phân tích đa thức ->nhân tử:
a)2x2+4x-70
b)x3-5x2+8x-4
c)x2-10+16
rút gọn:
(8x-8x3-10x2+3x4-5):(3x2-2x+1)
Phân tích các đa thức sau thành nhân tử:
a,x3+4x-5
b,x3-3x2+4
c,x3+2x2+3x+2
d,x2+2xy+y2+2x-2y-3
e,(x2+3x)2-2(x2+3x)-8
f,(x2+4x+10)2-7(x2+4x+11)+7
a) x3+4x-5 = x3-x2+x2+4x-5=(x3-x2)+(x2-x)+(5x-5)=x2(x-1)+x(x-1)+5(x-1)=(x2+x+5)(x-1)
b) x3-3x2+4=x3-2x2-x2+4=(x3-2x2)-(x2-4)=x2(x-2)-(x-2)(x+2)=(x2-x+2)(x-2)
c) x3+2x2+3x+2=x3+x2+x2+x+2x+2=(x3+x2)+(x2+x)+(2x+2)=x2(x+1)+x(x+1)+2(x+1)=(x2+x+2)(x+1)
d) bạn xem lại đề đúng ko
e) (x2+3x)2-2(x2+3x)-8=x4+6x3+9x2-2x2-6x-8=x4+6x3+7x2-6x-8=x4-x3+7x3-7x2+14x2-14x+8x-8=(x4-x3)+(7x3-7x2)+(14x2-14x)+(8x-8)=x3(x-1)+7x2(x-1)+14x(x-1)+8(x-1)=(x3+7x2+14x+8)(x-1)=(x3+x2+6x2+6x+8x+8)(x-1)=\(\left[\left(x^3+x^2\right)+\left(6x^2+6x\right)+\left(8x+8\right)\right]\left(x-1\right)\)\(=\left[x^2\left(x+1\right)+6x\left(x+1\right)+8\left(x+1\right)\right]\left(x-1\right)\)\(=\left(x^2+6x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left(x^2+2x+4x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\left(x+1\right)\left(x-1\right)\)\(=\left[x\left(x+2\right)+4\left(x+2\right)\right]\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)
f) (x2+4x+10)2-7(x2+4x+11)+7=(x2+4x+10)2-\(\left[7\left(x^2+4x+11\right)-7\right]\)\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)\)\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)
a) Ta có: \(x^3+4x-5\)
\(=x^3-x+5x-5\)
\(=x\left(x-1\right)\left(x+1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+5\right)\)
b) Ta có: \(x^3-3x^2+4\)
\(=x^3+x^2-4x^2+4\)
\(=x^2\left(x+1\right)-4\left(x-1\right)\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+4\right)\)
\(=\left(x+1\right)\cdot\left(x-2\right)^2\)
c) Ta có: \(x^3+2x^2+3x+2\)
\(=x^3+x^2+x^2+x+2x+2\)
\(=x^2\left(x+1\right)+x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+2\right)\)
d) Ta có: \(x^2+2xy+y^2+2x+2y-3\)
\(=\left(x+y\right)^2+2\left(x+y\right)-3\)
\(=\left(x+y\right)^2+3\left(x+y\right)-\left(x+y\right)-3\)
\(=\left(x+y\right)\left(x+y+3\right)-\left(x+y+3\right)\)
\(=\left(x+y+3\right)\left(x+y-1\right)\)
e) Ta có: \(\left(x^2+3x\right)^2-2\left(x^2+3x\right)-8\)
\(=\left(x^2+3x\right)^2-4\left(x^2+3x\right)+2\left(x^2+3x\right)-8\)
\(=\left(x^2+3x\right)\left(x^2+3x-4\right)+2\left(x^2+3x-4\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)
\(=\left(x+4\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
f) Ta có: \(\left(x^2+4x+10\right)^2-7\left(x^2+4x+11\right)+7\)
\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)-7+7\)
\(=\left(x^2+4x+10\right)\left(x^2+4x+10-7\right)\)
\(=\left(x^2+4x+3\right)\left(x^2+4x+10\right)\)
\(=\left(x+1\right)\left(x+3\right)\left(x^2+4x+10\right)\)
phân tích các đa thức sau thành nhân tử
a, 3x2 - 6
b, x2 - 2x +1 - y2
c, 9x3 - 9x2y - 4x + 4y
d, x3 - 2x2 - 8x
giúp emm
\(a,=3\left(x^2-2\right)\\ b,=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\\ c,=9x^2\left(x-y\right)-4\left(x-y\right)=\left(3x-2\right)\left(3x+2\right)\left(x-y\right)\\ d,=x\left(x^2-2x-8\right)=x\left(x^2+2x-4x-8\right)=x\left(x+2\right)\left(x-4\right)\)
i) (3xy3− 12x2y2+ 6xy)2:3xy − xy
j) (10x3− 19x2-4x + 4): (2x + 1)
k) (3x4− 8x3-10x2+ 8x − 5): (3x2− 2x + 1).
