f(x)=x^3+2x-1
timf nghieemj
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1: Khi m=3 thì pt sẽ là:
\(\left\{{}\begin{matrix}3x-2x=-1\\2x+3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\3y-2=1\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(-1;1\right)\)
2: THeo đề, ta có:
\(\left\{{}\begin{matrix}m\cdot\dfrac{-1}{2}-2\cdot\dfrac{-1}{2}=-1\\2\cdot\dfrac{-1}{2}+3\cdot\dfrac{2}{3}=1\end{matrix}\right.\Leftrightarrow m\cdot\dfrac{-1}{2}=-3\)
hay m=6
a: \(h\left(x\right)=7x^5+x^4-2x^3+4+x^4+6x^3-9x^2-2x-1=7x^5+2x^4+4x^3-9x^2-2x+3\)
b: \(h\left(x\right)=7x^5+x^4-2x^3+4-x^4-6x^3+9x^2+2x+1=7x^5-8x^3+9x^2+2x+5\)
a)f(x)+g(x)=\(x^5-4x^4-2x^2-7-2x^5+6x^4-2x^2+6.\)
=\(-x^5+2x^4-4x^2-1\)
f(x)-g(x)=\(x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)
=\(3x^5-10x^4-13\)
b)f(x)+g(x)=\(5x^4+7x^3-6x^2+3x-7-4x^4+2x^3-5x^2+4x+5\)
=\(x^4+9x^3-11x^2+7x-2\)
f(x)-g(x)=\(5x^4+7x^3-6x^2+3x-7+4x^4-2x^3+5x^2-4x-5\)
=\(9x^4+5x^3-x^2-x-12\)
a )
\(f\left(x\right)+g\left(x\right)=x^5-4x^4-2x^2-7+-2x^5+6x^4-2x^2+6\)
\(\Rightarrow f\left(x\right)+g\left(x\right)=\left(x^5-2x^5\right)+\left(6x^4-4x^4\right)-\left(2x^2+2x^2\right)+\left(6-7\right)\)
\(\Rightarrow f\left(x\right)+g\left(x\right)=-x^5+2x^4-4x^2-1\)
\(f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7-\left(-2x^5+6x^4-2x^2+6\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=\left(x^5+2x^5\right)-\left(4x^4+6x^4\right)+\left(2x^2-2x^2\right)-\left(6+7\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=3x^5-10x^4-13\)