Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{*{20}{l}}
\begin{array}{l}
a,{\rm{ }}2x{\rm{ }} - {\rm{ }}\left( { - 19} \right){\rm{ }} = {\rm{ }}15\\
= > 2{\rm{x}} = 15 + ( - 19)\\
= > 2{\rm{x}} = - 4\\
= > x = - 2\\
Vay:x = - 2
\end{array}\\
\begin{array}{l}
b,{\rm{ }}14{\rm{ }} - {\rm{ }}\left( {50{\rm{ }} - {\rm{ }}x} \right){\rm{ }} = 27\\
= > 50 - x = 14 - 27\\
= > 50 - x = - 13\\
= > x = 50 + 13\\
= > x = 63\\
Vay:x = 63
\end{array}\\
\begin{array}{l}
c,{\rm{ }}\left( {x + 1} \right){\rm{ }}\left( {{x^2} - {\rm{ }}4{\rm{ }}} \right){\rm{ }} = {\rm{ }}0\\
= > \left[ \begin{array}{l}
x + 1 = 0\\
{x^2} - 4 = 0
\end{array} \right.\\
= > \left[ \begin{array}{l}
x = - 1\\
{x^2} = 4
\end{array} \right.\\
= > \left[ \begin{array}{l}
x = - 1\\
x = 2\\
x = - 2
\end{array} \right.\\
vay:S = \{ - 1;2; - 2\}
\end{array}\\
\begin{array}{l}
d,x + \frac{1}{6} = \frac{{ - 1}}{2}\\
= > x = \frac{{ - 1}}{2} - \frac{1}{6}\\
= > x = \frac{{ - 2}}{3}\\
Vay:x = \frac{{ - 2}}{3}
\end{array}
\end{array}\]

\[\begin{array}{l}
e)Xet:A = \frac{{x + 3}}{{x - 2}}\left( {x \ne 2} \right)\\
= \frac{{x - 2 + 5}}{{x - 2}}\\
= 1 + \frac{5}{{x - 2}}\\
de:\left( {x + 3} \right) \vdots \left( {x - 2} \right)\\
= > \frac{5}{{x - 2}} \in Z = > x - 2 \in U(5) = \{ - 5; - 1;1;5\} \\
= > x \in \{ - 3;1;3;7\} \left( {tm} \right)\\
Vay:x \in \{ - 3;1;3;7\} \\
f)Xet:A = \frac{{2{\rm{x}} + 1}}{{x - 5}}\left( {x \ne 5} \right)\\
= \frac{{2{\rm{x}} - 10 + 11}}{{x - 5}}\\
= \frac{{2\left( {x - 5} \right)}}{{x - 5}} + \frac{{11}}{{x - 5}}\\
= 2 + \frac{{11}}{{x - 5}}\\
de:\left( {2{\rm{x}} + 1} \right) \vdots \left( {x - 5} \right)\\
= > \frac{{11}}{{x - 5}} \in Z = > x - 5 \in U(11) = \{ - 11; - 1;1;11\} \\
= > x \in \{ - 6;4;6;16\} \left( {tm} \right)\\
Vay:x \in \{ - 6;4;6;16\}
\end{array}\]

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Answer 2

\[\begin{array}{*{20}{l}}
\begin{array}{l}
a,{\rm{ }}2x{\rm{ }} - {\rm{ }}\left( { - 19} \right){\rm{ }} = {\rm{ }}15\\
= > 2{\rm{x}} = 15 + ( - 19)\\
= > 2{\rm{x}} = - 4\\
= > x = - 2\\
Vay:x = - 2
\end{array}\\
\begin{array}{l}
b,{\rm{ }}14{\rm{ }} - {\rm{ }}\left( {50{\rm{ }} - {\rm{ }}x} \right){\rm{ }} = 27\\
= > 50 - x = 14 - 27\\
= > 50 - x = - 13\\
= > x = 50 + 13\\
= > x = 63\\
Vay:x = 63
\end{array}\\
\begin{array}{l}
c,{\rm{ }}\left( {x + 1} \right){\rm{ }}\left( {{x^2} - {\rm{ }}4{\rm{ }}} \right){\rm{ }} = {\rm{ }}0\\
= > \left[ \begin{array}{l}
x + 1 = 0\\
{x^2} - 4 = 0
\end{array} \right.\\
= > \left[ \begin{array}{l}
x = - 1\\
{x^2} = 4
\end{array} \right.\\
= > \left[ \begin{array}{l}
x = - 1\\
x = 2\\
x = - 2
\end{array} \right.\\
vay:S = \{ - 1;2; - 2\}
\end{array}\\
\begin{array}{l}
d,x + \frac{1}{6} = \frac{{ - 1}}{2}\\
= > x = \frac{{ - 1}}{2} - \frac{1}{6}\\
= > x = \frac{{ - 2}}{3}\\
Vay:x = \frac{{ - 2}}{3}
\end{array}
\end{array}\]

\[\begin{array}{l}
e)Xet:A = \frac{{x + 3}}{{x - 2}}\left( {x \ne 2} \right)\\
= \frac{{x - 2 + 5}}{{x - 2}}\\
= 1 + \frac{5}{{x - 2}}\\
de:\left( {x + 3} \right) \vdots \left( {x - 2} \right)\\
= > \frac{5}{{x - 2}} \in Z = > x - 2 \in U(5) = \{ - 5; - 1;1;5\} \\
= > x \in \{ - 3;1;3;7\} \left( {tm} \right)\\
Vay:x \in \{ - 3;1;3;7\} \\
f)Xet:A = \frac{{2{\rm{x}} + 1}}{{x - 5}}\left( {x \ne 5} \right)\\
= \frac{{2{\rm{x}} - 10 + 11}}{{x - 5}}\\
= \frac{{2\left( {x - 5} \right)}}{{x - 5}} + \frac{{11}}{{x - 5}}\\
= 2 + \frac{{11}}{{x - 5}}\\
de:\left( {2{\rm{x}} + 1} \right) \vdots \left( {x - 5} \right)\\
= > \frac{{11}}{{x - 5}} \in Z = > x - 5 \in U(11) = \{ - 11; - 1;1;11\} \\
= > x \in \{ - 6;4;6;16\} \left( {tm} \right)\\
Vay:x \in \{ - 6;4;6;16\}
\end{array}\]

Số thứ tự	Loại hàng	Số lượng (quyển)	Giá đơn vị (đồng)	Tổng số tiền (đồng)
1	Vở loại 1	35	2000	...
2	Vở loại 2	42	1500	...
3	Vở loại 3	38	1200	...
Cộng:				...

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