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R(x) = 2x2 + 3x - 1
- M(x) = -x3 + x2
x3 + x2 + 3x - 1
Vậy R(x) - M(x) = x3 + x2 + 3x - 1
\(P\left(x\right)=3x^4+9x^2-2x-3\)
\(Q\left(x\right)=\left(3x^4-3x^4\right)+\left(x^2-4x^2+1.5x^2\right)+2x+1=-1.5x^2+2x+1\)
F(x)=62+5x+8+3x-3x2+3x3
=(36+8)+(5x+3x)-3x2+3x3
=3x3-3x2+8x+44
G(x)=12x2-6-9x2+3x3
=3x3+(12x2-9x2)-6
=3x3+3x2-6
F(x)+G(x)=3x3-3x2+8x+44+3x3+3x2-6
=(3x3+3x3)+(-3x2+3x2)+8x+(44-6)
=6x3+8x+38
\(F\left(x\right)=G\left(x\right)\\ \Rightarrow6^2-5x+8+3x-3x^2+3x^3=12x^2-6-9x^2+3x^3\\ \Leftrightarrow-3x^2-2x+44=3x^2-6\\ \Leftrightarrow6x^2+2x-50=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1+\sqrt{301}}{6}\\x=\dfrac{-1-\sqrt{301}}{6}\end{matrix}\right.\)
`P(x)=`\( 2x^4 + 3x^3 + 3x^2 - x^4 - 4x + 2 - 2x^2 + 6x\)
`= (2x^4-x^4)+3x^3+(3x^2-2x^2)+(-4x+6x)+2`
`= x^4+3x^3+x^2+2x+2`
`Q(x)=`\(x^4 + 3x^2 + 5x - 1 - x^2 - 3x + 2 + x^3\)
`= x^4+x^3+(3x^2-x^2)+(5x-3x)+(-1+2)`
`= x^4+x^3+2x^2+2x+1`
`P(x)+Q(x)=(x^4+3x^3+x^2+2x+2)+(x^4+x^3+2x^2+2x+1)`
`=x^4+3x^3+x^2+2x+2+x^4+x^3+2x^2+2x+1`
`=(x^4+x^4)+(3x^3+x^3)+(x^2+2x^2)+(2x+2x)+(2+1)`
`= 2x^4+4x^3+3x^2+4x+3`
`@`\(\text{dn inactive.}\)
P(x)=x^4+3x^3+x^2+2x+2
Q(x)=x^4+x^3+2x^2+2x+1
P(x)+Q(x)=2x^4+4x^3+3x^2+4x+3
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
c) \(2x\left(3x-1\right)-3x\left(5+2x\right)=0\)
\(\Rightarrow x\left[2\left(3x-1\right)-3\left(5+2x\right)\right]=0\)
\(\Rightarrow x\left(6x-2-15-6x\right)\)
\(\Rightarrow-16x=0\)
\(\Rightarrow x=0\)
d) \(\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\)
\(\Rightarrow9x^2-4-4x+4=0\)
\(\Rightarrow9x^2-4x=0\)
\(\Rightarrow x\left(9x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)
\(a,\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ b,\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
vậy thì;
3^x+3^x+2=90
=) 3^x .1+3^x . 3^2=90
=) 3^x.(1+ 3^2)=90
=) 3^x .(1+9)=90
=)3^x.10=90
=)3^x=90:10
=) 3^x=9
=) 3^x=3^2
=)x=2
Vậy x=2
mk làm chi tiết lắm đó chi
good luck
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