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

\(\Rightarrow\left(x-2\right)^3\left(\left(x-2\right)^2-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\\left(x-2\right)^2-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\\left(x-2\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x-2=1\\x-2=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\\x=1\end{matrix}\right.\)

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 

Bạn Nguyễn Đức Trí ơi, cảm ơn bạn đã trả lời giúp mik nhưng mik thấy cô mik dạy là x chỉ thuộc { 2 ; 3 } thôi nhé bạn!

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)5^x +5^x+2=650

=> 5^x+5^x. 5² =650

=> 5^x(1+5²)=650

=> 5^x. 26=650

=> 5^x=25

=> x= 2

b) 3^x-1 +5 x 3 ^x-1 = 162

=> 3 ^x-1 (5 +1) = 162

=> 3 ^x-1 x 6 = 162

=> 3 ^x-1 = 27

=> x-1 =3

=> x =2

   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) \(x^2-2x=24\)

\(\Leftrightarrow x^2-2x-24=0\)

\(\Leftrightarrow x^2+4x-6x-24=0\)

\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

Câu 3 số xấu rồi e

alo =0 nhé anh pop pop

\(\dfrac{3}{2}\)(\(x\) - \(\dfrac{5}{3}\)) - \(\dfrac{4}{5}\) = \(x\) + 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

     \(x\)                = \(\dfrac{43}{5}\)

\(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{4}{5}=x+1\\ \Rightarrow\dfrac{3.\left(x-\dfrac{5}{3}\right)}{2}-\dfrac{4}{5}=x+1\\ \Rightarrow\dfrac{3x-5}{2}-\dfrac{4}{5}=x+1\Rightarrow\dfrac{5\left(3x-5\right)}{10}-\dfrac{8}{10}=x+1\\ \Rightarrow\dfrac{15x-33}{10}=x+1\\ \Rightarrow\dfrac{15x-33}{10}-x=x+1\\ \Rightarrow\dfrac{15x-33}{10}=x+1-x\\ \Rightarrow5x-33=10\\ \Rightarrow5x=10+33\\\Rightarrow5x=43\\ \Rightarrow x=\dfrac{43}{5} \)

\(\dfrac{3}{2}\times\left(x-\dfrac{5}{3}\right)-\dfrac{4}{5}=x+1\)

\(\Leftrightarrow\dfrac{3}{2}\times x-\dfrac{5}{2}-\dfrac{4}{5}=x+1\)

\(\Leftrightarrow\dfrac{3}{2}\times x-x=1+\dfrac{5}{2}+\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}\times x=\dfrac{43}{10}\)

\(\Leftrightarrow x=\dfrac{43}{10}\div\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{43}{5}\)