Question

Tìm x, y biết:a, \(\left[\dfrac{1}{2}x^2\left(2x-1\right)^m-\dfrac{1}{2}x^{m+2}\right]:\dfrac{1}{2}x^2=0\) (m thuộc N)
b, \(\left(2x-3\right)^6=\left(2x-3\right)^8\)
c, \(4x^2-4x+y^2-\dfrac{2}{3}y+\dfrac{10}{9}=0\)

Answer 1

a: \(\left[\dfrac{1}{2}x^2\left(2x-1\right)^m-\dfrac{1}{2}x^{m+2}\right]:\dfrac{1}{2}x^2=0\)

\(\Leftrightarrow\left(2x-1\right)^m-x^m=0\)

\(\Leftrightarrow\left(2x-1\right)^m=x^m\)

=>2x-1=x

=>x=1

b: \(\left(2x-3\right)^8=\left(2x-3\right)^6\)

\(\Leftrightarrow\left(2x-3\right)^6\cdot\left(2x-4\right)\left(2x-2\right)=0\)

hay \(x\in\left\{\dfrac{3}{2};2;1\right\}\)

c: \(\Leftrightarrow4x^2-4x+1+y^2-\dfrac{2}{3}y+\dfrac{1}{9}+\dfrac{6}{9}=0\)

\(\Leftrightarrow\left(2x-1\right)^2+\left(y-\dfrac{1}{3}\right)^2+\dfrac{6}{9}=0\)(vô lý)