phan tich da thuc x2-5x+4 duoi dang: x2-5x+4=(x+1)2+b(x+1)+c
khi do b+c=?
Những câu hỏi liên quan
1.Viet bieu thuc sau duoi dang binh phuong cua mot tong:
2xy2 +x2y4 +1
2. Phan tich da thuc thanh nhan tu:
a) 3x2 - 7x -10
b) 2x2 -5x -7
HELP ME!!!!*!*
1. 2xy2 +x2y4+1 = (xy2+1)2
2. a)3x2+3x-10x-10=3x(x+1)-10(x+1)=(x+1)(3x-10)
b)2x2-5x-7=2x2+2x-7x-7=2x(x+1)-7(x+1)=(x+1)(2x-7)
Mong có thể giúp được bạn
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Viet bieu thuc sau duoi dang binh phuong cua mot tong:
2xy2 +x2y4 +1
Phan tich da thuc sau thanh nhan tu:
a) x2 -3x +2
b) 3x2 - 7x -10
c) 2x2 -5x -7
HELP ME!!!
x2 - 3x + 2
= x2 - x - 2x + 2
= x(x - 1) - 2(x - 1)
= (x - 1)(x - 2)
3x2 - 7x - 10
= 3x2 + 3x - 10x - 10
= 3x(x + 1) - 10(x + 1)
= (x + 1)(3x - 10)
2x2 - 5x - 7
= 2x2 + 2x - 7x - 7
= 2x(x + 1) - 7(x + 1)
= (x + 1)(2x - 7)
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Viet cac da thuc sau duoi dang tich :
a, 100x2−(x2+25)2
b, 1+(x−y+5)2−2(x−y+5)
c, (x2+4y2−5)2−16(x2y2+2xy+1)
d, (x2+8x−34)2−(3x2−8x−2)2
a) \(100x^2-\left(x^2+25\right)^2\)
\(=\left(10x\right)^2-\left(x^2+25\right)^2\)
\(=\left(10x+x^2+25\right)\left(10x-x^2-25\right)\)
\(=\left(x+5\right)^2\left(10x-x^2-25\right)\)
b) \(1+\left(x-y+5\right)^2-2\left(x-y+5\right)\)
\(=\left[\left(x-y+5\right)-1\right]^2\)
\(=\left(x-y+4\right)^2\)
c) \(\left(x^2+4y^2-5\right)^2-16\left(x^2y^2+2xy+1\right)\)
\(=\left(x^2+4y^2-5\right)^2-4^2\left(xy+1\right)^2\)
\(=\left[\left(x^2+4y^2-5\right)-4\left(xy+1\right)\right]\left[\left(x^2+4y^2-5\right)+4\left(xy+1\right)\right]\)
\(=\left(x^2+4y^2-5-4xy-4\right)\left(x^2+4y^2-5+4xy+4\right)\)
\(=\left(x^2-4xy+4y^2-9\right)\left(x^2+4xy+4y^2-1\right)\)
\(=\left[\left(x-2y\right)^2-3^2\right]\left[\left(x+2y\right)^2-1^2\right]\)
\(=\left(x-2y-3\right)\left(x-2y+3\right)\left(x+2y-1\right)\left(x+2y+1\right)\)
d) \(\left(x^2+8x-34\right)^2-\left(3x^2-8x-2\right)^2\)
\(=\left[\left(x^2+8x-34\right)-\left(3x^2-8x-2\right)\right]\left[\left(x^2+8x-34\right)+\left(3x^2-8x-2\right)\right]\)
\(=\left(x^2+8x-34-3x^2+8x+2\right)\left(x^2+8x-34+3x^2-8x-2\right)\)
\(=\left(-2x^2+16x-32\right)\left(4x^2-36\right)\)
\(=-2\left(x^2-8x+16\right)\left[\left(2x\right)^2-6^2\right]\)
\(=-2\left(x-4\right)^2\left(2x-6\right)\left(2x+6\right)\)
\(=-2\left(x-4\right)^24\left(x-3\right)\left(x+3\right)\)
\(=8\left(x-4\right)^2\left(x-3\right)\left(x+3\right)\)
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phan tich da thuc thanh nhan tu a. x^3+x+2
b, x^4+5x^3+10x-4
\(x^3+x+2=\left(x^3+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1+1\right)\)
\(=\left(x+1\right)\left(x^2-x+2\right)\)
\(b,x^4+5x^3+10x-4=\left(x^4-4\right)+\left(5x^3-10x\right)\)\(=\left(x^2+2\right)\left(x^2-2\right)+5x\left(x^2+2\right)\)
\(=\left(x^2+2\right)\left(x^2-2+5x\right)\)
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A(x)= (x^4 + 4x^2-5x +1)^1994
tinh tong cac he so cua hang tu cua da thuc nhan dc khi da khai trien va viet duoi dang thu gon.
tổng các hệ số trong đa thức một biến bằng giá trị của đa thức đó tại giá trị của biến bằng 1
A(1)=\(\left(1^4+4.1^2-5.1+1\right)^{1994}\)
\(\Rightarrow A\left(1\right)=\left(1+4-5+1\right)^{1994}=1^{1994}=1\)
vậy tổng các hệ số trong A(x) là 1
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X^4+4x^3+5x^2+2x+1
Phan tich da thuc thanh nhan tu
phan tich da thuc thanh nhan tu : a) 3x^2 - 22xy + 4x + 8y + 7x^2 + 1 ; b) 12x^2 + 5x - 12y^2 + 12y - 10xy - 3 ; c)x^4 + 6x^3 + 11x^2 + 6x + 1
phan tich da thuc thanh nhan tu
a) x^3+x+2
b) x^3+3x^2-4
c) x^4+x^3+6x^2+5x+5
a)\(3x^2-11x+6=3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(3x-2\right)\left(x-3\right)\)
b)\(8x^2+10x-3=8x^2+12x-2x-3=4x\left(2x+3\right)-\left(2x+3\right)=\left(4x-1\right)\left(2x+3\right)\)
c)\(8x^2-2x-1=8x^2+2x-4x-1=2x\left(4x+1\right)-\left(4x+1\right)=\left(2x-1\right)\left(4x+1\right)\)
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1. phan tich da thuc sau thanh nhan tu
a) x4 + 2x3 + x2
b) x2 - 5x - 6
a)x4+2x3+x2=x2(x2+2x+1)=x2.(x+1)2
b)x2+5x-6 =x2-x+6x-6 =x.(x-1)+6.(x-1) =(x-1).(x+6)
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