giúp mik vs mng oi ❤
Cho A = 1 + 2 + 22 + 23 + ... + 299
Tìm n thuộc N biết : A + 1 = 2n
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Có :
9 - x = 15
x = 9 - 15
x = - 6
Mà -6 \(\notin\) N ⇒ D = { }
9 - x = 15
=> x = 9 - 15
=> x = -6
Mà x ∈ N => K có x thỏa mãn
=> D = ∅
Ex1
2 Have you heard the news yet? - Yes, I have already heard it
3 Have you eaten this fruit yet? - Yes, I've already eaten it
4 Has Jill written the report yet? - Yes, she has already written it
5 Has Ted repair his bike yet? - Yes, he has already repaired it
6 Has Messi signed the contract yet? - Yes, he has already signed it
7 Have Vera and Tom got engaged yet? - Yes, they have already done it
8 Has Paul bought a new car yet? - Yes, he has already bought it
9 Has Yen passed her driving test yet? - Yes, she has already passed it
10 Have you sent the invitations yet? - Yes, I have already sent them
Ex1
2 have met
3 has knwon
4 have read
5 haven't seen
6 has never eaten
7 Have you told
8 hasn't rained
9 have had
10 have just sold
\(A=x\left(x^2-y\right)-x^2\left(x+y\right)+y\left(x^2-x\right)\\ =x^3-xy-x^3-x^2y+x^2y-xy\\ =-2xy\)
Thay `x=1/2;y=-100` vào A ta có:
\(A=-2\cdot\dfrac{1}{2}\cdot\left(-100\right)=100\)
\(B=\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)\\=x^3+3x^2-5x-15-x^3+4x+x^2-4x^2\\ =\left(x^3-x^3\right)+\left(3x^2-4x^2+x^2\right)+\left(-5x+4x\right)-15\\ =-x-15\)
\(\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|+8\dfrac{1}{5}=1,2\\\Rightarrow\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|=1,2-8\dfrac{1}{5}\\ \Rightarrow \left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|=-7\)
Nhận xét:
\(\left\{{}\begin{matrix}\left|2\dfrac{1}{5}-x\right|\ge0,\forall x\\\left|x-\dfrac{1}{5}\right|\ge0,\forall x\end{matrix}\right.\\
\Rightarrow\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|\ge0,\forall x\)
Mà \(-7< 0\) nên:
Không tìm được giá trị \(x\) thỏa mãn đề bài
Vậy...
\(\left\{{}\begin{matrix}2x+3y=-xy\\\dfrac{8}{x}-\dfrac{6}{y}=5\end{matrix}\right.\left(x;y\ne0\right)\Leftrightarrow\left\{{}\begin{matrix}2x+3y=-xy\\8y-6x=5xy\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3y=-xy\\6x-8y=-5xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=-xy\\6x-8y=5\left(2x+3y\right)\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3y=-xy\\6x-8y=10x+15y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=-xy\\-4x=23y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2\cdot\dfrac{-23}{4}y+3y=\dfrac{-23}{4}y\cdot y\\x=\dfrac{-23}{4}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}23y^2-43y=0\\x=\dfrac{-23}{4}y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y\left(23y-43\right)=0\\x=\dfrac{-23}{4}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}y=0\left(ktm\right)\\y=\dfrac{43}{23}\left(tm\right)\end{matrix}\right.\\x=\dfrac{-23}{4}\cdot\dfrac{43}{23}=\dfrac{-43}{4}\end{matrix}\right.\)
\(A=1+2+2^2+...+2^{99}\\ 2A=2+2^2+2^3+...+2^{100}\\ 2A-A=\left(2+2^2+2^3+...+2^{100}\right)-\left(1+2+2^2+...+2^{99}\right)\\ A=2^{100}-1\)
\(=>A+1=2^{100}-1+1=2^{100}\)
Mà: \(A+1=2^n=>2^n=2^{100}\)
\(=>n=100\)