Rút gọn các biểu thức:
a) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\) ( a <0 ; b # 0 )
b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\) ( x lớn hơn hoặc = 0)
c) \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\) ( x<3 tại x = 0,5)
d) \(\dfrac{x-1}{\sqrt{y}-1}.\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\) ( x khác 1; y >= 0, y khác 1)
e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\) ( x > -2 tại x = -\(\sqrt{2}\))
Đáp án:
d. \(\left[ \begin{array}{l}
D = \dfrac{1}{{x – 1}}\\
D = – \dfrac{1}{{x – 1}}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\dfrac{{4.{a^2}.{b^3}}}{{8\sqrt 2 .{{\left| a \right|}^3}.{b^3}}} = \dfrac{{4.{a^2}.{b^3}}}{{8\sqrt 2 .{{\left( { – a} \right)}^3}.{b^3}}} = – \dfrac{{{a^2}.{b^3}}}{{2\sqrt 2 .{a^3}.{b^3}}}\\
= – \dfrac{1}{{2\sqrt 2 a}}\\
b.B = \sqrt {\dfrac{{{{\left( {\sqrt x – 1} \right)}^2}}}{{{{\left( {\sqrt x + 1} \right)}^2}}}} = \dfrac{{\left| {\sqrt x – 1} \right|}}{{\left| {\sqrt x + 1} \right|}} = \dfrac{{\left| {\sqrt x – 1} \right|}}{{\sqrt x + 1}}\\
\to \left[ \begin{array}{l}
B = \dfrac{{\sqrt x – 1}}{{\sqrt x + 1}}\left( {DK:x \ge 1} \right)\\
B = – \dfrac{{\sqrt x – 1}}{{\sqrt x + 1}}\left( {DK:0 \le x < 1} \right)
\end{array} \right.\\
c.C = \dfrac{{\left| {x – 2} \right|}}{{\left| {3 – x} \right|}} + \dfrac{{{x^2} – 1}}{{x – 3}}\\
\to \left[ \begin{array}{l}
C = \dfrac{{x – 2}}{{x – 3}} + \dfrac{{{x^2} – 1}}{{x – 3}}\left( {DK:2 \le x < 3} \right)\\
C = – \dfrac{{x – 2}}{{x – 3}} + \dfrac{{{x^2} – 1}}{{x – 3}}\left( {DK:x < 2} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
C = \dfrac{{{x^2} + x – 3}}{{x – 3}}\\
C = \dfrac{{{x^2} – x + 1}}{{x – 3}}
\end{array} \right.\\
d.D = \dfrac{{x – 1}}{{\sqrt y – 1}}.\sqrt {\dfrac{{{{\left( {\sqrt y – 1} \right)}^2}}}{{{{\left( {x – 1} \right)}^4}}}} = \dfrac{{x – 1}}{{\sqrt y – 1}}.\dfrac{{\left| {\sqrt y – 1} \right|}}{{{{\left( {x – 1} \right)}^2}}}\\
\to \left[ \begin{array}{l}
D = \dfrac{{x – 1}}{{\sqrt y – 1}}.\dfrac{{\sqrt y – 1}}{{{{\left( {x – 1} \right)}^2}}}\left( {DK:y > 1} \right)\\
D = \dfrac{{x – 1}}{{\sqrt y – 1}}.\dfrac{{ – \left( {\sqrt y – 1} \right)}}{{{{\left( {x – 1} \right)}^2}}}\left( {DK:0 \le y < 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
D = \dfrac{1}{{x – 1}}\\
D = – \dfrac{1}{{x – 1}}
\end{array} \right.\\
e.E = 4x – 2\sqrt 2 + \dfrac{{\sqrt {{x^2}\left( {x + 2} \right)} }}{{\sqrt {x + 2} }} = 4x – 2\sqrt 2 + \left| x \right|\\
\to \left[ \begin{array}{l}
E = 4x – 2\sqrt 2 + x\left( {DK:x \ge 0} \right)\\
E = 4x – 2\sqrt 2 – x\left( {DK: – 2 < x < 0} \right)
\end{array} \right.\\
Thay:x = – \sqrt 2 \\
\to E = 4.\left( { – \sqrt 2 } \right) – 2\sqrt 2 – \left( { – \sqrt 2 } \right) = – 5\sqrt 2
\end{array}\)
a) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\)
\(=\dfrac{4a^2b^3}{8\sqrt{2}a^3b^3}\)
\(=\dfrac{1}{2\sqrt{2}a}\)
\(=\dfrac{\sqrt{2}}{4a}\)
b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
chịu đấy :v
c) \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{3-x}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{-\left(x-3\right)}+\dfrac{x^2-1}{x-3}\)
\(=-\dfrac{x-2}{x-3}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{-\left(x-2\right)+x^2-1}{x-3}\)
\(=\dfrac{-x+1+x^2}{x-3}\)
d) \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(x-1\right)^2}\)
\(=\dfrac{1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{x-1}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(\sqrt{y}-1\right)\left(x-1\right)}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{x\sqrt{y}-\sqrt{y}-x+1}\)
e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\dfrac{\sqrt{x^2\cdot\left(x+2\right)}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\sqrt{x^2}\)
\(=4x-2\sqrt{x}+x\)
\(=5x-2\sqrt{2}\)