bt: tìm x,y biết:
a, 2/7-(1/2-x)=3/7+(1/2x-1/3)
b, (x-1/7)^2020+|y+1/6|=0
c, (1/7x-1/8)^2020+(3/5y-1/2)^2020 ≤ 0
bt: tìm x,y biết:
a, 2/7-(1/2-x)=3/7+(1/2x-1/3)
b, (x-1/7)^2020+|y+1/6|=0
c, (1/7x-1/8)^2020+(3/5y-1/2)^2020 ≤ 0
Đáp án:
$\begin{array}{l}
a)\dfrac{2}{7} – \left( {\dfrac{1}{2} – x} \right) = \dfrac{3}{7} + \left( {\dfrac{1}{2}x – \dfrac{1}{3}} \right)\\
\Rightarrow \dfrac{2}{7} – \dfrac{1}{2} + x = \dfrac{3}{7} + \dfrac{1}{2}x – \dfrac{1}{3}\\
\Rightarrow x – \dfrac{1}{2}x = \dfrac{3}{7} – \dfrac{1}{3} – \dfrac{2}{7} + \dfrac{1}{2}\\
\Rightarrow \dfrac{1}{2}x = \dfrac{1}{7} + \dfrac{1}{6}\\
\Rightarrow \dfrac{1}{2}x = \dfrac{{13}}{{42}}\\
\Rightarrow x = \dfrac{{13}}{{21}}\\
Vay\,x = \dfrac{{13}}{{21}}\\
b){\left( {x – \dfrac{1}{7}} \right)^{2020}} + \left| {y + \dfrac{1}{6}} \right| = 0\\
Do:\left\{ \begin{array}{l}
{\left( {x – \dfrac{1}{7}} \right)^{2020}} \ge 0\\
\left| {y + \dfrac{1}{6}} \right| \ge 0
\end{array} \right.\\
\Rightarrow {\left( {x – \dfrac{1}{7}} \right)^{2020}} = \left| {y + \dfrac{1}{6}} \right| = 0\\
\Rightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{7}\\
y = – \dfrac{1}{6}
\end{array} \right.\\
c){\left( {\dfrac{1}{7}x – \dfrac{1}{8}} \right)^{2020}} + {\left( {\dfrac{3}{5}y – \dfrac{1}{2}} \right)^{2020}} \le 0\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {\dfrac{1}{7}x – \dfrac{1}{8}} \right)^{2020}} = 0\\
{\left( {\dfrac{3}{5}y – \dfrac{1}{2}} \right)^{2020}} = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{1}{7}x = \dfrac{1}{8}\\
\dfrac{3}{5}y = \dfrac{1}{2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = \dfrac{7}{8}\\
y = \dfrac{5}{6}
\end{array} \right.\\
Vay\,x = \dfrac{7}{8};y = \dfrac{5}{6}
\end{array}$