x^4+6x^3+17x^2-6x+1
tính tổng avf hiệu các đa thức sau
G(x) = 21x^2 + 1 + 17x và H(x) = -2+ 6x^3-12x^2-8
M(x) = 7x^5 + 1 + 17x^4 - 2 và N(x) = 6x^4 - 12x^2 - 23x^4 + x
`G(x)+H(x)=(21x^2+1+17x)+(-2+6x^3-12x^2-8)`
`=21x^2+1+17x-2+6x^3-12x^2-8`
`= 6x^3+(21x^2-12x^2)+17x+(1-2-8)`
`= 6x^3+9x^2+17x-9`
`G(x)-H(x)=(21x^2+1+17x)-(-2+6x^3-12x^2-8)`
`= 21x^2+1+17x+2-6x^3+12x^2+8`
`= -6x^3+(21x^2+12x^2)+17x+(1+2+8)`
`= -6x^3+33x^2+17x+11`
`----`
`M(x)+N(x)=(7x^5 + 1 + 17x^4 - 2)+(6x^4 - 12x^2 - 23x^4 + x)`
`= 7x^5 + 1 + 17x^4 - 2+6x^4 - 12x^2 - 23x^4 + x`
`= 7x^5+(17x^4+6x^4-23x^4)-12x^2+x+(1-2)`
`= 7x^5-12x^2+x-1`
`M(x)-N(x)=(7x^5 + 1 + 17x^4 - 2)-(6x^4 - 12x^2 - 23x^4 + x)`
`= 7x^5 + 1 + 17x^4 - 2-6x^4 + 12x^2 + 23x^4 - x`
`= 7x^5+(17x^4-6x^4+23x^4)+12x^2-x+(1-2)`
`= 7x^5+34x^4+12x^2-x-1`
tính tổng avf hiệu các đa thức sau
G(x) = 21x^2 + 1 + 17x và H(x) = -2+ 6x^3-12x^2-8
M(x) = 7x^5 + 1 + 17x^4 - 2 và N(x) = 6x^4 - 12x^2 - 23x^4 + x
Mình đã trl rồi nha!
(https://hoc24.vn/cau-hoi/tinh-tong-avf-hieu-cac-da-thuc-saugx-21x2-1-17x-va-hx-2-6x3-12x2-8mx-7x5-1-17x4-2-va-nx-6x4-12x2-23x4-x.7858748287383)
rút gọn phân thức
a) \(\dfrac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}\)
b) \(\dfrac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}\)
a)
\(\dfrac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}\)
\(=\dfrac{x^2\left(x+1\right)-4\left(x+1\right)}{x^3+2x^2+6x^2+12x+5x+10}\)
\(=\dfrac{\left(x+1\right)\left(x^2-4\right)}{x^2\left(x+2\right)+6x\left(x+2\right)+5\left(x+2\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x^2+6x+5\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left[x\left(x+5\right)+\left(x+5\right)\right]}\)
\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{x-2}{x+5}\)
b)
\(\dfrac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}\)
\(=\dfrac{x^4+3x^3+x^2+3x^3+9x^2+3x-x^2-3x-1}{x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1}\)
\(=\dfrac{x^2\left(x^2+3x+1\right)+3x\left(x^2+3x+1\right)-\left(x^2+3x+1\right)}{x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)}\)
\(=\dfrac{\left(x^2+3x+1\right)\left(x^2+3x-1\right)}{\left(x^2+3x-1\right)\left(x^2+3x-1\right)}\)
\(=\dfrac{x^2+3x+1}{x^2+3x-1}\)
phân tích thành đa thức nhân tử
x^3 + 8x^2 + 17x + 10
x^5 + x^4 + 1
x^3 - x^2 - 8x + 12
x^4 - 6x^3 + 11x^2 - 6x + 1
\(x^3-x^2-8x+12\)
\(=x^3+3x^2-4x^2-12x+4x+12\)
\(=x^2\left(x+3\right)-4x\left(x+3\right)+4\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-4x+4\right)\)
\(=\left(x+3\right)\left(x-2\right)^2\)
Rút gọn các biểu thức sau:
1. (6x+1)2 + (6x-1)2 - 2(1+6x)(6x-1)
2. 3x(x-2) - 5x(1-x) - 5(x2 - 3)
3. (7x-3)(2x+1) - (5x - 2)(x+4) - 9x2 + 17x
4. (6x -5)(x+8) - (3x-1)(2x+3) - 9(4x-3)
Tao không biết
Lưu lê thanh hạ rảnh lên à bạn ???
