Câu 1: Tính: (-2)^2-(-2)^2.2+(-2)^2.3-(-2)^2.4+…+(-2)^2.99-(-2)^2.100 Câu 2: Tìm x,y thuộc Z, biết: (x-2).y=-5 07/10/2021 Bởi Alaia Câu 1: Tính: (-2)^2-(-2)^2.2+(-2)^2.3-(-2)^2.4+…+(-2)^2.99-(-2)^2.100 Câu 2: Tìm x,y thuộc Z, biết: (x-2).y=-5
Đáp án: Giải thích các bước giải: 1)A=(−2)2−(−2)2.2+(−2)2.3−(−2)2.4+...+(−2)2.99−(−2)2.100=(−2)2.(1−2+3−4+...+99−100)=4.[(−1)+(−1)+...+(−1)]=4.(−1).50=−200C2)(x−2).y=−5=(−1).5=(−5).1⇒⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣{x−2=1y=−5{x−2=−5y=1{x−2=−1y=5{x−2=5y=−1⇒⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣{x=3y=−5{x=−3y=1{x=1y=5{x=7y=−1Vậy(x;y)={(3;−5);(−3;1);(1;5);(7;−1)} Bình luận
Đáp án: $\begin{array}{l}1)A = {\left( { – 2} \right)^2} – {\left( { – 2} \right)^2}.2 + {\left( { – 2} \right)^2}.3\\ – {\left( { – 2} \right)^2}.4 + … + {\left( { – 2} \right)^2}.99 – {\left( { – 2} \right)^2}.100\\ = {\left( { – 2} \right)^2}.\left( {1 – 2 + 3 – 4 + … + 99 – 100} \right)\\ = 4.\left[ {\left( { – 1} \right) + \left( { – 1} \right) + … + \left( { – 1} \right)} \right]\\ = 4.\left( { – 1} \right).50\\ = – 200\\C2)\\\left( {x – 2} \right).y = – 5 = \left( { – 1} \right).5 = \left( { – 5} \right).1\\ \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x – 2 = 1\\y = – 5\end{array} \right.\\\left\{ \begin{array}{l}x – 2 = – 5\\y = 1\end{array} \right.\\\left\{ \begin{array}{l}x – 2 = – 1\\y = 5\end{array} \right.\\\left\{ \begin{array}{l}x – 2 = 5\\y = – 1\end{array} \right.\end{array} \right. \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x = 3\\y = – 5\end{array} \right.\\\left\{ \begin{array}{l}x = – 3\\y = 1\end{array} \right.\\\left\{ \begin{array}{l}x = 1\\y = 5\end{array} \right.\\\left\{ \begin{array}{l}x = 7\\y = – 1\end{array} \right.\end{array} \right.\\Vậy\,\left( {x;y} \right) = \left\{ {\left( {3; – 5} \right);\left( { – 3;1} \right);\left( {1;5} \right);\left( {7; – 1} \right)} \right\}\end{array}$ Bình luận
Đáp án:
Giải thích các bước giải:
1)A=(−2)2−(−2)2.2+(−2)2.3−(−2)2.4+...+(−2)2.99−(−2)2.100=(−2)2.(1−2+3−4+...+99−100)=4.[(−1)+(−1)+...+(−1)]=4.(−1).50=−200C2)(x−2).y=−5=(−1).5=(−5).1⇒⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣{x−2=1y=−5{x−2=−5y=1{x−2=−1y=5{x−2=5y=−1⇒⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣{x=3y=−5{x=−3y=1{x=1y=5{x=7y=−1Vậy(x;y)={(3;−5);(−3;1);(1;5);(7;−1)}
Đáp án:
$\begin{array}{l}
1)A = {\left( { – 2} \right)^2} – {\left( { – 2} \right)^2}.2 + {\left( { – 2} \right)^2}.3\\
– {\left( { – 2} \right)^2}.4 + … + {\left( { – 2} \right)^2}.99 – {\left( { – 2} \right)^2}.100\\
= {\left( { – 2} \right)^2}.\left( {1 – 2 + 3 – 4 + … + 99 – 100} \right)\\
= 4.\left[ {\left( { – 1} \right) + \left( { – 1} \right) + … + \left( { – 1} \right)} \right]\\
= 4.\left( { – 1} \right).50\\
= – 200\\
C2)\\
\left( {x – 2} \right).y = – 5 = \left( { – 1} \right).5 = \left( { – 5} \right).1\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x – 2 = 1\\
y = – 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 2 = – 5\\
y = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 2 = – 1\\
y = 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 2 = 5\\
y = – 1
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 3\\
y = – 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x = – 3\\
y = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 1\\
y = 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 7\\
y = – 1
\end{array} \right.
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {3; – 5} \right);\left( { – 3;1} \right);\left( {1;5} \right);\left( {7; – 1} \right)} \right\}
\end{array}$