Phân tích thành nhân tử:
a, \(\left(x^2-x+2\right)^2+\left(x-2\right)^2\)
b, \(6x^5+15x^4+20x^3+15x^2+6x+1\)
Giúp hộ!!!
phân tích đa thức thành nhân tử
a)\(\left(x^2-x+2\right)^2+\left(x-2\right)^2\)
b)\(6x^5+15x^4+20x^3+15x^2+6x+1\)
PLEASE !!! GIÚP MK VS MK CẦN RẤT GẤP LÀM ƠN!!!
a, = [(x-2).(x+1)]^2+(x-2)^2
= (x-2)^2.(x+1)^2+(x-2)^2
= (x-2)^2.[(x+1)^2+1]
= (x-2)^2.(x^2+2x+2)
Tk mk nha
b) \(6x^5+15x^4+20x^3+15x^2+6x+1\)
\(=6x^5+3x^4+12x^4+6x^3+14x^3+7x^2+8x^2+4x+2x+1\)
\(=\left(2x+1\right)\left(3x^4+6x^3+7x^2+4x+1\right)\)
\(=\left(2x+1\right)\left(3x^4+3x^3+3x^2+3x^3+3x^2+3x+x^2+x+1\right)\)
\(=\left(2x+1\right)\left(x^2+x+1\right)\left(3x^2+3x+1\right)\)
Phân tích đa thức thành nhân tử :
b)\(B=\left(x^2+4x-3\right)^2-5x\left(x^2-4x-3\right)+6x^2\)
c)\(C=\left(x^2+x+4\right)^2+8x\left(x^2+x+4\right)+15x^2\)
b mk thấy nó sai đề sao ý
c) \(C=\left(x^2+x+4\right)^2+8x\left(x^2+x+4\right)+15x^2\)
\(=\left(x^2+x+4\right)^2+2.4x.\left(x^2+x+4\right)+16x^2-x^2\)
\(=\left(x^2+x+4+4x\right)^2-x^2\)
\(=\left(x^2+5x+4\right)^2-x^2\)
\(=\left(x^2+5x+4-x\right)\left(x^2+5x+4+x\right)=\left(x^2+4x+4\right)\left(x^2+6x+4\right)\)
Phân tích thành nhân tử:
a, (x2 – x +2)2 + (x-2)2
b, 6x5 +15x4 + 20x3 +15x2 + 6x +1
1.PTĐT thành nhân tử
a) \(x^5+4x+5\)
b) \(x^4+6x^3+11x^2+6x+1\)
c) \(64x^4+1\)
c) \(81x^4+4\)
d) \(4\left(x^2+15x+50\right)\left(x^2+18x+72\right)-3x^2\)
e) \(x^5-x^4-1\)
2.PTĐT thành nhân tử (PP hệ số bất định)
a) \(3x^2-22xy-4x+8y+7y^2+1=\left(3x+ay+b\right)\left(x+cy+d\right)\)
b) \(12x^2+5x-12y^2+12y-10xy-3=\left(ã+by-1\right)\left(dx+cy+3\right)\)
a) \(x^5+4x+5=\left(x^5+x^4\right)-\left(x^4+x^3\right)+\left(x^3+x^2\right)-\left(x^2+x\right)+\left(5x+5\right)=x^4\left(x+1\right)-x^3\left(x+1\right)+x^2\left(x+1\right)-x\left(x+1\right)+5\left(x+1\right)=\left(x^4-x^3+x^2-x+5\right)\left(x+1\right)\)
b) \(x^4+6x^3+11x^2+6x+1=\left(x^4+3x^3+x^2\right)+\left(3x^3+9x^2+3x\right)+\left(x^2+3x+1\right)=x^2\left(x^2+3x+1\right)+3x\left(x^2+3x+1\right)+\left(x^2+3x+1\right)=\left(x^2+3x+1\right)^2\)
c) \(64x^4+1=\left[\left(8x^2\right)^2+16x^2+1\right]-16x^2=\left(8x^2+1\right)^2-\left(4x\right)^2=\left(8x^2-4x+1\right)\left(8x^2+4x+1\right)\)d) \(81x^4+4=\left[\left(9x^2\right)^2+36x^2+2^2\right]-36x^2=\left(9x^2+2\right)^2-\left(6x\right)^2=\left(9x^2-6x+2\right)\left(9x^2+6x+2\right)\)
Câu 1:
\(e,x^5-x^4-1=x^5-x^4+x^3-x^3+x^2-x^2+x-x-1\\ =\left(x^5-x^4-x^3\right)+\left(x^3-x^2-x\right)+\left(x^2-x-1\right)\\ =x^3\left(x^2-x-1\right)+x\left(x^2-x-1\right)+\left(x^2-x-1\right)\\ =\left(x^2-x-1\right)\left(x^3+x+1\right)\)
Câu 2:
\(a,\left(3x+ay+b\right)\left(x+cy+d\right)\\ =3x^2+3xcy+3xd+axy+acy^2+ayd+bx+bcy+bd\\ =3x^2+xy\left(3c+a\right)+x\left(b+3d\right)+y\left(ad+bc\right)+acy^2+bd\\ \Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}3c+a=-22\\b+3d=-4\end{matrix}\right.\\ad+bc=8\\\left\{{}\begin{matrix}ac=7\\bd=1\end{matrix}\right.\end{matrix}\right.\)
Xét \(bd=1\Leftrightarrow\left[{}\begin{matrix}b=1;d=1\\b=-1;d=-1\end{matrix}\right.\)
Với \(b=1;d=1\Leftrightarrow b+3d=1+3\cdot1=4\left(ktm\right)\)
