\(\frac{x}{2016}+\frac{x-1}{2015}+\frac{x-2}{2014}+\frac{x-3}{2013}=4\)
\(\Leftrightarrow\left(\frac{x}{2016}-1\right)+\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)+\left(\frac{x-3}{2013}-1\right)=0\)
\(\Leftrightarrow\frac{x-2016}{2016}+\frac{x-2016}{2015}+\frac{x-2016}{2014}+\frac{x-2016}{2013}=0\)
\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}\right)=0\)
Dễ thấy cái vế sau > 0 nên x=2016
Câu b có cách nào hay hơn bằng cách phá ko ta,hóng quá:)
\(125x^3=\left(2x+1\right)^3+\left(3x-1\right)^3\)
\(\Leftrightarrow8x^3+12x^2+6x+1+27x^3-27x^2+9x-1=125x^3\)
\(\Leftrightarrow35x^3-15x^2+15x=125x^3\)
\(\Leftrightarrow90x^3+15x^2-15x=0\)
\(\Leftrightarrow x\left(90x^2+15x-15\right)=0\)
\(\Leftrightarrow x\left(3x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow x=0;x=-\frac{1}{2};x=\frac{1}{3}\)
Câu c có cách giải rất hay đó nha :)
\(\left(2x-5\right)^3+27\left(x-1\right)^3+\left(8-5x\right)^3=0\)
\(\Leftrightarrow\left(2x-5\right)^3+\left(3x-3\right)^3+\left(8-5x\right)^3=0\)
Đặt \(2x-5=a;3x-3=b;8-5x=c\Rightarrow a+b+c=0\)
\(\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^3=\left(-c\right)^3\)
\(\Leftrightarrow a^3+3ab\left(a+b\right)+b^3=-c^3\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)=-3ab\left(-c\right)=3abc\)
Khi đó:
\(\left(2x-5\right)^3+27\left(x-1\right)^3+\left(8-5x\right)^3=0\)
\(\Leftrightarrow3\left(2x-5\right)\left(3x-3\right)\left(8-5x\right)=0\)
\(\Leftrightarrow x=\frac{5}{2};x=1;x=\frac{8}{5}\)
