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\(\frac{x}{2016}+\frac{x-1}{2015}+\frac{x-2}{2014}+\frac{x-3}{2013}=4\)
\(\Leftrightarrow\left(\frac{x}{2016}-1\right)+\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)+\left(\frac{x-3}{2013}-1\right)=0\)
\(\Leftrightarrow\frac{x-2016}{2016}+\frac{x-2016}{2015}+\frac{x-2016}{2014}+\frac{x-2016}{2013}=0\)
\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}\right)=0\)
Dễ thấy cái vế sau > 0 nên x=2016
Câu b có cách nào hay hơn bằng cách phá ko ta,hóng quá:)
\(125x^3=\left(2x+1\right)^3+\left(3x-1\right)^3\)
\(\Leftrightarrow8x^3+12x^2+6x+1+27x^3-27x^2+9x-1=125x^3\)
\(\Leftrightarrow35x^3-15x^2+15x=125x^3\)
\(\Leftrightarrow90x^3+15x^2-15x=0\)
\(\Leftrightarrow x\left(90x^2+15x-15\right)=0\)
\(\Leftrightarrow x\left(3x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow x=0;x=-\frac{1}{2};x=\frac{1}{3}\)
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
a, \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right).\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)
\(\Leftrightarrow\frac{x^2-4x+4}{3}+\frac{9-4x^2}{8}+\frac{x^2-8x+16}{6}=0\)
\(\Leftrightarrow\frac{8\left(x^2-4x+4\right)+3\left(9-4x^2\right)+4\left(x^2-8x+16\right)}{24}=0\)
\(\Leftrightarrow\frac{8x^2-32x+32+27-12x^2+4x^2-32x+64}{24}=0\)
\(\Leftrightarrow\frac{123-64x}{24}=0\Leftrightarrow123-64x=0\Leftrightarrow x=\frac{123}{64}\)
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 ...