4*x^4 +4*x^3 +5*x^2 +2*x+1
phân tích thành nhân tử
x^2+2x-4+1
phân tích thành nhân tử
\(\left(x+1\right)^2-2^2=\left(x-1\right)\left(x+3\right)\)
\(=\left(x+1\right)^2-4=\left(x+1-2\right)\left(x+1+2\right)=\left(x-1\right)\left(x+3\right)\)
Phân tích thành nhân tử:
`4(x-2)(x+1)+(2x-4)^2 +(x+1)^2`
`x^9 -x^7 -x^6 -x^5 +x^4 +x^3 +x^2 -1`
a: =4(x-2)(x+1)+4(x-2)^2+(x+1)^2
=(2x-4)^2+2*(2x-4)(x+1)+(x+1)^2
=(2x-4+x+1)^2=(3x-3)^2=9(x-1)^2
b: =x^7(x^2-1)-x^5(x+1)+x^3(x+1)+(x^2-1)
=(x+1)[x^7(x-1)-x^5+x^3+x-1]
=(x+1)[x^7(x-1)-x^3(x-1)(x+1)+(x-1)]
=(x+1)(x-1)(x^7-x^4-x^3+1)
=(x+1)(x-1)(x^3-1)(x^4-1)
=(x+1)(x-1)^2*(x^2+x+1)(x^2+1)(x-1)(x+1)
=(x+1)^2*(x-1)^3*(x^2+1)(x^2+x+1)
Phân tích thành nhân tử
1/ x^5+x^4-x^3+x^2-x+2
2/ x^4+ 5x^3+ 10x- 4
`x^5 + x^4 +x^3 +x^2 +x+1`
phân tích thành nhân tử
=x^3(x^2+x+1)+(x^2+x+1)
=(x^2+x+1)(x^3+1)
=(x^2+x+1)(x+1)(x^2-x+1)
\(x^5+x^4+x^3+x^2+x+1\)
\(=\left(x^5+x^4\right)+\left(x^3+x^2\right)+\left(x+1\right)\)
\(=x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^4+x^2+1\right)\)
#Toru
phân tích thành nhân tử:(x+2)*(x+3)*(x+4)*(x+5)-24
(x+2).(x+3)(x+4)(x+5)-24
=(x+2)(x+5)(x+3)(x+4)-24
=(x2+7x+10)(x2+7x+12)-24
=(x2+7x+11-1)(x2+7x+11+1)-24
Đặt x2+7x+11=a thì
=(a-1)(a+1)-24
=a2-1-24=a2-25=a2-52
=(a+5)(a-5)
=(x2+7x+16)(x2+7x+6)
phân tích đa thức thành nhân tử :
(x + 2) (x + 3) (x + 4) (x + 5) - 24
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(x^2+7x+11\right)^2-1-24\\ =\left(x^2+7x+11\right)^2-25\\ =\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\\ =\left(x^2+7x+6\right)\left(x^2+7x+16\right)\\ =\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)+1-25\)
\(=\left(x^2+7x+11\right)^2-25=\left(x^2+7x+11+5\right)\left(x^2+7x+11-5\right)\)
\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)
=[(x+2)(x+5)][(x+3)(x+4)]−24=(x2+7x+10)(x2+7x+12)−24=(x2+7x+11)2−1−24=(x2+7x+11)2−25=(x2+7x+11−5)(x2+7x+11+5)=(x2+7x+6)(x2+7x+16)=(x+1)(x+6)(x2+7x+16)
Phân tích thành nhân tử ( x+3) (x+2) (x+4) (x+5) -24
( x+3) (x+2) (x+4) (x+5) -24
=(x+3)(x+4)(x+2)(x+5)-24
=(x2+7x+12)(x2+7x+10)-24
Đặt t=x2+7x+10 ta được:
(t+2)t-24
=t2+2t-24
=t2+4t-6t-24
=t.(t+4)-6.(t+4)
=(t+4)(t-6)
thay t=x2+7t+10 ta được:
(x2+7x+14)(x2+7+4)
Vậy ( x+3) (x+2) (x+4) (x+5) -24=(x2+7x+14)(x2+7x+4)
ban ghep (x+3)(x+4) thanh 1 cap va (x+2)(x+5) roi dat t=x^2+7x+10 ta co t(t+2)-24=(t^2 +2t+1) -25=(t+1)^2 - 5^2=(t+6)(t-4) thay t la xong . giup mk lam bai mk dang vs nhe!
(x+ 2)(x +3)(x+ 4)(x+ 5) -24 -> phân tích đa thức sau thành nhân tử?
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\\ =\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+11=y\)
\(\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(y+1\right)\left(y-1\right)-24\\ =y^2-1-24\\ =y^2-25\\ =\left(y-5\right)\left(y+5\right)\\ =\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\\ =\left(x^2+7x+6\right)\left(x^2+7x+16\right)\\ =\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(x^2+7x+11\right)^2-1-24\\ =\left(x^2+7x+11\right)^2-25\\ =\left(x^2+7x+16\right)\left(x^2+7x+6\right)\\ =\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)
Bài 1 : Phân tích các đa thức sau thành nhân tử :
1) 15x + 15y 2) 8x - 12y
3) xy - x 4) 4x^2- 6x
Bài 2 : Phân tích các đa thức sau thành nhân tử :
1) 2(x + y) - 5a(x + y) 2) a^2(x - 5) - 3(x - 5)
3) 4x(a - b) + 6xy(a - b) 4) 3x(x - 1) + 5(x -1)
Bài 3 : Tính giá trị của biểu thức :
1) A = 13.87 + 13.12 + 13
2) B = (x - 3).2x + (x - 3).y tại x = 13 và y = 4
Bài 4 : Tìm x :
1) x(x - 5) - 2(x - 5) = 0 2) 3x(x - 4) - x + 4 = 0
3) x(x - 7) - 2(7 - x) = 0 4) 2x(2x + 3) - 2x - 3 = 0
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)