Với \(b=-1;d=-1\Leftrightarrow b+3d=-1-3=-4\left(tm\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}3c+a=-22\\-a-c=8\\ac=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-1\\c=-7\end{matrix}\right.\)
Vậy \(3x^2-22xy-4x+8y+7y^2+1=\left(3x-y-1\right)\left(x-7y-1\right)\)
Cái chỗ ngoặc nhọn mà 5 dòng á a ko thấy trong cái phần công thức nên là ghi z chứ nó có 5 dòng đó nha
câu b tương tự, lười wa 😴
Phân tích thành nhân tử:
a, (x2 – x +2)2 + (x-2)2
b, 6x5 +15x4 + 20x3 +15x2 + 6x +1
phân tích đa thức thành nhân tử:
a)\(A=\left(x^2-2x\right)\left(x^2-2x-1\right)-6\)
b)\(B=\left(x^2+4x-3\right)^2-5x\left(x^2+4x-3\right)+6x^2\)
c)\(C=\left(x^2+x+1\right)+8x\left(x^2+x+4\right)+15x^2\)
a)\(A=\left(x^2-2x\right)\left(x^2-2x-1\right)-6=\left(x^2-2x\right)^2-\left(x^2-2x\right)-6\)
\(=\left(x^2-2x+2\right)\left(x^2-2x+3\right)\)
phân thích thành nhân tử
a) (x2 -x +2)2 +(x-2)2
b) 6x5 +15x4 +20x3 +15x2 +6x+1.hết =)
a)\(\left(x^2-x+2\right)^2+\left(x-2\right)^2\)
\(=x^4-2x^3+6x^2-8x+8\)
\(=x^4-2x^3+2x^2+4x^2-8x+8\)
\(=x^2\left(x^2-2x+2\right)+4\left(x^2-2x+2\right)\)
\(=\left(x^2+4\right)\left(x^2-2x+2\right)\)
b)\(6x^5+15x^4+20x^3+15x^2+6x+1\)
\(=6x^5+3x^4+12x^4+6x^3+14x^3+7x^2+8x^2+4x+2x+1\)
\(=3x^4\left(2x+1\right)+6x^3\left(2x+1\right)+7x^2\left(2x+1\right)+4x\left(2x+1\right)+\left(2x+1\right)\)
\(=\left(3x^4+6x^3+7x^2+4x+1\right)\left(2x+1\right)\)
\(=\left[3x^4+3x^3+x^2+3x^3+3x^2+x+3x^2+3x+1\right]\left(2x+1\right)\)
\(=\left[x^2\left(3x^2+3x+1\right)+x\left(3x^2+3x+1\right)+\left(3x^2+3x+1\right)\right]\left(2x+1\right)\)
\(=\left(x^2+x+1\right)\left(3x^2+3x+1\right)\left(2x+1\right)\)
B1: phân tích đa thức
a)\(x^3+4x^2-29x+24\)
b)\(x^4+6x^3+7x^2+6x+1\)
c)\(\left(x^2-x+2\right)^2+\left(x-2\right)^2\)
d)\(6x^5+15x^4+20x^3+15x^2+6x+1\)
e)\(x^6+3x^5+4x^4+4x^3+4x^2+3x+1\)
B2:phân tích đa thức
a)\(x^8+x^4+1\)
b)\(x^{10}+x^5+1\)
c)\(x^{12}+1\)
B3: phân tích đa thức
a)\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
b)\(\left(x+y+z\right)^5-x^5-y^5-z^5\)
AI LÀM ĐC PHẦN NÀO THÌ LÀM KO CẦN LÀM HẾT GIÚP MK MAI NỘP BÀI RỒI
\(x^8+x^4+1\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4+1\right)^2-x^4\)
\(=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)
\(=\left(x^4-x^2+1\right)\left(x^4+2x^2-x^2+1\right)\)
\(=\left(x^4-x^2+1\right)[\left(x^2+1\right)^2-x^2]\)
\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)
phân tích đa thức thành nhân tử
a) \(\left(x^2-x+2\right)^2+\left(x-2\right)^2\)
b) \(6x^5+15x^4+20x^3+15x^2+6x+1\)
PLEASE !!! GIÚP MK VS MK CẦN RẤT GẤP LÀM ƠN!!!
a) \(\left(x^2-x+2\right)^2+\left(x-2\right)^2\)
\(=x^4+x^2+4-2x^3-4x+4x^2+x^2-4x+4\)
\(=x^4-2x^3+6x^2-8x+8\)
\(=x^4-2x^3+2x^2+4x^2-8x+8\)
\(=x^2\left(x^2-2x+2\right)+4\left(x^2-2x+2\right)\)
\(=\left(x^2+4\right)\left(x^2-2x+2\right)\)
b) \(6x^5+15x^4+20x^3+15x^2+6x+1\)
\(=6x^5+3x^4+12x^4+6x^3+14x^3+7x^2+8x^2+4x+2x+1\)
\(=3x^4\left(2x+1\right)+6x^3\left(2x+1\right)+7x^2\left(2x+1\right)+4x\left(2x+1\right)+2x+1\)
\(=\left(2x+1\right)\left(4x^4+6x^3+7x^2+4x+1\right)\)
\(=\left(2x+1\right)\left(3x^4+3x^3+3x^2+3x^3+3x^2+3x+x^2+x+1\right)\)
\(=\left(2x+1\right)\left[\left(3x^2\right)\left(x^2+x+1\right)+3x\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]\)
\(=\left(2x+1\right)\left(x^2+x+1\right)\left(3x^2+3x+1\right)\)