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ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
a) M\(=\frac{x-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}:\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}-1}:\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)\(=\frac{x}{\sqrt{x}+1}\)
b) Khi \(x=7+4\sqrt{3}\Rightarrow\frac{7+4\sqrt{3}}{\sqrt{\left(2+\sqrt{3}\right)^2}+1}=\frac{7+4\sqrt{3}}{3+\sqrt{3}}\)
c)\(M=\frac{1}{2}\Leftrightarrow\frac{x}{\sqrt{x}+1}=\frac{1}{2}\Leftrightarrow\sqrt{x}=2x-1\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{1}{2}\\x^2=4x^2-4x+1\Leftrightarrow3x^2-4x+1=0\Leftrightarrow\left(3x-1\right)\left(x-1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{1}{2}\\\left[{}\begin{matrix}x=\frac{1}{3}\left(l\right)\\x=1\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
\(
1)Q = \left( {\dfrac{1}{{y - \sqrt y }} + \dfrac{1}{{\sqrt y - 1}}} \right):\left( {\dfrac{{\sqrt y + 1}}{{y - 2\sqrt y + 1}}} \right)\\
Q = \left( {\dfrac{1}{{\sqrt y \left( {\sqrt y - 1} \right)}} + \dfrac{1}{{\sqrt y - 1}}} \right).\dfrac{{y - 2\sqrt y + 1}}{{\sqrt y + 1}}\\
Q = \dfrac{{1 + \sqrt y }}{{\sqrt y \left( {\sqrt y - 1} \right)}}.\dfrac{{{{\left( {\sqrt y - 1} \right)}^2}}}{{\sqrt y + 1}}\\
Q = \dfrac{{\sqrt y - 1}}{{\sqrt y }}
\)
b) Thay \(y=3-2\sqrt{2}\) vào biểu thức ta được:
\(\dfrac{{\sqrt {3 - 2\sqrt 2 } - 1}}{{\sqrt {3 - 2\sqrt 2 } }} = \dfrac{{\sqrt {{{\left( {1 - \sqrt 2 } \right)}^2}} - 1}}{{\sqrt {{{\left( {1 - \sqrt 2 } \right)}^2}} }} = \dfrac{{ \sqrt 2 - 1-1}}{{\sqrt 2 -1}} \\= \dfrac{{\sqrt 2-2 }}{{ \sqrt 2 -1}} = \dfrac{{(\sqrt 2 -2)\left( { \sqrt 2+1 } \right)}}{{\left( { \sqrt 2-1 } \right)\left( {\sqrt 2+1 } \right)}} = - \sqrt 2 \)
\(2)B = \dfrac{{\sqrt y - 1}}{{{y^2} - y}}:\left( {\dfrac{1}{{\sqrt y }} - \dfrac{1}{{\sqrt y + 1}}} \right)\\ B = \dfrac{{\sqrt y - 1}}{{y\left( {y - 1} \right)}}:\dfrac{{\sqrt y + 1 - \sqrt y }}{{\sqrt y \left( {\sqrt y + 1} \right)}}\\ B = \dfrac{{\sqrt y - 1}}{{y\left( {\sqrt y - 1} \right)\left( {\sqrt y + 1} \right)}}:\dfrac{1}{{\sqrt y \left( {\sqrt y + 1} \right)}}\\ B = \dfrac{1}{{y\left( {\sqrt y + 1} \right)}}.\sqrt y \left( {\sqrt y + 1} \right)\\ B = \dfrac{{\sqrt y }}{y} \)
b) Thay \(y=3+2\sqrt{2}\) vào biểu thức ta được:
\(B = \dfrac{{\sqrt {3 + 2\sqrt 2 } }}{{3 + 2\sqrt 2 }} = \dfrac{{\sqrt {{{\left( {1 + \sqrt 2 } \right)}^2}} }}{{3 + 2\sqrt 2 }} = \dfrac{{\left( {1 + \sqrt 2 } \right)\left( {3 - 2\sqrt 2 } \right)}}{{\left( {3 + 2\sqrt 2 } \right)\left( {3 - 2\sqrt 2 } \right)}} = 3 - 2\sqrt 2 + 3\sqrt 2 - 4 = - 1 + \sqrt 2 \)
Nhiều quá @@
bình phương lên
\(Q^2=\frac{x-2\sqrt{\left(x-1\right)}+x+2\sqrt{\left(x-1\right)}+2\sqrt{\left(x-2\right)^2}}{x^2-4\left(x-1\right)}.\left(\frac{x-2}{x-1}\right)^2\)
\(=\frac{2x+2\left(x-2\right)}{\left(x-2\right)^2}.\frac{\left(x-2\right)^2}{\left(x-1\right)^2}=\frac{2\left(x+x-2\right)}{\left(x-1\right)^2}=\frac{4\left(x-1\right)}{\left(x-1\right)^2}=\frac{4}{x-1}\)
\(\Rightarrow Q=\frac{2}{\sqrt{x-1}}\)
Số hạng cuối cùng mẫu số là \(\frac{1}{\sqrt{x}-1}\) hay \(\frac{1}{\sqrt{x-1}}\) bạn?
do lỗi đánh nhầm ạ, phải là \(\frac{1}{\sqrt{x}-1}\)