j: \(\dfrac{10x^3-19x^2-4x+4}{2x+1}\)
\(=\dfrac{10x^3+5x^2-24x^2-12x+8x+4}{2x+1}\)
\(=5x^2-12x+4\)
Phân tích các đa thức sau thành nhân tử:
a) 5x-20xy
b) x2-9
c) x2-2xy+y2-z2
d) 5x.(x-1)-2.(x-1)
e) x2+4x+3
f) x3-x 3x2y+3xy2+y3-y
g) x2-x-y2-y
h) 16x-5x2-3
i) x3-4x
j) 2x2-6x
k) x3- 3x2-4x+12
l) x2-y2-5x+5y
Mn giúp em giải vs em cần gấp để lm bài kiểm tra.Em cảm ơn trc ạ
bài 1 phân tích các đa thức sau thành nhân tử
a) x2 + 4x +3 b) 16x - 5x2 - 3 c) 2x2 + 7x + 5
d) 2x2 + 3x -5 e) x3 - 3x2 + 1 - 3x f ) x2 - 4x - 5
g) (a2 + 1 )2 - 4a2 h) x3 - 3x2 - 4x + 12 i) x4 + x3 + x + 1
k) x4 - x3 - x2 + 1 l ) (2x + 1 )2 - ( x - 1 )
\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)
Phân tích các đa thức sau thành nhân tử [ làm các bước luôn ạ, chứ đừng có ra kết quả ko thoi ạ ( - _ - ) ]
a/ 3x2 - 6x
b/ x2 - 2x + 1 - y2
c/ 9x3 - 9x2y - 4x + 4y
d/ x3 - 2x2 - 8x
a) \(3x^2-6x\)
\(=3x\left(x-2\right)\)
b) \(x^2-2x+1-y^2\)
\(=\left(x-1\right)^2-y^2\)
\(=\left(x-y-1\right)\left(x+y-1\right)\)
c) \(9x^3-9x^2y-4x+4y\)
\(=\left(9x^3-9x^2y\right)-\left(4x-4y\right)\)
\(=9x^2\left(x-y\right)-4\left(x-y\right)\)
\(=\left(x-y\right)\left(9x^2-4\right)\)
\(=\left(x-y\right)\left(3x-2\right)\left(3x+2\right)\)
a)\(3x^2-6x=3x\left(x-2\right)\)
b)\(x^2-2x+1-y^2=\left(x^2-2x+1\right)-y^2=\left(x-1\right)^2-y^2=\left[\left(x-1\right)-y\right]\left[\left(x-1+y\right)\right]=\left(x-1-y\right)\left(x-1+y\right)\)
c)\(9x^3-9x^2y-4x+4y=\left(9x^3-9x^2y\right)-\left(4x-4y\right)=9x^2\left(x-y\right)-4\left(x-y\right)=\left(x-y\right)\left(9x^2-4\right)=\left(x-y\right)\left[\left(3x\right)^2-2^2\right]=\left(x-y\right)\left(3x-2\right)\left(3x+2\right)\)
d)\(x^3-2x^2-8x=x\left(x^2-2x-8\right)\)
phân tích đa thức thành nhân tử
a) 1+6x-6x2-x3
b) x3-4x2+8x-8
c) x3+2x2+2x+1
d) 8x3-12x2+6x-1
a) Ta có: \(1+6x-6x^2-x^3\)
\(=\left(1-x\right)\left(x^2+x+1\right)+6x\left(1-x\right)\)
\(=\left(1-x\right)\left(x^2+7x+1\right)\)
b:Ta có: \(x^3-4x^2+8x-8\)
\(=\left(x-2\right)\left(x^2+2x+4\right)-4x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-2x+4\right)\)
c: Ta có: \(x^3+2x^2+2x+1\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+1\right)\)
d: Ta có: \(8x^3-12x^2+6x-1\)
\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\)
\(=\left(2x-1\right)^3\)
bài 3 phân tích đa thức sau thành nhân tử
a 4x2 -16 + (3x +12) (4-2x)
b x3 + X2Y -15x -15y
c 3(x+8) -x2 -8x
d x3 -3x2 + 1 -3x
e 5x2 -5y2 -20x + 20y
kkk =0)
a) \(4x^2-16+\left(3x+12\right)\left(4-2x\right)\)
\(=\left(2x-4\right)\left(2x+4\right)-3\left(x+4\right)\left(2x-4\right)\)
\(=\left(2x-4\right)\left(2x+4-3x-12\right)\)
\(=-\left(2x-4\right)\left(x+8\right)\)
