Question

tim x;y biet:

a,(x-3)^2- xIx-3l =0

b,\(\frac{x^3+y^3}{6}=\frac{x^3-2y^3}{4}\)     vaf x^6.y^6=64

Answer 1

\(\left(x-3\right)^2-x\left|x-3\right|=0^{\left(1\right)}\)

\(\left(1\right)\Rightarrow\left(x-3\right)^2=x\left|x-3\right|^{\left(2\right)}\)

\(\left(x-3\right)^2\ge0,\left(2\right)\Rightarrow x\ge0\left(ĐK\right)\)

\(\left(x-3\right)^2-x\left|x-3\right|=0\)

\(\left(x-3\right)^2=x\left(x-3\right)\)

\(x^2-6x+9=x^2-3x\)

\(-6x+9=-3x\)

\(-6x+3x=-9\)

\(x\left(-6+3\right)=-9\)

\(x.\left(-3\right)=-9\)

\(\Rightarrow x=3\left(tm\right)\)