\(\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)
\(=\left(6x+1+6x-1\right)^2\)
\(=36x^2\)
\(3x\left(x-3\right)-5x\left(1-x\right)-5\left(x^2-3\right)\)
\(3x\left(x-3\right)-5\left(x-x^2+x^2-3\right)\)
\(3x\left(x-3\right)-5\left(x-3\right)\)
\(\left(x-3\right)\left(3x-5\right)\)
tim x
|9-7X|=5x-3
|17x-5|-|17x+5|=0
|3x+4|=2.|2x-9|
5^x+2=625
(x-1)^x+2=(x-1)6x+4
a: =>|7x-9|=5x-3
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(7x-9-5x+3\right)\left(7x-9+5x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(2x-6\right)\left(12x-12\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{3;1\right\}\)
b: =>|17x-5|=|17x+5|
=>17x-5=17x+5(vô lý) hoặc 17x-5=-17x-5
=>34x=0
hay x=0
c: =>|3x+4|=|4x-18|
=>4x-18=3x+4 hoặc 4x-18=-3x-4
=>x=22 hoặc 7x=14
=>x=22 hoặc x=2
giải phương trình : \(4x^3+6x^2-17x-22=4(x-2)\sqrt{6x+10}\)
4x3 +6x2 -17x-22 = 4(x-2)\(\sqrt{6x+10}\)
<=> (x-2)(4x2+14x+11)=4(x-2)\(\sqrt{6x+10}\)
<=> (x-2)(4x2+14x+11-4\(\sqrt{6x+10}\))=0
TH1:x=2
TH2:tự xét nha
Phân tích thành đa thức thành nhân tử
x^3 + 8x^2 + 17x + 10
( x^2 + x + 1)( x^2 + x + 2 ) -12
x^4 - 6x^3 + 11x^2 - 6x + 1
giải chi tiết giùm mình nha
\(x^3+8x^2+17x+10\)
\(=x^3+2x^2+x^2+5x^2+10x+5x+2x+10\)
\(=\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(5x^2+5x\right)+\left(10x+10\right)\)
\(=x^2\left(x+1\right)+2x\left(x+1\right)+5x\left(x+1\right)+10\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+2x+5x+10\right)\)
\(=\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]\)
\(=\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
Cộng các phân thức cùng mẫu thức :
a) \(\dfrac{1-2x}{6x^3y}+\dfrac{3+2y}{6x^3y}+\dfrac{2y-4}{6x^3y}\)
b) \(\dfrac{x^2-2}{x\left(x-1\right)^2}+\dfrac{2-x}{x\left(x-1\right)^2}\)
c) \(\dfrac{3x+1}{x^2-3x+1}+\dfrac{x^6-6x}{x^2-3x+1}\)
d) \(\dfrac{x^2+38x+4}{2x^2+17x+1}+\dfrac{3x^2-4x-2}{2x^2+17x+1}\)
a: \(=\dfrac{1-2x+3+2y+2y-4}{6x^3y}=\dfrac{-2x+4y}{6x^3y}=\dfrac{-2\left(x-2y\right)}{6x^3y}=\dfrac{-x+2y}{3x^3y}\)
b: \(=\dfrac{x^2-2+2-x}{x\left(x-1\right)^2}=\dfrac{x\left(x-1\right)}{x\left(x-1\right)^2}=\dfrac{1}{x-1}\)
c: \(=\dfrac{3x+1+x^6-3x}{x^2-3x+1}\)
\(=\dfrac{x^6+1}{x^2-3x+1}\)
d: \(=\dfrac{x^2+38x+4+3x^2-4x-2}{2x^2+17x+1}\)
\(=\dfrac{4x^2+34x+2}{2x^2+17x+1}=2\)