1. căn 3+2 căn2 – căn 3 – 2 căn 2 – căn 3-2 căn 2
2. phân tích thành nhân tử:
a- 5a căn b + 6a^2
3.cho P= 2 căn x -9/ x -5 căn x +6 – căn x + 3 / căn x -2 – 2 căn x +1 /3- căn x
a) rút gọn P
b) tìm x để P<1
1. căn 3+2 căn2 – căn 3 – 2 căn 2 – căn 3-2 căn 2
2. phân tích thành nhân tử:
a- 5a căn b + 6a^2
3.cho P= 2 căn x -9/ x -5 căn x +6 – căn x + 3 / căn x -2 – 2 căn x +1 /3- căn x
a) rút gọn P
b) tìm x để P<1
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
\sqrt {3 + 2\sqrt 2 } – \sqrt {3 – 2\sqrt 2 } \\
= \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} – \sqrt {{{\left( {\sqrt 2 – 1} \right)}^2}} \\
= \left( {\sqrt 2 + 1} \right) – \left( {\sqrt 2 – 1} \right)\\
= 2\\
2,\\
a – 5a\sqrt b + 6{a^2}\\
= a.\left( {1 – 5\sqrt b + 6a} \right)\\
3,\\
P = \dfrac{{2\sqrt x – 9}}{{x – 5\sqrt x + 6}} – \dfrac{{\sqrt x + 3}}{{\sqrt x – 2}} – \dfrac{{2\sqrt x + 1}}{{3 – \sqrt x }}\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
x \ge 0\\
x \ne 4\\
x \ne 9
\end{array} \right)\\
= \dfrac{{2\sqrt x – 9}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}} – \dfrac{{\sqrt x + 3}}{{\sqrt x – 2}} + \dfrac{{2\sqrt x + 1}}{{\sqrt x – 3}}\\
= \dfrac{{\left( {2\sqrt x – 9} \right) – \left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right) + \left( {2\sqrt x + 1} \right)\left( {\sqrt x – 2} \right)}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}}\\
= \dfrac{{\left( {2\sqrt x – 9} \right) – \left( {x – 9} \right) + \left( {2x – 3\sqrt x – 2} \right)}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}}\\
= \dfrac{{x – \sqrt x – 2}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}}\\
= \dfrac{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x – 3}}\\
b,\\
P < 1 \Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x – 3}} < 1\\
\Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x – 3}} – 1 < 0\\
\Leftrightarrow \dfrac{{\left( {\sqrt x + 1} \right) – \left( {\sqrt x – 3} \right)}}{{\sqrt x – 3}} < 0\\
\Leftrightarrow \dfrac{4}{{\sqrt x – 3}} < 0\\
\Leftrightarrow \sqrt x < 3\\
\Leftrightarrow x < 9\\
\Rightarrow \left\{ \begin{array}{l}
0 \le x < 9\\
x \ne 4
\end{array} \right.
\end{array}\)