Tổng hợp bài tập trắc nghiệm rèn luyện tư duy chuyên đề Căn bậc hai và ứng dụng của nó

  1. Tác giả: LTTK CTV
    Đánh giá: ✪ ✪ ✪ ✪ ✪
  2. Tác giả: LTTK CTV
    Đánh giá: ✪ ✪ ✪ ✪ ✪
  3. Tác giả: LTTK CTV
    Đánh giá: ✪ ✪ ✪ ✪ ✪
    Rút gọn biểu thức \(\dfrac{a-b}{\sqrt{a}-\sqrt{b}}+\dfrac{\sqrt{a}^3+\sqrt{b}^3}{a-b}\)
    • \(\dfrac{2a-\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
    • \(\sqrt{a}\)
    • \(2\sqrt{a}\)
    • \(\sqrt{a}+\sqrt{b}\)
    Hướng dẫn giải:

    \(\dfrac{a-b}{\sqrt{a}-\sqrt{b}}+\dfrac{\sqrt{a}^3+\sqrt{b}^3}{a-b}\)
    \(=\sqrt{a}+\sqrt{b}+\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
    \(=\sqrt{a}+\sqrt{b}+\dfrac{a-\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\)
    \(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}+\dfrac{a-\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\)
    \(=\dfrac{a-b+a-\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\)
    \(=\dfrac{2a-\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
     
  4. Tác giả: LTTK CTV
    Đánh giá: ✪ ✪ ✪ ✪ ✪
  5. Tác giả: LTTK CTV
    Đánh giá: ✪ ✪ ✪ ✪ ✪
    Rút gọn biểu thức \(2\sqrt{3a}-\sqrt{75a}+a\sqrt{\dfrac{6}{5}.\dfrac{5}{2a}}-\dfrac{2}{5}\sqrt{300a^3}\) (a > 0)
    • \(-2\sqrt{3a}\left(1+2a\right)\)
    • \(2\sqrt{3a}\left(1+2a\right)\)
    • \(-2\sqrt{3a}\left(1-2a\right)\)
    • \(-2\sqrt{3a}\)
    Hướng dẫn giải:

    \(2\sqrt{3a}-\sqrt{75a}+a\sqrt{\dfrac{6}{5}.\dfrac{5}{2a}}-\dfrac{2}{5}\sqrt{300a^3}\)
    \(=2\sqrt{3a}-5\sqrt{3a}+a\sqrt{\dfrac{3}{a}}-\dfrac{2}{5}.10.a\sqrt{3a}\)
    \(=-3\sqrt{3a}+\sqrt{\dfrac{3}{a}.a^2}-4a\sqrt{3a}\)
    \(=-3\sqrt{3a}+\sqrt{3a}-4a\sqrt{3a}\)
    \(=-2\sqrt{3a}-4a\sqrt{3a}\)
    \(=-2\sqrt{3a}\left(1+2a\right)\).
     
  6. Tác giả: LTTK CTV
    Đánh giá: ✪ ✪ ✪ ✪ ✪
    Giải phương trình : \(\sqrt{36x+72}-\sqrt{\dfrac{1}{4}x+\dfrac{1}{2}}+\dfrac{3}{2}\sqrt{x+2}=14\).
    • \(x=2\)
    • \(x=1\)
    • \(x=\dfrac{1}{2}\)
    • \(x=3\)
    Hướng dẫn giải:

    Đkxđ: \(x\ge-2\).
    \(\sqrt{36x+72}-\sqrt{\dfrac{1}{4}x+\dfrac{1}{2}}+\dfrac{3}{2}\sqrt{x+2}=14\)
    \(\Leftrightarrow\sqrt{36\left(x+2\right)}-\sqrt{\dfrac{1}{4}\left(x+2\right)}+\dfrac{3}{2}\sqrt{x+2}=14\)
    \(\Leftrightarrow6\sqrt{x+2}-\dfrac{1}{2}\sqrt{x+2}+\dfrac{3}{2}\sqrt{x+2}=14\)
    \(\Leftrightarrow7\sqrt{x+2}=14\)
    \(\Leftrightarrow\sqrt{x+2}=2\)
    \(\Leftrightarrow x+2=4\)
    \(\Leftrightarrow x=2\)
     
  7. Tác giả: LTTK CTV
    Đánh giá: ✪ ✪ ✪ ✪ ✪
    Rút gọn biểu thức \(\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\)
    • \(4a\)
    • \(-4a\)
    • \(a+1\)
    • \(a-1\)
    Hướng dẫn giải:

    \(\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\)
    \(=\left(\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+4\sqrt{a}\right)\left(\dfrac{a-1}{\sqrt{a}}\right)\)
    \(=\left(\dfrac{a+2\sqrt{a}+1-\left(a-2\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+4\sqrt{a}\right).\dfrac{a-1}{\sqrt{a}}\)
    \(=\left(\dfrac{4\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+4\sqrt{a}\right).\dfrac{a-1}{\sqrt{a}}\)
    \(=4\sqrt{a}\left(\dfrac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+1\right).\dfrac{a-1}{\sqrt{a}}\)
    \(=4\left(\dfrac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right).\left(a-1\right)\)
    \(=4\left[1+\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\right]\)
    \(=4a\)
     
  8. Tác giả: LTTK CTV
    Đánh giá: ✪ ✪ ✪ ✪ ✪
    Biết \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}=2\). Tìm x.
    • \(x=8\)
    • \(x=\dfrac{1}{4}\)
    • \(x=2\sqrt{2}\)
    • \(x=16\)
    Hướng dẫn giải:

    Đkxđ: \(x\ne4\)
    \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}=2\)
    \(\Leftrightarrow\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=2\)
    \(\Leftrightarrow\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=2\)
    \(\Leftrightarrow\dfrac{\sqrt{x}\cdot\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=2\)
    \(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=2\)
    \(\Leftrightarrow\sqrt{x}=2\left(\sqrt{x}-2\right)\)
    \(\Leftrightarrow\sqrt{x}=4\)
    \(\Leftrightarrow x=16\).
     
  9. Tác giả: LTTK CTV
    Đánh giá: ✪ ✪ ✪ ✪ ✪
    Tính giá trị của biểu thức \(\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
    • 1
    • \(\sqrt{2}\)
    • \(\sqrt{3}\)
    • 2
    Hướng dẫn giải:

    \(\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
    \(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{20}+2}\)
    \(=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
    \(=\dfrac{\left(\sqrt{5}-1\right).\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
    \(=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\left(\sqrt{5}+1\right)}\)
    \(=\dfrac{2+2\sqrt{5}}{2\left(\sqrt{5}+1\right)}=1\)