Ta cos\(\frac{16}{5}x=178-125x\)

=> \(\frac{16}{5}x+125x=178\)

=> \(x\left(\frac{16}{5}+125\right)=178\)

=> \(x.128,2=178\)

=> x = \(\frac{890}{641}\)

phân tích thành nhân tử ở mẫu và tử sau đó ta rút gọn vậy là ra đáp số

a) \(=\frac{5x\left(16x^2-25\right)}{\left(x-3\right)\left(4x-5\right)}\)\(\)

\(=\frac{5x\cdot\left(4x-5\right)\left(4x+5\right)}{\left(x-3\right)\left(4x-5\right)}\)

\(=\frac{5x\left(4x+5\right)}{x-3}\)

b) \(=\frac{3^2-\left(x+5\right)^2}{\left(x+2\right)^2}\)

\(=\frac{\left(3-x-5\right)\left(3+x+5\right)}{\left(x+2\right)^2}\)

\(=\frac{\left(x+2\right)\left(8+x\right)}{\left(x+2\right)^2}\)

\(=\frac{8+x}{x+2}\)

\(125^3+\left(\frac{1}{5}y\right)^3=\left(5x\right)^3+\left(\frac{1}{5}y\right)^3=\left(5x+\frac{1}{5}y\right)\left(25x^2-xy+\frac{1}{25}y^2\right)\)

\(\frac{x}{2016}+\frac{x-1}{2015}+\frac{x-2}{2014}+\frac{x-3}{2013}=4\)

\(\Leftrightarrow\left(\frac{x}{2016}-1\right)+\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)+\left(\frac{x-3}{2013}-1\right)=0\)

\(\Leftrightarrow\frac{x-2016}{2016}+\frac{x-2016}{2015}+\frac{x-2016}{2014}+\frac{x-2016}{2013}=0\)

\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}\right)=0\)

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:

\(\left(2x-5\right)^3+27\left(x-1\right)^3+\left(8-5x\right)^3=0\)

\(\Leftrightarrow3\left(2x-5\right)\left(3x-3\right)\left(8-5x\right)=0\)

\(\Leftrightarrow x=\frac{5}{2};x=1;x=\frac{8}{5}\)

a. \(\frac{x}{2016}+\frac{x-1}{2015}+\frac{x-2}{2014}+\frac{x-3}{2013}=4\)

\(\rightarrow\left(\frac{x}{2016}-1\right)+\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)+\left(\frac{x-3}{2013}-1\right)=0\)

\(\rightarrow\frac{x-2016}{2016}+\frac{x-2016}{2015}+\frac{x-2016}{2014}+\frac{x-2016}{2013}=0\)

\(\rightarrow\left(x-2016\right).\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{1014}+\frac{1}{2013}\right)=0\)

Vì \(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}\ne0\)

\(\rightarrow x-2016=0\)

\(\rightarrow x=2016\)

Vậy ...