12 bài tập Rút gọn biểu thức có chứa căn thức bậc hai 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 = $\frac{1}{{\sqrt x - 4}}$.

Hướng dẫn giải

Với x > 0, x ≠ 9, ta có:

$B = \frac{{2\sqrt x }}{{\sqrt x + 3}} - \frac{{\sqrt x + 1}}{{\sqrt x - 3}} - \frac{{7\sqrt x + 3}}{{9 - x}}$

$B = \frac{{2\sqrt x \left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} - \frac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} + \frac{{7\sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}$

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

$B = \frac{{x - 3\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} = \frac{{\left( {\sqrt x - 3} \right)\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} = \frac{{\sqrt x }}{{\sqrt x + 3}}$.

Hướng dẫn giải

Với x > 0, x ≠ 9, ta có:

$A = \left( {\frac{{2\sqrt x }}{{\sqrt x + 3}} + \frac{{\sqrt x }}{{\sqrt x - 3}} + \frac{{3x + 3}}{{9 - x}}} \right):\left( {\frac{{2\sqrt x - 2}}{{\sqrt x - 3}} - 1} \right)$

$A = \left[ {\frac{{2\sqrt x \left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} + \frac{{\sqrt x \left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} - \frac{{3x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}} \right]:\left( {\frac{{\sqrt x + 1}}{{\sqrt x - 3}}} \right)$

$A = \left[ {\frac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}} \right].\left( {\frac{{\sqrt x - 3}}{{\sqrt x + 1}}} \right)$

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

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

$A = \frac{{ - 3}}{{\sqrt x + 3}}$.

Vậy với x > 0, x ≠ 9 thì $A = \frac{{ - 3}}{{\sqrt x + 3}}$.

Với x ≥ 0, ta có:

$P = \left( {1 + \frac{{\sqrt x }}{{x + \sqrt x + 1}}} \right):\frac{{\sqrt x + 1}}{{x\sqrt x - 1}}$

$P = \frac{{x + \sqrt x + 1 + \sqrt x }}{{x + \sqrt x + 1}}.\frac{{x\sqrt x - 1}}{{\sqrt x + 1}}$

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

$P = \frac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{x + \sqrt x + 1}}.\frac{{\left( {\sqrt x - 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\sqrt x + 1}}$

$P = \frac{{\left( {\sqrt x + 1} \right)}}{{x + \sqrt x + 1}}.\left( {\sqrt x - 1} \right)\left( {x - \sqrt x + 1} \right)$

$P = \frac{{\left( {x - 1} \right)\left( {x - \sqrt x + 1} \right)}}{{x + \sqrt x + 1}}$.

Với x ≥ 0, x ≠ 9, ta có:

$A = \left( {\frac{{x + 3}}{{x - 9}} + \frac{1}{{\sqrt x + 3}}} \right):\frac{{\sqrt x }}{{\sqrt x + 3}}$

$A = \left[ {\frac{{\left( {x + 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} + \frac{{\sqrt x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}} \right].\frac{{\sqrt x + 3}}{{\sqrt x }}$

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

$A = \frac{{\sqrt x + 1}}{{\sqrt x - 3}}.$