Đặt \(\left\{{}\begin{matrix}x-1=a\\x+3=b\end{matrix}\right.\) \(\Rightarrow a+b=2x+2\)
PT đã cho có dạng là :
\(a^3+b^3=\left(a+b\right)^3\)
\(\Leftrightarrow3ab\left(a+b\right)=0\)
\(\Leftrightarrow3\left(x-1\right)\left(x+3\right)\left(2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\\x=-1\end{matrix}\right.\) ( thỏa mãn )
giải pt:
a) x^5 + 2x^4 + 3x^3 + 3x^2 + 2x +1=0
b) x^4 + 3x^3 - 2x^2 + x - 3 = 0
Giải pt sau
\(\left(\dfrac{x-1}{x+2}\right)^2-\left(\dfrac{2x+4}{x-3}\right)^2+3\left(\dfrac{x-1}{x-3}\right)=0\)
Giải PT sau:
a, 3x - 7 = 0
b, 8 - 5x = 0
c, 3x - 2 = 5x + 8
d, \(\dfrac{3x-2}{3}\) = \(\dfrac{1-x}{2}\)
e, ( 5x + 1)(x - 3) = 0
f, (x + 1)(2x - 3) = 0
g, 4x(x + 3) - 5(x + 3) = 0
h, 8(x - 6) - 2x(6 - x) = 0
i, \(\dfrac{2}{x-1}\) + \(\dfrac{1}{x}\) = \(\dfrac{2x+5}{x^2-x}\)
k, \(\dfrac{3}{x+2}\) - \(\dfrac{2}{x-2}\) = \(\dfrac{2-x}{x^2-4}\)
m, \(\dfrac{3}{x}\) - \(\dfrac{2}{x-3}\) = \(\dfrac{4-x}{x^2-3}\)
n,\(\dfrac{3}{2x+10}\)+ \(\dfrac{2x}{x^2-25}\) = \(\dfrac{3}{x-5}\)
u, \(\dfrac{2}{x+3}\) - \(\dfrac{3}{x-2}\) = \(\dfrac{x+4}{\left(x+3\right)\left(x-2\right)}\)
\(\left(x-1^{ }\right)^3+\left(x+3\right)^3=\left(2x+2^{ }\right)^3\)
giải pt
giải pt: a. (x - 2)(x+1)(x+3) = (x+3)(x+1)(2x-5)
b. \(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)
giải pt:
(2x-3)2+(x+1)2=3x(x-2)
Giải pt: \(\frac{9x}{2x^2+x+3}-\frac{x}{2x^2-x+3}=8\)
1. C/m pt sau vô nghiệm
x^4 - 2x^3 + 3x^2 - 2x + 1 =0
2.giải pt
(x^2-4)^2=8x + 1
