Giải phương trình
bài 1: \(\left(5x+3\right)^3-\left(2x+4\right)^3=\left(3x-1\right)^3\)
bài 2: \(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}-\dfrac{x-3}{2011}=\dfrac{x-4}{2010}\)
bài 3: \(\left(2x-5\right)^3-\left(x-2\right)^3=\left(x-3\right)^3\)
bài 4: \(\dfrac{x+43}{57}+\dfrac{x+46}{54}=\dfrac{x+49}{51}+\dfrac{x+52}{48}\)
bài 5: \(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}=4\)
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Bài 1:
Đặt \(\left\{\begin{matrix} 5x+3=a\\ 2x+4=b\end{matrix}\right.\) \(\Rightarrow 3x-1=a-b\)
PT trở thành:
\(a^3-b^3=(a-b)^3\)
\(\Leftrightarrow (a-b)(a^2+ab+b^2)=(a-b)^3\)
\(\Leftrightarrow (a-b)[a^2+ab+b^2-(a^2-2ab+b^2)]=0\)
\(\Leftrightarrow 3ab(a-b)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\b=0\\a=b\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{5}\\x=-2\\5x+3=2x+4\Leftrightarrow x=\dfrac{1}{3}\end{matrix}\right.\)
Thử lại thấy đều thỏa mãn
Vậy \(x\in\left\{\frac{-3}{5};-2;\frac{1}{3}\right\}\)
Bài 2:
\(\frac{x-1}{2013}+\frac{x-2}{2012}-\frac{x-3}{2011}=\frac{x-4}{2010}\)
\(\Leftrightarrow \frac{x-1}{2013}-1+\frac{x-2}{2012}-1-\left(\frac{x-3}{2011}-1\right)=\frac{x-4}{2010}-1\)
\(\Leftrightarrow \frac{x-2014}{2013}+\frac{x-2014}{2012}-\frac{x-2014}{2011}=\frac{x-2014}{2010}\)
\(\Leftrightarrow (x-2014)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\) (1)
Thấy rằng \(2013> 2011; 2012> 2010\Rightarrow \frac{1}{2013}< \frac{1}{2011}; \frac{1}{2012}< \frac{1}{2010}\)
\(\Rightarrow \frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}< 0\) (2)
Từ (1),(2) suy ra \(x-2014=0\Leftrightarrow x=2014\)
Bài 3:
Đặt \(\left\{\begin{matrix} 2x-5=a\\ x-2=b\end{matrix}\right.\Rightarrow x-3=a-b\)
PT trở thành: \(a^3-b^3=(a-b)^3\)
\(\Leftrightarrow (a-b)(a^2+ab+b^2)-(a-b)(a^2-2ab+b^2)=0\)
\(\Leftrightarrow 3ab(a-b)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\b=0\\a-b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=2\\x-3=0\Leftrightarrow x=3\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{5}{2}; 2; 3\right\}\)
Bài 4:
\(\frac{x+43}{57}+\frac{x+46}{54}=\frac{x+49}{51}+\frac{x+52}{48}\)
\(\Leftrightarrow \frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)
\(\Leftrightarrow \frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)
\(\Leftrightarrow (x+100)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\) (1)
Dễ thấy: \(\frac{1}{57}< \frac{1}{51}; \frac{1}{54}< \frac{1}{48}\Rightarrow \frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}< 0\) (2)
Từ \((1);(2)\Rightarrow x+100=0\Leftrightarrow x=-100\)
Bài 5:
\(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)
\(\Leftrightarrow \frac{x-17}{33}-1+\frac{x-21}{29}-1+\frac{x}{25}-2=0\)
\(\Leftrightarrow \frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)
\(\Leftrightarrow (x-50)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)
Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\neq 0\) suy ra \(x-50=0\Leftrightarrow x=50\)
