729 x 99 + 729
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72,9 x 99 + 729 x 0,1
=72,9 x 99 + 72,9
=72,9 x (99+1)
=72,9 x 100
=7290
72,9 x 99 + 729 x 0,1
72,9 x 99 + 72,9
72,9 x 99 + 72,9 x 1
72,9 x ( 99 + 1 )
72,9 x 100
7290
`Answer:`
\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{9}\right)+\left(x+\frac{1}{27}\right)+...+\left(x+\frac{1}{729}\right)=\frac{4209}{729}\)
\(\Leftrightarrow\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{3^2}\right)+\left(x+\frac{1}{3^3}\right)+...+\left(x+\frac{1}{3^6}\right)=\frac{4209}{729}\)
\(\Leftrightarrow6x+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\right)=\frac{4209}{729}\text{(*)}\)
Đặt \(N=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\)
\(\Leftrightarrow3N=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
\(\Leftrightarrow3N-N=\left(1+\frac{1}{3}+\frac{1}{3^2}+..+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\right)\)
\(\Leftrightarrow2N=1-\frac{1}{3^6}\)
\(\Leftrightarrow2N=\frac{728}{729}\)
\(\Leftrightarrow N=\frac{364}{729}\)
\(\text{(*)}\Leftrightarrow6x+\frac{364}{729}=\frac{4209}{729}\)
\(\Leftrightarrow6x=\frac{3845}{729}\)
\(\Leftrightarrow x=\frac{3845}{4374}\)
Ta có: \(A=\left[6.\left(\frac{-1}{3}\right)^2-\left(-\frac{1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)
\(\Rightarrow A=\left[6.\frac{1}{9}+\frac{1}{3}+1\right]:\left(\frac{-1}{3}-\frac{3}{3}\right)\)
\(\Rightarrow A=\left[\frac{2}{3}+\frac{1}{3}+1\right]:\frac{-4}{3}\)
\(\Rightarrow A=\left[1+1\right].\frac{-3}{4}=2.\frac{-3}{4}=\frac{-3}{2}\)
Mà \(B=\left(729-1^3\right)\left(729-2^3\right)\left(729-3^3\right)...\left(729-125^3\right)\)
\(=\left(729-1^3\right)\left(729-2^3\right)...\left(729-9^3\right)...\left(729-125^3\right)\)
\(=\left(729-1^3\right)\left(729-2^3\right)...0...\left(729-125^3\right)=0\)
Vì \(\frac{-3}{2}< 0\)nên A < B
729 x 99 + 729
= 729 x ( 99 + 1 )
= 729 x 100
= 72900 nha bạn ^^
= 72171 + 729
= 72900