Câu hỏi của Nguyen Thu Ngoc

Question

Tinh :P= 1/2 +1/2^2 + ... + 1/ 2^99ai lam duoc minh tich cho

Thắng Nguyễn · Accepted Answer

$2P=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)$$2P=1+\frac{1}{2}+...+\frac{1}{2^{98}}$$2P-P=\left(1+\frac{1}{2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)$$P=1-\frac{1}{2^{99}}$Câu trả lời: $2P=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)$$2P=1+\frac{1}{2}+...+\frac{1}{2^{98}}$$2P-P=\left(1+\frac{1}{2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)$$P=1-\frac{1}{2^{99}}$

VICTORY_ Trần Thạch Thảo · Answer

$P=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}$$2P=2.\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)$$2P=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99-1}}$$2P=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}$$P=2P-P=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)$$P=1-\frac{1}{2^{99}}$$P=\frac{2^{99}-1}{2^{99}}$Câu trả lời: $P=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}$$2P=2.\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)$$2P=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99-1}}$$2P=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}$$P=2P-P=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)$$P=1-\frac{1}{2^{99}}$$P=\frac{2^{99}-1}{2^{99}}$

x	x<-1	-1</x<0	x>/0
\|x\|	-x	-x	x
\|x+1\|	-x-1	x+1	x+1
\|x\|+\|x+1\|=2	-2x-1=2=>x=-3/2 (TM)	1=2 loại	2x+1=2=>x=1/2 (TM)

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