phan tich thanh nhan tu
a) 9x2+6x-8
b)x2-7xy +10y2
c) x8+x7+1
Phan tich da thuc thanh nhan tu
3x^2-11x+6
x^2-6x+5
x^4+x^2+1
x^4-4x^2+3
6x^2+7xy+2y^2
(*)\(3x^2-11x+6=3x^2-2x-9x+6=x\left(3x-2\right)-3\left(3x-2\right)=\left(x-3\right)\left(3x-2\right)\)
(*)\(x^2-6x+5=x^2-x-5x+5=x\left(x-1\right)-5\left(x-1\right)=\left(x-5\right)\left(x-1\right)\)
(*)\(x^4+x^2+1=x^4+2x^2+1-x^2=\left(x^2+1\right)^2-x^2=\left(x^2+1+x\right)\left(x^2+1-x\right)\)
(*)\(x^4-4x^2+3=x^4-x^2-3x^2+3=x^2\left(x^2-1\right)-3\left(x^2-1\right)=\left(x+1\right)\left(x-1\right)\left(x^2-3\right)\)
(*)\(6x^2+7xy+2y^2=6x^2+4xy+3xy+2y^2=2x\left(3x+2y\right)+y\left(3x+2y\right)=\left(2x+y\right)\left(3x+2y\right)\)
a, \(3x^2-11x+6=3x^2-2x-9x+6=x\left(3x-2\right)-3\left(3x-2\right)=\left(3x-2\right)\left(x-3\right)\)
b, \(x^2-6x+5=x^2-x-5x+5=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)
c, \(x^4+x^2+1=x^4+2x^2+1-x^2=\left(x^2+1\right)^2-x^2=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
d, \(x^4-4x^2+3=x^4-4x^2+4-1=\left(x^2-2\right)^2-1=\left(x^2-1\right)\left(x^2-3\right)=\left(x+1\right)\left(x-1\right)\left(x^2-3\right)\)
e, \(6x^2+7xy+2y^2=6x^2+3xy+4xy+2y^2=3x\left(2x+y\right)+2y\left(2x+y\right)=\left(2x+y\right)\left(3x+2y\right)\)
Phan tich da thuc thanh nhan tu
a) x^2 - 7xy + 10y^2
b) 4x^2 + 12x - 27
c) 81x^4 + 4
d) 3a^2 - 6ab + 3b^2 - 12c^2
e) x^3 + 6x^2 +11x +6
phan tich da thuc thanh nhan tu
a) 7xy^2+5x^2y
b) x^2+2xy+y^2-11x-11y
a) 7xy^2+5x^2y
= xy(7y+5x)
b) x^2+2xy+y^2-11x-11y
= (x^2+2xy+y^2)-(11x+11y)
=(x+y)^2-11(x+y)
=(x+y)(x+y-11)
a) 7xy2 + 5x2y
= xy ( 7y + 5x )
b) x2 + 2xy + y2 - 11x - 11y
= ( x2 + 2xy + y2 ) - ( 11x + 11y )
= ( x + y )2 - 11 ( x + y )
= ( x + y )( x + y - 11 )
6x^2+5x+1
phan tich da thuc thanh nhan tu
Phân tích thành đa nhân tử
đc kết quả :
(2x+1)(3x+1)
:)
Phan tich da thuc thanh nhan tu : \(2x^2-6x+1\)
Bài này ko thể phân tích theo kiểu lớp 8 được (chưa học căn thức)
\(2x^2-6x+1=\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.\frac{3\sqrt{2}}{2}+\left(\frac{3\sqrt{2}}{2}\right)^2-\frac{7}{2}\)
\(=\left(\sqrt{2}x-\frac{3\sqrt{2}}{2}\right)^2-\left(\frac{\sqrt{14}}{2}\right)^2\)
\(=\left(\sqrt{2}x-\frac{3\sqrt{2}}{2}+\frac{\sqrt{14}}{2}\right)\left(\sqrt{2}x-\frac{3\sqrt{2}}{2}-\frac{\sqrt{14}}{2}\right)\)
\(=\left(\sqrt{2}x+\frac{\sqrt{14}-3\sqrt{2}}{2}\right)\left(\sqrt{2}x-\frac{\sqrt{14}+3\sqrt{2}}{2}\right)\)
\(2x^2-6x+1=2\left(x^2-3x+\frac{9}{4}-\frac{7}{4}\right)=2\left[\left(x-\frac{3}{2}\right)^2-\left(\frac{\sqrt{7}}{2}\right)^2\right]=2\left(x-\frac{3}{2}-\frac{\sqrt{7}}{2}\right)\left(x-\frac{3}{2}+\frac{\sqrt{7}}{2}\right)\)
\(=2\left(x-\frac{3+\sqrt{7}}{2}\right)\left(x-\frac{3-\sqrt{7}}{2}\right)\)
Phan tich da thuc sau thanh nhan tu
6x^3+x^2+x+1
\(6x^3+x^2+x+1=\left(6x^3+3x^2\right)+\left(-2x^2-x\right)+\left(2x+1\right)\)
\(=3x^2.\left(2x+1\right)-x.\left(2x+1\right)+\left(2x+1\right)=\left(2x+1\right)\left(3x^2-x+1\right)\)
K sai dau
giao an truong Tran dai nghia do
8x^3-12x^2+6x-1
phan tich da thuc thanh nhan tu chung
cái này dễ mà
= (2x)^3-3(2x)^2*1+2*3x*1^2-1^3
= (2x-1)^3
(2x+5)^2-6x-15 phan tich thanh nhan tu
\(\left(2x+5\right)^2-6x-15=\left(2x+5\right)^2-3\left(2x-5\right)=\left(2x-5\right)^2-3\left(2x-5\right)\)
\(=\left(2x-5\right)\left(2x-5-3\right)=\left(2x-5\right)\left(2x-2\right)=2\left(2x-5\right)\left(x-1\right)\)
(2x+5)^2-6x-15
=4x2+20x+25-6x-15
=4x2+14x+10
=(2x+1)2+9 (đề bài có nhầm không)
ღ๖ۣۜLinh's ๖ۣۜLinh'sღ] ★we are one★ hình như bạn làm sai rồi thì phải
đề là phân tích thành nhân tử kia mà
phan tich da thuc thanh nhan tu
B=\(^{x^4}+6x^3+7x^2-6x+1\)
\(B=x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)
\(=\left(x^2+3x-1\right)^2\)