Toán Lớp 7: Tìm x biết:
a, -10x . ( x – 1/3 ) = 0
b, -2x . ( x + 1/2 ) . ( x – 1/7 ) = 0
Tìm các số nguyên x; y biết:
x/4 – 1/y = 1/2
Tính: -4/11 . -3/11 . -2/11 . ….. . 3/11 . 4/11
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 7: Tìm x biết:
a, -10x . ( x – 1/3 ) = 0
b, -2x . ( x + 1/2 ) . ( x – 1/7 ) = 0
Tìm các số nguyên x; y biết:
x/4 – 1/y = 1/2
Tính: -4/11 . -3/11 . -2/11 . ….. . 3/11 . 4/11
Comments ( 1 )
a) – 10x\left( {x – \dfrac{1}{3}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x – \dfrac{1}{3} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{1}{3}
\end{array} \right.\\
Vậy\,x = 0;x = \dfrac{1}{3}\\
b) – 2x\left( {x + \dfrac{1}{2}} \right).\left( {x – \dfrac{1}{7}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x + \dfrac{1}{2} = 0\\
x – \dfrac{1}{7} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = – \dfrac{1}{2}\\
x = \dfrac{1}{7}
\end{array} \right.\\
Vậy\,x = 0;x = – \dfrac{1}{2};x = \dfrac{1}{7}\\
B2)\\
\dfrac{x}{4} – \dfrac{1}{y} = \dfrac{1}{2}\\
\Leftrightarrow \dfrac{x}{4} = \dfrac{1}{y} + \dfrac{1}{2}\\
\Leftrightarrow \dfrac{x}{4} = \dfrac{{y + 2}}{{2y}}\\
\Leftrightarrow 2xy = 4y + 8\\
\Leftrightarrow xy = 2y + 4\\
\Leftrightarrow xy – 2y = 4\\
\Leftrightarrow y\left( {x – 2} \right) = 4 = 1.4 = 2.2\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y = 4\\
x – 2 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
y = 1\\
x – 2 = 4
\end{array} \right.\\
\left\{ \begin{array}{l}
y = 2\\
x – 2 = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
y = – 4\\
x – 2 = – 1
\end{array} \right.\\
\left\{ \begin{array}{l}
y = – 1\\
x – 2 = – 4
\end{array} \right.\\
\left\{ \begin{array}{l}
y = – 2\\
x – 2 = – 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
y = 4;x = 3\\
y = 1;x = 6\\
y = 2;x = 4\\
y = – 4;x = 1\\
y = – 1;x = – 2\\
y = – 2;x = 0
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {4;3} \right);\left( {1;6} \right);\left( {2;4} \right);\left( { – 4;1} \right);\left( { – 1; – 2} \right);\left( { – 2;0} \right)} \right\}\\
B3)\\
\dfrac{{ – 4}}{{11}}.\dfrac{{ – 3}}{{11}}.\dfrac{{ – 2}}{{11}}….\dfrac{3}{{11}}.\dfrac{4}{{11}}\\
= \dfrac{{ – 4}}{{11}}.\dfrac{{ – 3}}{{11}}.\dfrac{{ – 2}}{{11}}.\dfrac{{ – 1}}{{11}}.0.\dfrac{1}{{11}}.\dfrac{2}{{11}}\dfrac{3}{{11}}.\dfrac{4}{{11}}\\
= 0
\end{array}$