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\(\left(x-2\right)^5-\left(x-2\right)^3=0\)
\(\Rightarrow\left(x-2\right)^3\left(\left(x-2\right)^2-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\\left(x-2\right)^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\\left(x-2\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x-2=1\\x-2=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{1;2;3\right\}\)
⇒ ( x - 2)3 . (x - 2)2 - (x - 2)3 . 1 = 0 ⇒ ( x - 2)3 . [( x - 2)2 - 1] = 0
1. áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x+2}{3}=\frac{y-7}{5}=\frac{x+y-5}{3+5}=\frac{16}{8}=2\Rightarrow\hept{\begin{cases}x+2=6\\y-7=10\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\\y=17\end{cases}}}\)
2. áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x+5}{2}=\frac{y-2}{3}=\frac{x+5-y+2}{2-3}=\frac{-10+7}{-1}=3\Rightarrow\hept{\begin{cases}x+5=6\\y-2=9\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=11\end{cases}}\)
\(\dfrac{4}{5}\left(\dfrac{2}{3}x-\dfrac{9}{5}\right)-\dfrac{2}{5}\cdot\dfrac{-3}{7}=\dfrac{15}{4}\)
\(\Rightarrow\dfrac{2}{5}\left(\dfrac{2}{3}x-\dfrac{9}{5}\right)-\dfrac{-6}{35}=\dfrac{15}{4}\)
\(\Rightarrow\dfrac{2}{5}\left(\dfrac{2}{3}x-\dfrac{9}{5}\right)+\dfrac{6}{35}=\dfrac{15}{4}\)
\(\Rightarrow\dfrac{2}{5}\left(\dfrac{2}{3}x-\dfrac{9}{5}\right)=\dfrac{15}{4}-\dfrac{6}{35}=\dfrac{501}{140}\)
\(\Rightarrow\dfrac{2}{3}x-\dfrac{9}{5}=\dfrac{501}{140}:\dfrac{2}{5}=\dfrac{501\cdot5}{28\cdot5\cdot2}=\dfrac{501}{56}\)
\(\Rightarrow\dfrac{2}{3}x=\dfrac{501}{56}+\dfrac{9}{5}=\dfrac{2001}{280}\)
\(\Rightarrow x=\dfrac{2001}{280}:\dfrac{2}{3}=\dfrac{2001\cdot3}{280\cdot2}=\dfrac{6003}{560}\)
`4x(x-5)-(x-1) (4x-3)-5=0`
`=> 4x*x - 4x*5 - ( x*4x-3*x-1*4x+ 1*3) -5=0`
`=> 4x^2 - 20x-(4x^2 -3x-4x+3)-5=0`
`=> 4x^2 - 20x-4x^2+3x+4x-3-5=0`
`=>-13x-8=0`
`=> -13x=8`
`=> x=-8/13`
Vậy `x=-8/13`
`4x(x-5)-(x-1)(4x-3)-5 = 0`
`=> 4x^2 - 20x - (4x^2 -3x-4x+3)= 5`
`=> 4x^2 - 20x - 4x^2 + 3x + 4x -3 = 5`
`=> (4x^2 - 4x^2) - (20x - 3x - 4x) = 8`
`=> -13x = 8`
`=> x = -8/13`
a) 4x + 5(x - 3) = 3
<=> 4x + 5x - 15 = 3
<=> 9x = 3 + 15
<=> 9x = 18
<=> x = 2
b) -3(x - 5) + 6(x + 2) = 9
<=> -3x + 15 + 6x + 12 = 9
<=> 3x + 27 = 9
<=> 3x = -18
<=> x = -6
1) \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3-2\right)\left(x-3+2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
2) \(x^2-2x=24\)
\(\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow x^2+4x-6x-24=0\)
\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
\(\dfrac{3}{2}\)(\(x\) - \(\dfrac{5}{3}\)) - \(\dfrac{4}{5}\) = \(x\) + 1
\(\dfrac{3}{2}\) \(x\) - \(\dfrac{15}{6}\) - \(\dfrac{4}{5}\) = \(x\) + 1
\(\dfrac{3}{2}\)\(x\) - \(x\) = 1 + \(\dfrac{15}{6}\) + \(\dfrac{4}{5}\)
\(\dfrac{1}{2}\)\(x\) =\(\dfrac{43}{10}\)
\(x\) = \(\dfrac{43}{10}\) \(\times\) 2
\(x\) = \(\dfrac{43}{5}\)
\(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{4}{5}=x+1\\ \Rightarrow\dfrac{3.\left(x-\dfrac{5}{3}\right)}{2}-\dfrac{4}{5}=x+1\\ \Rightarrow\dfrac{3x-5}{2}-\dfrac{4}{5}=x+1\Rightarrow\dfrac{5\left(3x-5\right)}{10}-\dfrac{8}{10}=x+1\\ \Rightarrow\dfrac{15x-33}{10}=x+1\\ \Rightarrow\dfrac{15x-33}{10}-x=x+1\\ \Rightarrow\dfrac{15x-33}{10}=x+1-x\\ \Rightarrow5x-33=10\\ \Rightarrow5x=10+33\\\Rightarrow5x=43\\ \Rightarrow x=\dfrac{43}{5} \)
\(\left(x+\frac{2}{5}\right)^5=\left(x+\frac{2}{5}\right)^3\)
\(\Leftrightarrow\left(x+\frac{2}{5}\right)^3\left[\left(x+\frac{2}{5}\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{2}{5}=0\\x+\frac{2}{5}=\pm1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{7}{5};x=-\frac{3}{5}\end{cases}}\)