`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`

\(a,\frac{1}{3}+\frac{1}{2}:x=\frac{1}{5}\)

\(\Leftrightarrow\frac{1}{2}:x=\frac{1}{5}-\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{2}:x=\frac{3}{15}-\frac{5}{15}\)

\(\Leftrightarrow\frac{1}{2}:x=\frac{-2}{15}\)

\(\Leftrightarrow x=\frac{1}{2}:\frac{-15}{2}=\frac{-15}{4}\)

\(b,\frac{1}{3}x+\frac{2}{5}\left[x+1\right]=0\)

\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)

\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\Leftrightarrow x=\frac{-2}{5}:\frac{11}{15}=\frac{-2}{5}\cdot\frac{15}{11}=\frac{-2}{1}\cdot\frac{3}{11}=\frac{-6}{11}\)

\(c,\frac{11}{12}-\left[\frac{2}{5}+x\right]=\frac{2}{3}\)

\(\Leftrightarrow\frac{11}{12}-\frac{2}{3}=\frac{2}{5}+x\)

\(\Leftrightarrow\frac{11}{12}-\frac{8}{12}=\frac{2}{5}+x\)

\(\Leftrightarrow\frac{1}{4}=\frac{2}{5}+x\)

\(\Leftrightarrow\frac{1}{4}-\frac{2}{5}=x\Leftrightarrow x=-\frac{3}{20}\)

\(\left(x-\dfrac{3}{2}\right)\times\left(2x+1\right)>0\)

Th1:

\(x-\dfrac{3}{2}>0\Leftrightarrow x>\dfrac{3}{2}\)

\(2x+1>0\Leftrightarrow2x>1\Leftrightarrow x>\dfrac{1}{2}\)

( 1 )

Th2: 

\(x-\dfrac{3}{2}< 0\Leftrightarrow x< \dfrac{3}{2}\)

\(2x+1< 0\Leftrightarrow2x< -1\Leftrightarrow x< -\dfrac{1}{2}\)

( 2 )

Từ ( 1 ) và ( 2 ), ta có:

\(\Rightarrow x< -\dfrac{1}{2};x>\dfrac{3}{2}\)

\(\left(2-x\right)\times\left(\dfrac{4}{5}-x\right)< 0\)

Th1:

\(2-x>0\Leftrightarrow x>2\)

\(\dfrac{4}{5}-x< 0\Leftrightarrow x< \dfrac{4}{5}\)

( Loại )

Th2:

\(2-x< 0\Leftrightarrow x< 2\)

\(\dfrac{4}{5}-x>0\Leftrightarrow x>\dfrac{4}{5}\)

=> \(\dfrac{4}{5}< x< 2\)

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

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

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

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

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

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

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!

a/ Ta có: (x² + 1)(2x - 5) = 0

    => 2x³ - 5x² + 2x - 5 = 0 

    => (2x - 5)(x² + 1)  = 0    (chỗ này dùng sơ đồ Hoc-ne)

    => 2x - 5 = 0 => x = 5/2

   hoặc x² + 1 = 0 => x² = -1 (vô nghiệm)

          Vậy x = 5/2

b/ \(\left(x+\frac{1}{2}\right)^2=0\Rightarrow x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)

          Vậy x = -1/2

a) \(\left(x^2+1\right)\left(2x-5\right)=0\)

=>\(2x-5=0\)=>\(x=\frac{5}{2}\)

Vậy \(x=\frac{5}{2}\)

b) \(\left(x+\frac{1}{2}\right)^2=0\)

=>\(x+\frac{1}{2}=0\)=>\(x=-\frac{1}{2}\)

Chúc bạn học tốt!!!

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

Làm rõ từng ý nhé các bạn !

Cảm ơn các bạn rất nhiều!