Từ gt =>X2 = lXl2
=>X TÙY Ý
mình nghĩ zậy(ko chắc)
Đặt \(t=\left|x\right|\) , \(t\ge0\)
PT đã cho trở thành :
\(t^2\left(2t-3\right)=t^2.\left(2t-3\right)\)
Vì \(x^2=\left|x\right|^2\Rightarrow t^2=t^2\) nên từ đó suy ra pt trên vô số nghiệm.
Từ gt =>X2 = lXl2
=>X TÙY Ý
mình nghĩ zậy(ko chắc)
Đặt \(t=\left|x\right|\) , \(t\ge0\)
PT đã cho trở thành :
\(t^2\left(2t-3\right)=t^2.\left(2t-3\right)\)
Vì \(x^2=\left|x\right|^2\Rightarrow t^2=t^2\) nên từ đó suy ra pt trên vô số nghiệm.
tìm x biết :
\(\frac{1}{\left(x-1\right)x}+\frac{1}{\left(x-2\right)\left(x-1\right)}+\frac{1}{\left(x-3\right)\left(x-2\right)}+\frac{1}{\left(x-4\right)\left(x-3\right)}=\frac{x}{x^2-4x}\)
Tìm x : \(\left(x-1\right)^3-2\left(x-2\right)^2=\left(2+3x\right)^3-3\left(x+1\right)^2-\left(x-1\right)\left(x-2\right)\)
Tìm x :
a ) \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6.\)
b ) \(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
Tìm x
a) \(\left(2x-1\right).\left(2x+1\right)-4x^2=3\)
b) \(5x.\left(x-3\right)^2-5.\left(x-1\right)^3+15.\left(x+2\right).\left(x-2\right)=5\)
Tìm x: \(\left(2x-1\right)^3-3\left(3x+1\right)^2=\left(3+2x\right)^3-2\left(2-x\right)^2-\left(x-1\right)\left(x+2\right)\)
Tìm x :
a ) \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)
Rút gọn
a) \(x.\left(x+4\right).\left(x-4\right)-\left(x^2+1\right).\left(x-1\right)\)
b) \(\left(y-3\right).\left(y+3\right).\left(y^2+9\right)-\left(y^2+2\right).\left(y^2-2\right)\)
Tim x
a) \(\left(x+3\right)^3-x.\left(3x+1\right)^2+\left(2x+1\right).\left(4x^2-2x+1-3x^2\right)=54\)
b) \(\left(x-3\right)^3-\left(x-3\right).\left(x^2+3x+9\right)+6.\left(x+1\right)^2+3x^2=-33\)
Tính
a) \(\left(x+3\right).\left(x^2-3x+9\right)-x.\left(x-2\right)\left(x+2\right)\)
b) \(\left(x+y\right)^3-x.\left(x+3y\right)^2+y\left(y-3x\right)^2\)