Toán Lớp 7: tìm x
A=x+1/x−2(x khác2 )
B= 2x-1 / x+5 (x khác -5)
C= 10x – 9/ 2x – 3
( chia hết)
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Toán Lớp 7: tìm x
A=x+1/x−2(x khác2 )
B= 2x-1 / x+5 (x khác -5)
C= 10x – 9/ 2x – 3
( chia hết)
Comments ( 1 )
A = \dfrac{{x + 1}}{{x – 2}} = \dfrac{{x – 2 + 3}}{{x – 2}} = 1 + \dfrac{3}{{x – 2}}\\
\Leftrightarrow \dfrac{3}{{x – 2}} \in Z\\
\Leftrightarrow \left( {x – 2} \right) \in Ư\left( 3 \right)\\
\Leftrightarrow x – 2 \in \left\{ { – 3; – 1;1;3} \right\}\\
\Leftrightarrow x \in \left\{ { – 1;1;3;5} \right\}\left( {tmdk} \right)\\
Vậy\,x \in \left\{ { – 1;1;3;5} \right\}\\
B = \dfrac{{2x – 1}}{{x + 5}}\\
= \dfrac{{2x + 10 – 11}}{{x + 5}}\\
= \dfrac{{2\left( {x + 5} \right)}}{{x + 5}} – \dfrac{{11}}{{x + 5}}\\
= 2 – \dfrac{{11}}{{x + 5}}\\
\Leftrightarrow \left( {x + 5} \right) \in Ư\left( {11} \right)\\
\Leftrightarrow x + 5 \in \left\{ { – 11; – 1;1;11} \right\}\\
\Leftrightarrow x \in \left\{ { – 16; – 6; – 4;6} \right\}\\
Vậy\,x \in \left\{ { – 16; – 6; – 4;6} \right\}\\
C = \dfrac{{10x – 9}}{{2x – 3}} = \dfrac{{10x – 15 + 6}}{{2x – 3}}\\
= \dfrac{{5\left( {2x – 3} \right) + 6}}{{2x – 3}}\\
= 5 + \dfrac{6}{{2x – 3}}\\
\Leftrightarrow \left( {2x – 3} \right) \in Ư\left( 6 \right)\\
\Leftrightarrow \left( {2x – 3} \right) \in \left\{ { – 6; – 3; – 2; – 1;1;2;3;6} \right\}\\
\Leftrightarrow 2x \in \left\{ { – 3;0;1;2;4;5;6;9} \right\}\\
\Leftrightarrow x \in \left\{ { – \dfrac{3}{2};0;\dfrac{1}{2};1;2;\dfrac{5}{2};3;\dfrac{9}{2}} \right\}\\
Vậy\,x \in \left\{ { – \dfrac{3}{2};0;\dfrac{1}{2};1;2;\dfrac{5}{2};3;\dfrac{9}{2}} \right\}
\end{array}$