b) \(x^3+x^2y-15x-15y\)
\(=x^2\left(x+y\right)-15\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-15\right)\)
c) \(3\left(x+8\right)-x^2-8x\)
\(=3\left(x+8\right)-x\left(x+8\right)\)
\(=\left(x+8\right)\left(3-x\right)\)
d) \(x^3-3x^2+1-3x\)
\(=x^3+1-3x^2-3x\)
\(=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
d) \(5x^2-5y^2-20x+20y\)
\(=5\left(x^2-y^2\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y-4\right)\)
Tìm phân thức P biết : 2 x 2 + 4 x + 8 x 3 - 3 x 2 - x + 3 : P = x 3 - 8 x + 1 x - 3
2 x 2 + 4 x + 8 x 3 - 3 x 2 - x + 3 : P = x 3 - 8 x + 1 x - 3
Bài 4. Tính tổng và hiệu của các đa thức sau:
a) P(x) = 5x4 + 3x2 - 3x5 + 2x - x2 - 4 +2x5 và Q(x) = x5 - 4x4 + 7x - 2 + x2 - x3 + 3x4 - 2x2
b) H (x) = ( 3x5 - 2x3 + 8x + 9) - ( 3x5 - x4 + 1 - x2 + 7x) và R( x) = x4 + 7x3 - 4 - 4x ( x2 + 1) + 6x
ai giúp mình với
`@` `\text {Ans}`
`\downarrow`
`a)`
Thu gọn:
`P(x)=`\(5x^4 + 3x^2 - 3x^5 + 2x - x^2 - 4 +2x^5\)
`= (-3x^5 + 2x^5) + 5x^4 + (3x^2 - x^2) + 2x - 4`
`= -x^5 + 5x^4 + 2x^2 + 2x - 4`
`Q(x) =`\(x^5 - 4x^4 + 7x - 2 + x^2 - x^3 + 3x^4 - 2x^2\)
`= x^5 + (-4x^4 + 3x^4) - x^3 + (x^2 - 2x^2) + 7x - 2`
`= x^5 - x^4 - x^3 - x^2 + 7x - 2`
`@` Tổng:
`P(x)+Q(x)=`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) + (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)
`= -x^5 + 5x^4 + 2x^2 + 2x - 4 + x^5 - x^4 - x^3 - x^2 + 7x - 2`
`= (-x^5 + x^5) - x^3 + (5x^4 - x^4) + (2x^2 - x^2) + (2x + 7x) + (-4-2)`
`= 4x^4 - x^3 + x^2 + 9x - 6`
`@` Hiệu:
`P(x) - Q(x) =`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) - (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)
`= -x^5 + 5x^4 + 2x^2 + 2x - 4 - x^5 + x^4 + x^3 + x^2 - 7x + 2`
`= (-x^5 - x^5) + (5x^4 + x^4) + x^3 + (2x^2 + x^2) + (2x - 7x) + (-4+2)`
`= -2x^5 + 6x^4 + x^3 + 3x^2 - 5x - 2`
`b)`
`@` Thu gọn:
\(H (x) = ( 3x^5 - 2x^3 + 8x + 9) - ( 3x^5 - x^4 + 1 - x^2 + 7x)\)
`= 3x^5 - 2x^3 + 8x + 9 - 3x^5 + x^4 - 1 + x^2 - 7x`
`= (3x^5 - 3x^5) + x^4 - 2x^3 - x^2 + (8x + 7x) + (9+1)`
`= x^4 - 2x^3 - x^2 + 15x + 10`
\(R( x) = x^4 + 7x^3 - 4 - 4x ( x^2 + 1) + 6x\)
`= x^4 + 7x^3 - 4 - 4x^3 - 4x + 6x`
`= x^4 + (7x^3 - 4x^3) + (-4x + 6x) - 4`
`= x^4 + 3x^3 + 2x - 4`
`@` Tổng:
`H(x)+R(x)=` \((x^4 - 2x^3 - x^2 + 15x + 10)+(x^4 + 3x^3 + 2x - 4)\)
`= x^4 - 2x^3 - x^2 + 15x + 10+x^4 + 3x^3 + 2x - 4`
`= (x^4 + x^4) + (-2x^3 + 3x^3) - x^2 + (15x + 2x) + (10-4)`
`= 2x^4 + x^3 - x^2 + 17x + 6`
`@` Hiệu:
`H(x) - R(x) =`\((x^4 - 2x^3 - x^2 + 15x + 10)-(x^4 + 3x^3 + 2x - 4)\)
`=x^4 - 2x^3 - x^2 + 15x + 10-x^4 - 3x^3 - 2x + 4`
`= (x^4 - x^4) + (-2x^3 - 3x^3) - x^2 + (15x - 2x) + (10+4)`
`= -5x^3 - x^2 + 13x + 14`
`@` `\text {# Kaizuu lv u.}`