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}` Reply
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 $3$ $↔ \sqrt[]{x}+1∈\{-3;-1;1;3\}$ $↔ \sqrt[]{x}∈\{-4;-2;0;2\}$ $→ x∈\{0;4\}$ Reply
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}`
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 $3$
$↔ \sqrt[]{x}+1∈\{-3;-1;1;3\}$
$↔ \sqrt[]{x}∈\{-4;-2;0;2\}$
$→ x∈\{0;4\}$