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mk ko chép đề mà tách luôn nha
M = x2x2 + x2x2 + x2y2 + x2y2 + x2y2 + y2y2 + y2
= ( x2x2 + x2y2 ) + ( x2x2 + x2y2 ) + ( x2y2 + y2y2 ) + y2
= x2( x2 + y2 ) + x2( x2 + y2 ) + y2( x2 + y2 ) + y2
= ( x2 + y2 ) (x2 + x2 + y2 ) + y2
= 1( x2 + 1) + y2
= x2 + y2 +1 = 2
\(B=\left[\dfrac{1}{100}-1^2\right]\left[\dfrac{1}{100}-\left(\dfrac{1}{2}\right)^2\right]\cdot...\cdot\left[\dfrac{1}{100}-\left(\dfrac{1}{10}\right)^2\right]\cdot...\cdot\left[\dfrac{1}{100}-\left(\dfrac{1}{120}\right)^2\right]\)
\(=\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}-\dfrac{1}{4}\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{100}\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{14400}\right)\)
=0
a) \(\left(x-3\right)\left(x-2\right)< 0\)
Ta có : \(x-2>x-3\)
\(\Rightarrow\left\{{}\begin{matrix}x-3< 0\\x-2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 3\\x>2\end{matrix}\right.\Rightarrow2< x< 3\)
Vậy \(2< x< 3\)
b) \(3x+x^2=0\)
\(x\left(3+x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\3+x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
Vậy \(x\in\left\{-3;0\right\}\)
ta có : \(\left(x-3\right)^{x+2}-\left(x-3\right)^{x+8}=0\)
\(\Rightarrow\left(x-3\right)^{x+2}-\left(x-3\right)^{x+2+6}=0\)
\(\Rightarrow\left(x-3\right)^{x+2}-\left(x-3\right)^{x+2}.\left(x-3\right)^6=0\)
\(\Rightarrow\left(x-3\right)^{x+2}.[1-\left(x-3\right)^6]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^{x+2}=0\\1-\left(x-3\right)^6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^6=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x-3=1\\x-3=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=4\\x=2\end{matrix}\right.\)
Cái câu hỏi bn để ở trước câu này mk có thể làm dc, nhưng mik thấy làm nhiều lần rồi ngán nên ko trả lời nữa (lười chính hiệu:))))
\(3x^2y^4\)-\(5xy^3\)-\(\dfrac{3}{2}x^2y^4\)+\(3xy^3\)+\(2xy^3\)+1=1,5\(x^2y^4\)+1>0
????????
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\(\left\{{}\begin{matrix}P\left(x\right)=x+x^2-x^3+2x^3+2=x^3+x^2+x+2\\Q\left(x\right)=1+3x-x^2-4x+x^3=x^3-x^2-x+1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}P\left(x\right)+Q\left(x\right)=2x^3+3\\P\left(x\right)-Q\left(x\right)=2x^2+2x+1\end{matrix}\right.\)
P=x3+x2y-2x2-xy-y2+3y+x+2017 với x+y=2
P=x3+x2y-2x2-xy-y2+2y+y+x+2017
\(P=\left(x^3+x^2y-2x^2\right)-\left(xy+y^2-2y\right)+\left(x+y-2\right)+2019\)
\(P=x^2\left(x+y-2\right)-y\left(x+y-2\right)+\left(x+y-2\right)+2019\)
có x+y=2 suy ra x+y-2=0
suy ra \(P=2019\)