Toán Lớp 7: a)x^4+(y-4/7)^20=0
b)(1/3x-4)^10+(y^2-1/9)^20=0
c)(x-5)^x+1-(x-5)^x+21=0
giai giup minh vs dang voi
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Toán Lớp 7: a)x^4+(y-4/7)^20=0
b)(1/3x-4)^10+(y^2-1/9)^20=0
c)(x-5)^x+1-(x-5)^x+21=0
giai giup minh vs dang voi
Comments ( 1 )
a){x^4} + {\left( {y – \dfrac{4}{7}} \right)^{20}} = 0\\
Do:\left\{ \begin{array}{l}
{x^4} = {\left( {{x^2}} \right)^2} \ge 0\\
{\left( {y – \dfrac{4}{7}} \right)^{20}} = {\left\{ {{{\left( {y – \dfrac{4}{7}} \right)}^{10}}} \right\}^2} \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^4} = 0\\
{\left( {y – \dfrac{4}{7}} \right)^{20}} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 0\\
y = \dfrac{4}{7}
\end{array} \right.\\
Vậy\,x = 0;y = \dfrac{4}{7}\\
b){\left( {\dfrac{1}{3}x – 4} \right)^{10}} + {\left( {{y^2} – \dfrac{1}{9}} \right)^{20}} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{3}x – 4 = 0\\
{y^2} – \dfrac{1}{9} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{3}x = 4\\
{y^2} = \dfrac{1}{9}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 12\\
y = \dfrac{1}{3};y = – \dfrac{1}{3}
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left[ {\left( {12;\dfrac{1}{3}} \right);\left( {12; – \dfrac{1}{3}} \right)} \right]\\
c){\left( {x – 5} \right)^{x + 1}} – {\left( {x – 5} \right)^{x + 21}} = 0\\
\Leftrightarrow {\left( {x – 5} \right)^{x + 1}}.\left( {1 – {{\left( {x – 5} \right)}^{20}}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\left( {x – 5} \right)^{x + 1}} = 0\\
{\left( {x – 5} \right)^{20}} = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x – 5 = 0\\
x – 5 = 1\\
x – 5 = – 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = 6\\
x = 4
\end{array} \right.\\
Vậy\,x = 4;x = 5;x = 6
\end{array}$