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17 tháng 10 2018
\(a)\)\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt{x-3}}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}-3}{\sqrt{x}-3}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\frac{3\sqrt{x}+3}{\sqrt{x}+3}.\frac{\sqrt{x}-3}{\sqrt{x+1}}\)
\(R=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(R=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
\(b)\) Ta có : \(R< -1\)
\(\Leftrightarrow\)\(\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}< -1\)
\(\Leftrightarrow\)\(\frac{\sqrt{x}-3}{\sqrt{x}+3}< \frac{-1}{3}\)
\(\Leftrightarrow\)\(3\sqrt{x}-9< -\sqrt{x}-3\)
\(\Leftrightarrow\)\(4\sqrt{x}< 6\)
\(\Leftrightarrow\)\(\sqrt{x}< \frac{3}{2}\)
\(\Leftrightarrow\)\(x< \frac{9}{4}\)
Chúc bạn học tốt ~
16 tháng 9 2018
ĐKXĐ: \(x\ne3;x\ne-3\)
Biểu thức = \(\frac{2\sqrt{x}.\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
=\(\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\) = \(\frac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\) =\(\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
=\(\frac{3\sqrt{x}}{\sqrt{x}-3}\)
B
28 tháng 7 2019
\(a,E=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\left(Đk:x\ge0;x\ne\pm1\right)\)(Đề như này mới đúng!)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{2x-2\sqrt{x}+3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{7\sqrt{x}-2-5x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{5\sqrt{x}+2\sqrt{x}-2-5x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\left(5\sqrt{x}-5x\right)+\left(2\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
Vậy...
\(b,\)Ta có:\(\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{-15+17-5\sqrt{x}}{\sqrt{x}+3}=\frac{\left(-15-5\sqrt{x}\right)+17}{\sqrt{x}+3}=\frac{-5\left(\sqrt{x}+3\right)+17}{\sqrt{x}+3}=-5+\frac{17}{\sqrt{x}+3}\)
Vì \(\sqrt{x}\ge0\forall x\Rightarrow\sqrt{x}+3\ge3\forall x\Rightarrow\frac{17}{\sqrt{x}+3}\le\frac{17}{3}\Rightarrow-5+\frac{17}{\sqrt{x}+3}\le\frac{2}{3}\Rightarrow E\le\frac{2}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
H9
HT.Phong (9A5)
CTVHS
23 tháng 8 2023
a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)
\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(2-5\right)\)
\(=-\left(-3\right)\)
\(=3\)
b) Ta có:
\(x^2-x\sqrt{3}+1\)
\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)
Dấu "=" xảy ra:
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)
Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)
H
23 tháng 8 2023
a)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
\(\frac{x^2-3}{x+\sqrt{3}}\)
\(=\frac{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}{x+\sqrt{3}}\)
\(=x-\sqrt{3}\)
#H
\(\frac{x^2-3}{x+\sqrt{3}}=\frac{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}{x+\sqrt{3}}=x-\sqrt{3}\)