(x^2+8x+7)(x^2+8x+15)+15
hãy phân tích đa thức trên thành nhân tử:
PHÂN TÍCH ĐA THỨC THÀNH NHÂN TỬ:
\(\left(x^2+8x+8\right)\left(x^2+8x+15\right)+15\)
=(x^2+8x)^2+23(x^2+8x)+135
Cái này ko phân tích được nha bạn
\(\left(x^2+8x+8\right)\left(x^2+8x+15\right)+15\\ \Leftrightarrow\left(x^4+8x^3+15x^2+8x^3+64x^2+120x+8x^2+64x+120\right)+15\\ \Leftrightarrow x^4+16x^3+87x^2+184x+135\)
Gọi `A=(x^2+8x+8)(x^2+8x+15)+15`
Đặt `t=x^2+8x+11,5`
`=>A=(t-3,5)(t+3,5)+15=t^2-3,5^2+15=t^2-2,75=(t-sqrt(2,75))(t+sqrt(2,75))=(x^2+8x+11,5-(sqrt11)/2)(x^2+8x+11,5+(sqrt11)/2)=(x^2+8x+(23-\sqrt11)/2)(x^2+8x+(23+\sqrt11)/2)`
Phân tích đa thức thành nhân tử
x^2-y^2+8x+6y+7
\(x^2-y^2+8x+6y+7\)
\(=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)+x-y+7\)
\(=\left(x+y\right)\left(x-y+7\right)+\left(x-y+7\right)\)
\(=\left(x+y+1\right)\left(x-y+7\right)\)
Ta có x2 - y2 + 8x + 6y + 7
= x2 + 8x + 16 - y2 + 6y - 9
= \(x^2+4x+4x+16-y^2+3y+3y-9\)
= x(x + 4) + 4(x + 4) - y(y - 3) + 3(y - 3)
= (x + 4)2 - (y - 3)2
= (x + 4 + y - 3)(x + 4 - y + 3)
= (x + y + 1)(x - y + 7)
phân tích đa thức thành nhân tử
\(x^2-8x +15\)
\(x^2-8x+15\)
\(=x^2-3x-5x+15\)
\(=x\left(x-3\right)-5 \left(x-3\right)\)
\(=\left(x-5\right)\left(x-3\right)\)
\(x^2-8x+15\)
\(=\left(x^2-3x\right)-\left(5x-15\right)\)
\(=x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(x-5\right)\)
Tham khảo nhé~
Phân tích đa thức thành nhân tử:
x^2-8x+7.
x^2-8x+7= x^2-x-7x-7=x(x-1)-7(x-1)=(x-7)(x-1)
phân tích đa thức thành nhân tử
1)(x2+8x+7)(x+3)(x+5)+15
2)x2-2xy+3x-3y-10+y2
1) \(\left(x^2+8x+7\right).\left(x+3\right).\left(x+5\right)+15\)
\(=\left(x^2+8x+7\right).\left(x^2+5x+3x+15\right)+15\)
\(=\left(x^2+8x+7\right).\left(x^2+8x+15\right)+15\)
Ta đặt: \(x^2+8x+7=n\)
\(=n.\left(n+8\right)+15\)
\(=n^2+8n+15\)
\(=n^2+3n+5n+15\)
\(=\left(n^2+3n\right)+\left(5n+15\right)\)
\(=n.\left(n+3\right)+5.\left(n+3\right)\)
\(=\left(n+3\right).\left(n+5\right)\)
\(=\left(x^2+8x+7+3\right).\left(x^2+8x+7+5\right)\)
\(=\left(x^2+8x+10\right).\left(x^2+8x+12\right)\)
\(=\left(x^2+8x+10\right).\left(x^2+2x+6x+12\right)\)
\(=\left(x^2+8x+10\right).[x.\left(x+2\right)+6.\left(x+2\right)]\)
\(=\left(x^2+8x+10\right).\left(x+2\right).\left(x+6\right)\)
2) \(x^2-2xy+3x-3y-10+y^2\)
\(=\left(x-y\right)^2+3.\left(x-y\right)-10\)
Ta đặt: \(x-y=n\)
\(=n^2+3n-10\)
\(=n^2-2n+5n-10\)
\(=\left(n^2-2n\right)+\left(5n-10\right)\)
\(=n.\left(n-2\right)+5.\left(n-2\right)\)
\(=\left(n-2\right).\left(n+5\right)\)
\(=\left(x-y-2\right).\left(x-y+5\right)\)
phân tích các đa thức sau thành nhân tử x^4 -4x^3-8x^2+8x
x4 - 4x3 - 8x2 + 8x
= x(x3 - 4x2 - 8x + 8)
= x[x3 + 8 - 4x(x + 2)]
= x[(x + 2)(x2 - 2x + 4) - 4x(x + 2)]
= x(x + 2)(x2 - 6x + 4)
= x(x + 2)(x2 - 6x + 9 - 5)
= \(x\left(x+2\right)\left[\left(x-3\right)^2-5\right]=x\left(x+2\right)\left(x-3+\sqrt{5}\right)\left(x-3-\sqrt{5}\right)\)
\(x^4-4x^3-8x^2+8x\)
\(=x\left(x^3-4x^2-8x+8\right)\)
\(=x\left(x^3-6x^2+2x^2+4x-12x+8\right)\)
\(=x\left[\left(x^3-6x^2+4x\right)+\left(2x^2-12x+8\right)\right]\)
\(=x\left[x\left(x^2-6x+4\right)+2\left(x^2-6x+4\right)\right]\)
\(=x\left(x^2-6x+4\right)\left(x+2\right)\)
\(=x\left[\left(x-3\right)^2-\left(\sqrt{5}\right)^2\right]\left(x+2\right)\)
\(=x\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\left(x+2\right)\)
$x^2+8x+15$
Phân tích đa thức thảnh nhân tử
\(=\left(x+4\right)^2-1=\left(x+4-1\right)\left(x+4+1\right)=\left(x+3\right)\left(x+5\right)\)
\(x^2+8x+15=x\left(x+3\right)+5\left(x+3\right)=\left(x+3\right)\left(x+5\right)\)
\(=x^2+3x+5x+15=\left(x+3\right)\left(x+5\right)\)
Phân tích đa thức thành nhân tử
\(x^2-8x+12\)
\(x^2-8x+12=\left(x^2-6x\right)-\left(2x-12\right)=x\left(x-6\right)-2\left(x-6\right)=\left(x-2\right)\left(x-6\right)\)
Phân tích đa thức thành nhân tử
\(x^3-x^2-4x^2+8x-4\)
\(=x^2\left(x-1\right)-4\left(x-1\right)^2=\left(x-1\right)\left[x^2-4\left(x-1\right)\right]\\ =\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2\)
\(x^3-x^2-4x^2+8x-4\)
\(=x^3-4x^2-4x-x^2+4x-4\)
\(=x\left(x^2-4x+4\right)-\left(x^2-4x+4\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)\)
\(=\left(x-1\right)\left(x-2\right)^2\)