Nhờ các bạn giúp mình

Nhờ các bạn giúp mình
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  1. Đáp án:

    a/ $B=\dfrac{3}{7}$ khi $x=36$

    b/ $A=\dfrac{3\sqrt{x}}{\sqrt{x}-3}$

    c/ `x={0; 4}`

    Giải thích các bước giải:

    a/ $B=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}$

    $\text{ĐKXĐ: $x \geq 0$}$

    $\text{Thay $x=36$ vào B ta được:}$

    $B=\dfrac{\sqrt{36}-3}{\sqrt{36}+1}$

    $=\dfrac{6-3}{6+1}$

    $=\dfrac{3}{7}$

    b/ $A=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}$

    $\text{ĐKXĐ: $x \geq 0$ và $x \neq 9$}$

    $⇔ A=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{(3-\sqrt{x})(3+\sqrt{x})}$

    $=\dfrac{2\sqrt{x}(3-\sqrt{x})-(\sqrt{x}+1)(\sqrt{x}+3)+3-11\sqrt{x}}{(3-\sqrt{x})(3+\sqrt{x})}$

    $=\dfrac{6\sqrt{x}-2x-x-\sqrt{x}-3\sqrt{x}-3+3-11\sqrt{x}}{(3-\sqrt{x})(3+\sqrt{x})}$

    $=\dfrac{-3x-9\sqrt{x}}{(3-\sqrt{x})(3+\sqrt{x})}$

    $=\dfrac{-3\sqrt{x}(\sqrt{x}+3)}{(3-\sqrt{x})(3+\sqrt{x})}$

    $=\dfrac{-3\sqrt{x}}{3-\sqrt{x}}$

    $=\dfrac{3\sqrt{x}}{\sqrt{x}-3}$

    c/ $P=A.B=\dfrac{3\sqrt{x}}{\sqrt{x}-3}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}$

    $=\dfrac{3\sqrt{x}}{\sqrt{x}+1}$

    $=\dfrac{3\sqrt{x}+3-3}{\sqrt{x}+1}$

    $=3-\dfrac{3}{\sqrt{x}+1}$

    $\text{Để P là số nguyên thì $\dfrac{3}{\sqrt{x}+1}$ nguyên}$

    $⇔ (\sqrt{x}+1) ∈ Ư_{3}=$ `{1; 3}` $\text{(Vì $\sqrt{x}+1 \geq 1$)}$

    $· \sqrt{x}+1=1 ⇔ \sqrt{x}=0 ⇔ x=0$

    $· \sqrt{x}+1=3 ⇔ \sqrt{x}=2 ⇔ x=4$

    $\text{Vậy để P nguyên thì}$ `x={0; 4}`

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  2. Đáp án:

    a) $B=\dfrac{3}{7}$

    b) $A=\dfrac{3\sqrt[]{x}}{\sqrt[]{x}-3}$

    c) $x∈\{0;4\}$

    Giải thích các bước giải:

    a) Thay $x=36$ vào $B$, ta có:

    $B=\dfrac{\sqrt[]{36}-3}{\sqrt[]{36}+1}$

    $=\dfrac{6-3}{6+1}$

    $=\dfrac{3}{7}$

    b) $A=\dfrac{2\sqrt[]{x}}{\sqrt[]{x}+3}+\dfrac{\sqrt[]{x}+1}{\sqrt[]{x}-3}+\dfrac{3-11\sqrt[]{x}}{9-x}$

    $=\dfrac{2\sqrt[]{x}}{\sqrt[]{x}+3}+\dfrac{\sqrt[]{x}+1}{\sqrt[]{x}-3}-\dfrac{3-11\sqrt[]{x}}{(\sqrt[]{x}+3)(\sqrt[]{x}-3)}$

    $=\dfrac{2\sqrt[]{x}(\sqrt[]{x}-3)+(\sqrt[]{x}+1)(\sqrt[]{x}+3)-3+11\sqrt[]{x}}{(\sqrt[]{x}+3)(\sqrt[]{x}-3)}$

    $=\dfrac{2x-6\sqrt[]{x}+x+4\sqrt[]{x}+3-3+11\sqrt[]{x}}{(\sqrt[]{x}+3)(\sqrt[]{x}-3)}$

    $=\dfrac{3x+9\sqrt[]{x}}{(\sqrt[]{x}+3)(\sqrt[]{x}-3)}$

    $=\dfrac{3\sqrt[]{x}(\sqrt[]{x}+3)}{(\sqrt[]{x}+3)(\sqrt[]{x}-3)}$

    $=\dfrac{3\sqrt[]{x}}{\sqrt[]{x}-3}$

    c) $P=A.B$

    $=\dfrac{3\sqrt[]{x}}{\sqrt[]{x}-3}.\dfrac{\sqrt[]{x}-3}{\sqrt[]{x}+1}$

    $=\dfrac{3\sqrt[]{x}}{\sqrt[]{x}+1}$

    $=\dfrac{3(\sqrt[]{x}+1)}{\sqrt[]{x}+1}-\dfrac{3}{\sqrt[]{x}+1}$

    $=3-\dfrac{3}{\sqrt[]{x}+1}$

    Để $P∈\mathbb{Z}$ thì $\sqrt[]{x}+1$ phải là ước nguyên của $3$

    $↔ \sqrt[]{x}+1∈\{-3;-1;1;3\}$

    $↔ \sqrt[]{x}∈\{-4;-2;0;2\}$

    $→ x∈\{0;4\}$

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