x^4+6x^3+17x^2-6x+1
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2 tháng 4 2023
`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`
2 tháng 4 2023
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)
HV
23 tháng 11 2017
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)
\(\dfrac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}\)
\(=\dfrac{x^4+3x^3+x^2+3x^3+9x^2+3x-x^2-3x-1}{x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1}\)
\(=\dfrac{x^2\left(x^2+3x+1\right)+3x\left(x^2+3x+1\right)-\left(x^2+3x+1\right)}{x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)}\)
\(=\dfrac{\left(x^2+3x+1\right)\left(x^2+3x-1\right)}{\left(x^2+3x-1\right)\left(x^2+3x-1\right)}\)
\(=\dfrac{x^2+3x+1}{x^2+3x-1}\)
NT
1
23 tháng 5 2022
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
TM
1
M
1 tháng 11 2015
\(x^3+8x^2+17x+10\)
\(=x^3+2x^2+x^2+5x^2+10x+5x+2x+10\)
\(=\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(5x^2+5x\right)+\left(10x+10\right)\)
\(=x^2\left(x+1\right)+2x\left(x+1\right)+5x\left(x+1\right)+10\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+2x+5x+10\right)\)
\(=\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]\)
\(=\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
16 tháng 5 2022
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: \(=\dfrac{x^2+38x+4+3x^2-4x-2}{2x^2+17x+1}\)
\(=\dfrac{4x^2+34x+2}{2x^2+17x+1}=2\)
\(x^4+6x^3+17x^2-6x+1=0\) (1)
Với \(x=0\), (1) vô nghiệm
Với \(x\ne0\), chia 2 vế cho \(x^2\) ta được:
\(x^2+6x+17-\frac{6}{x}+\frac{1}{x^2}=0\)
\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)+6\left(x-\frac{1}{x}\right)+17=0\) (1')
Đặt \(x-\frac{1}{x}=y\Rightarrow x^2+\frac{1}{x^2}=y^2+2\), khi đó (1') trở thành:
\(y^2+2+6y+17=0\Leftrightarrow y^2+6y+19=0\)
\(\Delta'=3^2-19=-10< 0\)
Vậy (1') vô nghiệm tương đương (1) vô nghiệm.