-2}-1}\) với x=23-4\(\sqrt{5}\)
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14 tháng 8 2023
a: \(P=\dfrac{15\sqrt{x}-11+\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11+3x+7\sqrt{x}-6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+21\sqrt{x}-14}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
b: Khi x=9 thì \(P=\dfrac{9+21\cdot3-14}{\left(3+3\right)\left(3-1\right)}=\dfrac{29}{6}\)
27 tháng 8 2021
a) \(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right).\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\left(đk:x>0\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\dfrac{1-x}{2\sqrt{x}}\right)^2=\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}.\dfrac{\left(x-1\right)^2}{4x}=\dfrac{-4\sqrt{x}\left(x-1\right)}{4x}=\dfrac{1-x}{\sqrt{x}}\)
b) \(P-\left(-2\sqrt{x}\right)=\dfrac{1-x}{\sqrt{x}}+2\sqrt{x}=\dfrac{1-x+2x}{\sqrt{x}}=\dfrac{1+x}{\sqrt{x}}>0\)
\(\Rightarrow P>-2\sqrt{x}\)
HP
27 tháng 8 2021
a, ĐK: \(x\ge0;x\ne1\)
\(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(2-2x\right)^2}{16x}\)
\(=\dfrac{-4\sqrt{x}}{x-1}.\dfrac{4\left(x-1\right)^2}{16x}\)
\(=-\dfrac{x-1}{\sqrt{x}}\)
23 tháng 5 2021
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
4 tháng 8 2017
ĐK \(x\ne\left\{2;3\right\}\)
Ta có \(A=\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}-3.\frac{\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4-\left(x-2\sqrt{x}-3\right)-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4-x+2\sqrt{x}+3-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=-\frac{1}{\sqrt{x}-3}\)
\(B=\frac{23-4\sqrt{5}-3}{\sqrt{23-4\sqrt{5}-2}-1}\)
\(=\frac{20-4\sqrt{5}}{\sqrt{21-4\sqrt{5}}-1}=\frac{20-4\sqrt{5}}{\sqrt{\left(2\sqrt{5}-1\right)^2}-1}\)
\(=\frac{20-4\sqrt{5}}{2\sqrt{5}-1-1}=\frac{20-4\sqrt{5}}{2\sqrt{5}-2}=\frac{10-2\sqrt{5}}{\sqrt{5}-1}\)