Bài tập Rút gọn biểu thức có chứa căn thức bậc hai lớp 9 (có lời giải)

Hướng dẫn giải

Với x > 0, x ≠ 4, ta có:

\(A = \left( {\frac{1}{{\sqrt x - 2}} + \frac{1}{{\sqrt x + 2}}} \right).\frac{{x - 4}}{{\sqrt x }}\)

\(A = \left[ {\frac{{\sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} + \frac{{\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}} \right].\frac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\sqrt x }}\)

\(A = \frac{{\left( {\sqrt x + 2 + \sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\frac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\sqrt x }}\)

\(A = \frac{{2\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\frac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\sqrt x }} = 2\).

Vậy với x > 0, x ≠ 4 thì A = 2.

Hướng dẫn giải

Với a ≥ 0, a ≠ 1, ta có:

\(A = \left( {\frac{{\sqrt a - 1}}{{\sqrt a + 1}} + \frac{{\sqrt a + 1}}{{\sqrt a - 1}}} \right).\left( {1 - \frac{2}{{a + 1}}} \right)\)

\(A = \left[ {\frac{{{{\left( {\sqrt a - 1} \right)}^2}}}{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}} + \frac{{{{\left( {\sqrt a + 1} \right)}^2}}}{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}} \right].\frac{{a - 1}}{{a + 1}}\)

\(A = \frac{{\left( {a - 2\sqrt a + 1 + a + 2\sqrt a + 1} \right)}}{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}.\frac{{\left( {a - 1} \right)}}{{\left( {a + 1} \right)}}\)

\(A = \frac{{2\left( {a + 1} \right)}}{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}.\frac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}{{\left( {a + 1} \right)}} = 2\).

Vậy với a ≥ 0, a ≠ 1 thì a = 2.

Với a > 0, a ≠ 4, ta có:

\(A = \left( {\frac{{\sqrt a - 2}}{{\sqrt a + 2}} - \frac{{\sqrt a + 2}}{{\sqrt a - 2}}} \right).\left( {\sqrt a - \frac{2}{{\sqrt a }}} \right)\)

\(A = \left[ {\frac{{{{\left( {\sqrt a - 2} \right)}^2}}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 2} \right)}} - \frac{{{{\left( {\sqrt a + 2} \right)}^2}}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 2} \right)}}} \right].\left( {\frac{{a - 2}}{{\sqrt a }}} \right)\)

\(A = \frac{{{{\left( {\sqrt a - 2} \right)}^2} - {{\left( {\sqrt a + 2} \right)}^2}}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 2} \right)}}.\left( {\frac{{a - 2}}{{\sqrt a }}} \right)\)

\(A = \frac{{\left( {\sqrt a - 2 - \sqrt a - 2} \right)\left( {\sqrt a - 2 + \sqrt a + 2} \right)}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 2} \right)}}.\left( {\frac{{a - 2}}{{\sqrt a }}} \right)\)

\(A = \frac{{ - 4.2\sqrt a }}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 2} \right)}}.\left( {\frac{{a - 2}}{{\sqrt a }}} \right)\)

\(A = \frac{{ - 8\left( {a - 2} \right)}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 2} \right)}}\).

Hướng dẫn giải

Với x ≥ 0, x ≠ 4, ta có:

\(A = \frac{{5\sqrt x - 3}}{{\sqrt x - 2}} + \frac{{3\sqrt x + 1}}{{\sqrt x + 2}} - \frac{{x + 2\sqrt x - 8}}{{x - 4}}\)

\(A = \frac{{\left( {5\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} + \frac{{\left( {3\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} - \frac{{x + 2\sqrt x - 8}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\)

\(A = \frac{{5x + 7\sqrt x - 6}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} + \frac{{3x - 5\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} - \frac{{x + 2\sqrt x - 8}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\)

\(A = \frac{{5x + 7\sqrt x - 6 + 3x - 5\sqrt x - 2 - x - 2\sqrt x + 8}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\)

\(A = \frac{{7x}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\).

Hướng dẫn giải

Với x ≥ 0, x ≠ 16, ta có:

\(A = \frac{{5 - 5\sqrt x }}{{x - 16}} - \frac{2}{{4 - \sqrt x }} + \frac{3}{{\sqrt x + 4}}\)

\(A = \frac{{5 - 5\sqrt x }}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}} + \frac{{2\left( {\sqrt x + 4} \right)}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}} + \frac{{3\left( {\sqrt x - 4} \right)}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}}\)

\(A = \frac{{5 - 5\sqrt x + 2\sqrt x + 8 + 3\sqrt x - 12}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}}\)

\(A = \frac{1}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}} = \frac{1}{{x - 16}}\).

