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a: M(x)=3x^3-7x^2+4/5x-1/5
N(x)=3x^3-7x^2-x+3/2
b: M(x)+N(x)=6x^3-14x^2-1/5x+13/10
c: H(x)=M(x)-N(x)
=3x^3-7x^2+4/5x-1/5-3x^3+7x^2+x-3/2
=9/5x-17/10
a/\(M\left(x\right)+N\left(x\right)=\left(3x^2+5x-x^3+4\right)+\left(x^3-5+4x^2+6x\right)\)
\(=3x^2+5x-x^3+4+x^3-5+4x^2+6x\)
\(=\left(3x^2+4x^2\right)+\left(5x+6x\right)-\left(x^3-x^3\right)-\left(-4+5\right)\)
\(=7x^2+11x-0-1\)
\(=7x^2+11x-1\)
b/\(M\left(x\right)-Q\left(x\right)=\left(3x^2+5x-x^3+4\right)-\left(x^3-5+4x^2+6x\right)\)
\(=3x^2+5x-x^3+4-x^3+5-4x^2-6x\)
\(=\left(3x^2-4x^2\right)+\left(5x-6x\right)-\left(x^3+x^3\right)+\left(4+5\right)\)
\(=-x^2+-x-2x^3+9\)
\(=-x^2-x-2x^3+9\)
#DatNe
a/\(F\left(x\right)=-4x^2+7-6x+5x^3\)
\(=5x^3-4x^2-6x+7\)
\(G\left(x\right)=4x^2+9x-2x^4+4x^3-12\)
\(=-2x^4+4x^3+4x^2+9x-12\)
b/\(F\left(x\right)+G\left(x\right)=\left(5x^3-4x^2-6x+7\right)+\left(-2x^4+4x^3+4x^2+9x-12\right)\)
\(=5x^3-4x^2-6x+7-2x^4+4x^3+4x^2+9x-12\)
\(=\left(5x^3+4x^3\right)-\left(4x^2-4x^2\right)+\left(-6x+9x\right)-2x^4-\left(-7+12\right)\)
\(=9x^3-0+3x-2x^4-5\)
\(=9x^3+3x-2x^4-5\)
2:
a: \(A\left(x\right)=3x^3-2x^2-5x+3\)
\(B\left(x\right)=5x^3+x^2+2x-1\)
b: A(x)+B(x)=8x^3-x^2-3x+2
c: A(x)-B(x)
=3x^3-2x^2-5x+3-5x^3-x^2-2x+1
=-2x^3-3x^2-7x+4
a: Xét ΔAIB và ΔAID có
AI chung
\(\widehat{BAI}=\widehat{DAI}\)
AB=AD
Do đó: ΔAIB=ΔAID
a) Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{x+y+z}{3+4+5}=\dfrac{360}{60}=6\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=6\\\dfrac{y}{4}=6\\\dfrac{z}{5}=6\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=18\\y=24\\z=30\end{matrix}\right.\)
b) Áp dụng tính chất dãy tỉ số bằng nhau:\(\dfrac{x}{-2}=-\dfrac{y}{4}=\dfrac{z}{5}=-\dfrac{2y}{8}=\dfrac{3z}{15}=\dfrac{x-2y+3z}{-2+8+15}=\dfrac{1200}{21}=\dfrac{400}{7}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-2}=\dfrac{400}{7}\\-\dfrac{y}{4}=\dfrac{400}{7}\\\dfrac{z}{5}=\dfrac{400}{7}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{800}{7}\\y=-\dfrac{1600}{7}\\z=\dfrac{2000}{7}\end{matrix}\right.\)
c) Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{z}{-2}=\dfrac{-2z}{4}=\dfrac{x+y-2z}{5+1+4}=\dfrac{160}{10}=16\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=16\\\dfrac{y}{1}=16\\\dfrac{z}{-2}=16\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=80\\y=16\\z=-32\end{matrix}\right.\)
Bài `2`
`a, P(x)=12-6x^3+9x-x^2`
`= -6x^3 -x^2 +9x+12`
__
`Q(x)=4+10x^2+29x^3-6x`
`= 29x^3 +10x^2 -6x+4`
`b,P(x)+Q(x)= -6x^3 -x^2 +9x+12 + 29x^3 +10x^2 -6x+4`
`= (-6x^3+ 29x^3)+(-x^2+10x^2) +(9x -6x) +(12+4)`
`= 23x^3+ 9x^2 + 3x + 16`
`c,P(x)-Q(x)=(-6x^3 -x^2 +9x+12) - (29x^3 +10x^2 -6x+4)`
`= -6x^3 -x^2 +9x+12- 29x^3-10x^2 +6x-4`
`= (-6x^3- 29x^3 )+(-x^2-10x^2)+(9x+6x) +(12-4)`
`= -35x^3 -11x^2 + 15x + 8`
em xem lại P(x)+Q(x) nì :v