x(x+2)(x2+2x+2)+1=0
⇔(x2+2x)(x2+2x+2)=−1
Đặt x2+2x+1=t.
Ta có: (t−1)(t+1)=−1
⇔t2−1=−1
⇔t2=0⇔t=0
Do đó (x+1)2=0
⇔x+1=0⇔x=−1
Vậy x = -1.
x(x+2)(x2+2x+2)+1=0
⇔(x2+2x)(x2+2x+2)=−1
Đặt x2+2x+1=t.
Ta có: (t−1)(t+1)=−1
⇔t2−1=−1
⇔t2=0⇔t=0
Do đó (x+1)2=0
⇔x+1=0⇔x=−1
Vậy x = -1.
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 \(\left(\dfrac{x+1}{x-2}\right)^2-3\left(\dfrac{2x-4}{x-4}\right)^2+\dfrac{x+1}{x-4}=0\)
1)Giải pt: \(2\cdot\left(x+\dfrac{1}{x}\right)^2+\left(x^2+\dfrac{1}{x^2}\right)^2-\left(x^2+\dfrac{1}{x^2}\right)\cdot\left(x+\dfrac{1}{x}\right)^2=\left(x+2\right)^2\)
2)Giải pt: \(\dfrac{|3-2x|-|x|}{|2+3x|+x-2}=5\)
3)tìm tất cả các cặp số nguyên tố(x,y) là nghiệm của pt: x2 - 2y2 - 1=0
Giải PT sau:
1)\(\left(2x+7\right)^2=9\left(x+2\right)^2\)
2)\(\left(x^2-16\right)^2-\left(x-4\right)^2=0\)
3)\(\left(5x^2-2x+10\right)^2=\left(3x^2+10x-8\right)^2\)
giải pt sau
a)\(\left(x-2\right)\left(x-3\right)+2x=\left(x-2\right)^2-2\)
b) \(\left(x-1\right)^2+3x\left(x-1\right)+7=\left(2x-1\right)^2+5\left(x-3\right)\)
c)\(5\left(x^1-2x-1\right)+2\left(3x-2\right)=5\left(x+1\right)^2\)
d)\(\left(x-1\right)\left(x^2+x+1\right)-2x=x\left(x-1\right)\left(x+1\right)\)
giải pt:
\(\left(x^2-1\right)^2-x\left(x^2-1\right)-2x^2=0\)
p/s: Đặt biến phụ dạng đẳng cấp bậc hai
Giải PT
a)\(\left|x^2+4x-5\right|=x+2\)
b)\(\left|2x-4\right|-x^2+x-6=0\)
\(\left(x-1^{ }\right)^3+\left(x+3\right)^3=\left(2x+2^{ }\right)^3\)
giải pt
Giải PT: \(x\left(x+2\right)\left(x^3+3x^2+3x+1\right)+1=0\)