A=2×√X+1/x-7√X+12-√x+3/√x-4-2√x+1/3-√x a)RG…..A b)tính A tại x=2√7-4√3 02/08/2021 Bởi Savannah A=2×√X+1/x-7√X+12-√x+3/√x-4-2√x+1/3-√x a)RG…..A b)tính A tại x=2√7-4√3
Giải thích các bước giải: Ta có: \(\begin{array}{l}a,\\A = \dfrac{{2\sqrt x + 1}}{{x – 7\sqrt x + 12}} – \dfrac{{\sqrt x + 3}}{{\sqrt x – 4}} – \dfrac{{2\sqrt x + 1}}{{3 – \sqrt x }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}x \ge 0\\x \ne 9\\x \ne 12\end{array} \right)\\ = \dfrac{{2\sqrt x + 1}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 4} \right)}} + \dfrac{{\sqrt x + 3}}{{\sqrt x – 4}} + \dfrac{{2\sqrt x + 1}}{{\sqrt x – 3}}\\ = \dfrac{{2\sqrt x + 1 + \left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right) + \left( {2\sqrt x + 1} \right)\left( {\sqrt x – 4} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 4} \right)}}\\ = \dfrac{{2\sqrt x + 1 + \left( {x – 9} \right) + \left( {2x – 7\sqrt x – 4} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 4} \right)}}\\ = \dfrac{{3x – 5\sqrt x – 12}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 4} \right)}}\\ = \dfrac{{\left( {3\sqrt x + 4} \right)\left( {\sqrt x – 3} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 4} \right)}}\\ = \dfrac{{3\sqrt x + 4}}{{\sqrt x – 4}}\\b,\\x = 2\sqrt {7 – 4\sqrt 3 } = 2.\sqrt {4 – 2.2.\sqrt 3 + 3} = 2\sqrt {{{\left( {2 – \sqrt 3 } \right)}^2}} = 2.\left( {2 – \sqrt 3 } \right) = 4 – 2\sqrt 3 \\ \Rightarrow \sqrt x = \sqrt {4 – 2\sqrt 3 } = \sqrt {3 – 2.\sqrt 3 .1 + 1} = \sqrt {{{\left( {\sqrt 3 – 1} \right)}^2}} = \sqrt 3 – 1\\A = \dfrac{{3\sqrt x + 4}}{{\sqrt x – 4}} = \dfrac{{3.\left( {\sqrt 3 – 1} \right) + 4}}{{\left( {\sqrt 3 – 1} \right) – 4}} = \dfrac{{3\sqrt 3 + 1}}{{\sqrt 3 – 5}}\\ = \dfrac{{\left( {3\sqrt 3 + 1} \right)\left( {\sqrt 3 + 5} \right)}}{{\left( {\sqrt 3 – 5} \right)\left( {\sqrt 3 + 5} \right)}} = \dfrac{{9 + 16\sqrt 3 + 5}}{{3 – 25}} = \dfrac{{14 + 16\sqrt 3 }}{{ – 22}}\\ = – \dfrac{{7 + 8\sqrt 3 }}{{11}}\end{array}\) Bình luận
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = \dfrac{{2\sqrt x + 1}}{{x – 7\sqrt x + 12}} – \dfrac{{\sqrt x + 3}}{{\sqrt x – 4}} – \dfrac{{2\sqrt x + 1}}{{3 – \sqrt x }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
x \ge 0\\
x \ne 9\\
x \ne 12
\end{array} \right)\\
= \dfrac{{2\sqrt x + 1}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 4} \right)}} + \dfrac{{\sqrt x + 3}}{{\sqrt x – 4}} + \dfrac{{2\sqrt x + 1}}{{\sqrt x – 3}}\\
= \dfrac{{2\sqrt x + 1 + \left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right) + \left( {2\sqrt x + 1} \right)\left( {\sqrt x – 4} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 4} \right)}}\\
= \dfrac{{2\sqrt x + 1 + \left( {x – 9} \right) + \left( {2x – 7\sqrt x – 4} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 4} \right)}}\\
= \dfrac{{3x – 5\sqrt x – 12}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 4} \right)}}\\
= \dfrac{{\left( {3\sqrt x + 4} \right)\left( {\sqrt x – 3} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 4} \right)}}\\
= \dfrac{{3\sqrt x + 4}}{{\sqrt x – 4}}\\
b,\\
x = 2\sqrt {7 – 4\sqrt 3 } = 2.\sqrt {4 – 2.2.\sqrt 3 + 3} = 2\sqrt {{{\left( {2 – \sqrt 3 } \right)}^2}} = 2.\left( {2 – \sqrt 3 } \right) = 4 – 2\sqrt 3 \\
\Rightarrow \sqrt x = \sqrt {4 – 2\sqrt 3 } = \sqrt {3 – 2.\sqrt 3 .1 + 1} = \sqrt {{{\left( {\sqrt 3 – 1} \right)}^2}} = \sqrt 3 – 1\\
A = \dfrac{{3\sqrt x + 4}}{{\sqrt x – 4}} = \dfrac{{3.\left( {\sqrt 3 – 1} \right) + 4}}{{\left( {\sqrt 3 – 1} \right) – 4}} = \dfrac{{3\sqrt 3 + 1}}{{\sqrt 3 – 5}}\\
= \dfrac{{\left( {3\sqrt 3 + 1} \right)\left( {\sqrt 3 + 5} \right)}}{{\left( {\sqrt 3 – 5} \right)\left( {\sqrt 3 + 5} \right)}} = \dfrac{{9 + 16\sqrt 3 + 5}}{{3 – 25}} = \dfrac{{14 + 16\sqrt 3 }}{{ – 22}}\\
= – \dfrac{{7 + 8\sqrt 3 }}{{11}}
\end{array}\)