Toán Lớp 7: Tìm x ∈ Z để các biểu thức sau là số nguyên: A=3x-5/3x+2
B= √x-3= √x+2
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Toán Lớp 7: Tìm x ∈ Z để các biểu thức sau là số nguyên: A=3x-5/3x+2
B= √x-3= √x+2
Comments ( 1 )
A = \dfrac{{3x – 5}}{{3x + 2}}\\
= \dfrac{{3x + 2 – 7}}{{3x + 2}}\\
= 1 – \dfrac{7}{{3x + 2}}\\
A \in Z\\
\Leftrightarrow \dfrac{7}{{3x + 2}} \in Z\\
\Leftrightarrow \left( {3x + 2} \right) \in \left\{ { – 7; – 1;1;7} \right\}\\
\Leftrightarrow 3x \in \left\{ { – 9; – 3; – 1;5} \right\}\\
\Leftrightarrow x \in \left\{ { – 3; – 1; – \dfrac{1}{3};\dfrac{5}{3}} \right\}\\
Do:x \in Z\\
\Leftrightarrow x \in \left\{ { – 3; – 1} \right\}\\
b)Dkxd:x \ge 0\\
B = \dfrac{{\sqrt x – 3}}{{\sqrt x + 2}}\\
= \dfrac{{\sqrt x + 2 – 5}}{{\sqrt x + 2}}\\
= 1 – \dfrac{5}{{\sqrt x + 2}}\\
\Leftrightarrow \sqrt x + 2 \in \left\{ { – 5; – 1;1;5} \right\}\\
\Leftrightarrow \sqrt x \in \left\{ { – 7; – 3; – 1;3} \right\}\\
Do:\sqrt x \ge 0\\
\Leftrightarrow \sqrt x = 3\\
\Leftrightarrow x = 9\left( {tm} \right)\\
Vậy\,x = 9
\end{array}$