tìm nghiệm đa thức: (x+2) * (x+3)*(x-5)*(x-6) = 180
chieuf nay cần rùi giúp mình nha
thanks nhiều
a)\(6x^4+6x^3-7x^3-7x^2+x+1=6x^3\left(x+1\right)-7x^2\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(6x^3-7x^2+1\right)=\left(x+1\right)\left(6x^3-6x^2-x^2+x-x+1\right)\)
\(=\left(x+1\right)\left[6x^2\left(x-1\right)-x\left(x-1\right)-\left(x-1\right)\right]=\left(x+1\right)\left(x-1\right)\left(6x^2-x-1\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(6x^2-3x+2x-1\right)=\left(x+1\right)\left(x-1\right)\left[3x\left(2x-1\right)+\left(2x-1\right)\right]\)
\(=\left(x+1\right)\left(x-1\right)\left(3x+1\right)\left(2x-1\right)\)
Do bạn ko ghi phường trình = gì nên nếu phương trình = 0 thì
x-1=0=>x=1
x+1=0=>x=-1
3x+1=0=>x=-1/3
2x-1=0=>x=1/2
Chúc bạn học tốt 1 cái t i c k nha cảm ơn
ta có
\(\left(x+2\right)\left(x-5\right)=x^2-5x+2x-10=x^2-3x-10\)
\(\left(x+3\right)\left(x-6\right)=x^2-6x+3x-18=x^2-3x-18\)
\(\Rightarrow\left(x+2\right)\left(x+3\right)\left(x-5\right)\left(x-6\right)=180\)sẽ bằng \(\left(x^2-3x-10\right)\left(x^2-3x-18\right)=180\)
Đặt \(y=x^2-3x-10\Rightarrow y-8=x^2-3x-18\)
\(\Rightarrow y\left(y-8\right)-180=0\Rightarrow y^2-8y-180=0\Rightarrow y^2-18y+10y-180=0\)
\(\Rightarrow y\left(y-18\right)+10\left(y-18\right)=0\Rightarrow\left(y-18\right)\left(y+10=0\right)\)
\(\Rightarrow\left(x^2-3x-10-18\right)\left(x^2-3x-10+10\right)=0\Rightarrow\left(x^2-7x+4x-28\right)x\left(x-3\right)=0\)
\(\Rightarrow\left[x\left(x-7\right)+4\left(x-7\right)\right].x.\left(x-3\right)=0\Rightarrow x\left(x+4\right)\left(x-7\right)\left(x-3\right)=0\)
=> x=0
x+4=0=>x=-4
x-7=0=>x=7
x-3=0=>x=3
