Tìm x, biết :
a, 4x.(x - 2017 ) - x + 2017 = 0
b, ( x + 1 ) 2 = x + 1
c, x(x - 5) - ( 4x + 20 ) = 0
d, x4 - 2x3 + 10x2 -20x = 0
e, x4 - 3x2 -x + 3 = 0
f, (2x -3 ) 2 = ( x + 5 )2
g, 2x2 - x - 12 = 0
h, 2x2 - x -12 = 0
i, x3 = 16 x
j, ( 2x -1 )2 - ( 4x2 - 1 ) = 0
k, ( x + 2 ) ( x + 3 ) ( x + 4 ) ( x + 5 ) = 24 .
Giúp mình với mai phải nộp rồi 
a, 4x.(x - 2017 ) - x + 2017 = 0
\(\Leftrightarrow\) 4x ( x - 2017 ) - ( x - 2017 ) = 0
\(\Leftrightarrow\) ( x - 2017 ) ( 4x - 1 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-2017=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy phương trình có nghiệm x = 2017 hoặc x = \(\dfrac{1}{4}\) .
b) \(\left(x+1\right)^2=x+1\)
\(\left(x+1\right)^2-\left(x+1\right)=0\)
\(\left(x+1\right)\left(x+1-x-1\right)=0\)
\(x+1=0\)
x = -1
c) \(x\left(x-5\right)-\left(4x-20\right)=0\)
\(x\left(x-5\right)-4\left(x-5\right)=0\)
\(\left(x-5\right)\left(x-4\right)=0\)
\(\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)
d, x4 - 2x3 + 10x2 - 20x = 0
\(\Leftrightarrow\) ( x4 - 2x3 ) + ( 10x2 - 20x ) = 0
\(\Leftrightarrow\) x3 ( x - 2 ) + 10x ( x - 2 ) = 0
\(\Leftrightarrow\) x ( x - 2 ) ( x2 + 10 ) = 0
Vì x2 + 10 luôn luôn lớn hơn không nên \(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy x = 0 hoặc x = 2 .
f , ( 2x - 3 )2 = ( x + 5 )2
\(\Leftrightarrow\) ( 2x - 3 )2 - ( x + 5 )2 = 0
\(\Leftrightarrow\) ( 2x - 3+ x + 5 ) ( 2x - 3 - x - 5 ) = 0
\(\Leftrightarrow\) ( 3x + 2 ) ( x - 8 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2}{3}\\x=8\end{matrix}\right.\)
Vậy phương trình có nghiệm x = 8 hoặc x = \(-\dfrac{2}{3}\)
j ) Tương tự ý f .
k ) Tương tự : " https://olm.vn/hoi-dap/question/98533.html "
