Ta có : \(\left(x-3\right)^2\ge0\forall x\)
\(\left|y^2-25\right|\ge0\forall x\)
Mà : \(\left(x-3\right)^2+\left|y^2-25\right|=0\)
Nên : \(\hept{\begin{cases}\left(x-3\right)^2=0\\\left|y^2-25\right|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-3=0\\y^2-25=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=3\\y^2=25\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=3\\y=-5;5\end{cases}}\)
Ta có : 3300 = (33)100 = 27100
5200 = (52)100 = 25100
Vì : 27100 > 25100
Nên : 3300 > 5200
1 ) a Ta có : \(2^{225}=\left(2^3\right)^{75}=8^{75};3^{150}=\left(3^2\right)^{75}=9^{75}\)
Do \(8^{75}< 9^{75}\) nên \(2^{225}< 3^{150}\)
b ) \(2^{12}=\left(2^4\right)^3=16^3;4^{18}=\left(4^2\right)^9=16^9\)
2 a) \(3^{300}=\left(3^3\right)^{100}=27^{100};5^{200}=\left(5^2\right)^{100}=25^{100}\)
Vì \(27^{100}>25^{100}\) nên \(3^{300}>5^{200}\)
b ) \(\frac{37-x}{x+13}=\frac{3}{7}\Leftrightarrow259-7x=3x+39\Leftrightarrow-7x-3x=39-259\Leftrightarrow-10x=-220\Rightarrow x=22\)
c ) Vì \(\hept{\begin{cases}\left(x-3\right)^2\ge0\\\left|y^2-25\right|\ge0\end{cases}}\) nên \(\left(x-3\right)^2+\left|y^2-25\right|\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-3=0\\y^2-25=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=+-5\end{cases}}}\)
Vậy ...........
Sorry mk làm rở thì có đứa nghịch mình lam f tiếp nhé :
\(\frac{37-x}{x+13}=\frac{3}{7}\)
\(\Rightarrow7\left(37-x\right)=3\left(x+13\right)\)
\(\Rightarrow259-7x=3x+39\)
\(\Rightarrow-7x-3x=39-259\)
\(\Rightarrow-10x=-220\)
=> x = 22