by a) Giải phương trình sau: \(\sqrt{4{{x}^{2}}-4x+1}=\left| x-2 \right|\)
b) Thu gọn biểu thức sau: \(A=\frac{5+\sqrt{5}}{\sqrt{5}+2}+\frac{\sqrt{5}}{\sqrt{5}-1}-\frac{3\sqrt{5}}{3+\sqrt{5}}\)
Lời giải chi tiết:
a) Cách 1:
\(\begin{array}{l}\sqrt {4{x^2} – 4x + 1} = \left| {x – 2} \right|\\ \Leftrightarrow \sqrt {{{\left( {2x – 1} \right)}^2}} = \left| {x – 2} \right|\\ \Leftrightarrow \left| {2x – 1} \right| = \left| {x – 2} \right|\\ \Leftrightarrow \left[ \begin{array}{l}2x – 1 = x – 2\\2x – 1 = – x + 2\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}2x – x = – 2 + 1\\2x + x = 2 + 1\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = – 1\\3x = 3\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = – 1\\x = 1\end{array} \right.\end{array}\)
Cách 2:
\(\begin{array}{l}\sqrt {4{x^2} – 4x + 1} = \left| {x – 2} \right|\\ \Leftrightarrow {\left( {\sqrt {4{x^2} – 4x + 1} } \right)^2} = {\left( {x – 2} \right)^2}\\ \Leftrightarrow 4{x^2} – 4x + 1 = {x^2} – 4x + 4\\ \Leftrightarrow 4{x^2} – {x^2} = 4 – 1\\ \Leftrightarrow 3{x^2} = 3\\ \Leftrightarrow {x^2} = 1\\ \Leftrightarrow x = \pm 1\end{array}\)
Vậy: Nghiệm phương trình là: \(S=\left\{ -1;1 \right\}\)
b) \(\begin{array}{l}A = \frac{{5 + \sqrt 5 }}{{\sqrt 5 + 2}} + \frac{{\sqrt 5 }}{{\sqrt 5 – 1}} – \frac{{3\sqrt 5 }}{{3 + \sqrt 5 }}\\A = \frac{{\left( {5 + \sqrt 5 } \right)\left( {\sqrt 5 – 2} \right)}}{{\left( {\sqrt 5 + 2} \right)\left( {\sqrt 5 – 2} \right)}} + \frac{{\sqrt 5 \left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 – 1} \right)\left( {\sqrt 5 + 1} \right)}} – \frac{{3\sqrt 5 \left( {3 – \sqrt 5 } \right)}}{{\left( {3 + \sqrt 5 } \right)\left( {3 – \sqrt 5 } \right)}}\\A = \frac{{ – 5 + 3\sqrt 5 }}{1} + \frac{{5 + \sqrt 5 }}{4} – \frac{{ – 15 + 9\sqrt 5 }}{4}\\A = \frac{{ – 20 + 12\sqrt 5 }}{4} + \frac{{5 + \sqrt 5 }}{4} – \frac{{ – 15 + 9\sqrt 5 }}{4}\\A = \frac{{ – 20 + 12\sqrt 5 + 5 + \sqrt 5 + 15 – 9\sqrt 5 }}{4}\\A = \frac{{4\sqrt 5 }}{4}\\A = \sqrt 5 \end{array}\)