Hướng dẫn giải

Với x ≥ 0, x ≠ \(\frac{9}{4}\), ta có:

\(A = \frac{{\sqrt x }}{{2\sqrt x - 3}} + \frac{{\sqrt x - 2}}{{2\sqrt x + 3}} + \frac{{15 - 4\sqrt x }}{{9 - 4x}}\)

\(A = \frac{{\sqrt x \left( {2\sqrt x + 3} \right)}}{{\left( {2\sqrt x - 3} \right)\left( {2\sqrt x + 3} \right)}} + \frac{{\left( {\sqrt x - 2} \right)\left( {2\sqrt x - 3} \right)}}{{\left( {2\sqrt x - 3} \right)\left( {2\sqrt x + 3} \right)}} - \frac{{15 - 4\sqrt x }}{{\left( {2\sqrt x - 3} \right)\left( {2\sqrt x + 3} \right)}}\)

\(A = \frac{{2x + 3\sqrt x }}{{\left( {2\sqrt x - 3} \right)\left( {2\sqrt x + 3} \right)}} + \frac{{2x - 7\sqrt x + 6}}{{\left( {2\sqrt x - 3} \right)\left( {2\sqrt x + 3} \right)}} - \frac{{15 - 4\sqrt x }}{{\left( {2\sqrt x - 3} \right)\left( {2\sqrt x + 3} \right)}}\)

\(A = \frac{{2x + 3\sqrt x + 2x - 7\sqrt x + 6 - 15 + 4\sqrt x }}{{\left( {2\sqrt x - 3} \right)\left( {2\sqrt x + 3} \right)}}\)

\(A = \frac{{4x - 9}}{{\left( {2\sqrt x - 3} \right)\left( {2\sqrt x + 3} \right)}}\)

\(A = \frac{{\left( {2\sqrt x - 3} \right)\left( {2\sqrt x + 3} \right)}}{{\left( {2\sqrt x - 3} \right)\left( {2\sqrt x + 3} \right)}} = 1\).

Vậy với x ≥ 0, x ≠ \(\frac{9}{4}\), thì A = 1.

Hướng dẫn giải

Với x > 0, ta có:

\(B = \frac{{2x + \sqrt x - 4}}{{x + 2\sqrt x }} - \frac{{\sqrt x + 1}}{{\sqrt x + 2}}\)

\(B = \frac{{2x + \sqrt x - 4}}{{\sqrt x \left( {\sqrt x + 2} \right)}} - \frac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x + 2} \right)}}\)

\(B = \frac{{2x + \sqrt x - 4 - x - \sqrt x }}{{\sqrt x \left( {\sqrt x + 2} \right)}}\)

\(B = \frac{{x - 4}}{{\sqrt x \left( {\sqrt x + 2} \right)}}\)

\(B = \frac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}{{\sqrt x \left( {\sqrt x + 2} \right)}} = \frac{{\sqrt x - 2}}{{\sqrt x }}\).

Ta có: \(P = \frac{A}{B}\)

Suy ra P = \(\frac{{x - 2\sqrt x + 2}}{{\sqrt x }}\): \(\frac{{\sqrt x - 2}}{{\sqrt x }}\) = \(\frac{{x - 2\sqrt x + 2}}{{\sqrt x }}\).\(\frac{{\sqrt x }}{{\sqrt x - 2}}\) = \(\frac{{x - 2\sqrt x + 2}}{{\sqrt x - 2}}\).

Vậy với x > 0 thì P = \(\frac{{x - 2\sqrt x + 2}}{{\sqrt x - 2}}\).

Hướng dẫn giải

Với x ≥ 0, x ≠ 9, x ≠ 16, ta có:

\(B = \frac{{2\sqrt x + 3}}{{\sqrt x - 3}} + \frac{{\sqrt x + 3}}{{4 - \sqrt x }} - \frac{{x - 6\sqrt x }}{{x - 7\sqrt x + 12}}\)

\(B = \frac{{\left( {2\sqrt x + 3} \right)\left( {\sqrt x - 4} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 4} \right)}} - \frac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 4} \right)}} - \frac{{x - 6\sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 4} \right)}}\)

\(B = \frac{{2x - 5\sqrt x - 12}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 4} \right)}} - \frac{{x - 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 4} \right)}} - \frac{{x - 6\sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 4} \right)}}\)

\(B = \frac{{2x - 5\sqrt x - 12 - x + 9 - x + 6\sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 4} \right)}}\)

\(B = \frac{{\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 4} \right)}} = \frac{1}{{\sqrt x - 4}}\).

Vậy với x ≥ 0, x ≠ 9, x ≠ 16 thì B = \(\frac{1}{{\sqrt x - 4}}\).