Tìm x :
a) ( x - 15 ) . 35 = 0
x - 15 = 0 : 35
x - 15 = 0
x = 0 + 15
x = 15
b) 32 ( x - 10 ) = 32
x - 10 = 32 : 32
x - 10 = 1
x = 1 + 10
x = 11
\(\frac{1}{x^3+y^3}+\frac{1}{xy}=\frac{1}{x^2-xy+y^2}+\frac{1}{x}+\frac{1}{y}=1+\frac{3xy}{x^3+y^3}+1+\frac{x}{y}+1+\frac{y}{x}\ge5\)
Ta có: \(x^3+y^3=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]\)
Áp dụng BĐT CAuchy-Schwarz ta có:
\(VT=\frac{1}{x^3+y^3}+\frac{1}{xy}=\frac{x+y}{\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]}+\frac{1}{xy}\)
\(=\frac{1}{1-3xy}+\frac{3}{3xy}=\frac{1}{1-3xy}+\frac{\sqrt{3}^2}{3xy}\)
\(\ge\frac{\left(1+\sqrt{3}\right)^2}{1-3xy+3xy}=\left(1+\sqrt{3}\right)^2=4+2\sqrt{3}\)
Ta có:
\(3xy=3xy\left(x+y\right)\)
\(\Rightarrow\frac{1}{x^3+y^3}+\frac{1}{xy}=\frac{1}{x^3+y^3}+\frac{3}{3xy\left(x+y\right)}\ge\frac{\left(1+\sqrt{3}\right)^2}{\left(x+y\right)^3}=4+2\sqrt{3}\)
