3(x-2)+3=x+1
x+1/2=3/y+2
xy-3y+x=3
1/x+2/y=1
/x-2/+/y-1/^100+/2z+4/^209=0
(x-2)^2020+3/xy+4/^20=0
(Mik cần gấp mn ,giúp mik vs
3(x-2)+3=x+1
x+1/2=3/y+2
xy-3y+x=3
1/x+2/y=1
/x-2/+/y-1/^100+/2z+4/^209=0
(x-2)^2020+3/xy+4/^20=0
(Mik cần gấp mn ,giúp mik vs
Đáp án:
$\begin{array}{l}
a)3\left( {x – 2} \right) + 3 = x + 1\\
\Rightarrow 3x – 6 + 3 = x + 1\\
\Rightarrow 3x – x = 1 + 6 – 3\\
\Rightarrow 2x = 4\\
\Rightarrow x = 2\\
Vậy\,x = 2\\
b)\frac{{x + 1}}{2} = \frac{3}{{y + 2}}\\
\Rightarrow \left( {x + 1} \right).\left( {y + 2} \right) = 3.2 = 1.6\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 = 3\\
y + 2 = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 = 2\\
y + 2 = 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 = 1\\
y + 2 = 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 = 6\\
y + 2 = 1
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 2;y = 0\\
x = 1;y = 1\\
x = 0;y = 4\\
x = 5;y = – 1
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {2;0} \right);\left( {1;1} \right);\left( {0;4} \right);\left( {5; – 1} \right)} \right\}\\
c)xy – 3y + x = 3\\
\Rightarrow y.\left( {x – 3} \right) + x – 3 = 0\\
\Rightarrow \left( {x – 3} \right).\left( {y + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 3\\
y = – 1
\end{array} \right.\\
Vậy\,x = 3;x = – 1\\
d)\left| {x – 2} \right| + {\left| {y – 1} \right|^{100}} + {\left| {2z + 4} \right|^{209}} = 0\\
Do:\left| a \right| \ge 0\forall a\\
\Rightarrow \left\{ \begin{array}{l}
\left| {x – 2} \right| = 0\\
\left| {y – 1} \right| = 0\\
\left| {2z + 4} \right| = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 2\\
y = 1\\
z = – 2
\end{array} \right.\\
Vậy\,x = 2;y = 1;z = – 2\\
e){\left( {x – 2} \right)^{2020}} + 3{\left| {xy + 4} \right|^{20}} = 0\\
\Rightarrow \left\{ \begin{array}{l}
x – 2 = 0\\
xy + 4 = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 2\\
xy = – 4
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 2\\
y = – 2
\end{array} \right.\\
Vậy\,x = 2;y = – 2
\end{array}$