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Vì \(\left|2x+1\right|\ge0;\left|x+y-\frac{1}{2}\right|\ge0\)
Mà \(\left|2x+1\right|+\left|x+y-\frac{1}{2}\right|\le0\Rightarrow\orbr{\begin{cases}2x+1=0\\x+y-\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\y=\frac{1}{4}\end{cases}}\)(1)
Thế (1) vào A
\(\Rightarrow A=4.\left(-\frac{1}{2}\right)^3.\left(\frac{1}{4}\right)^2-\frac{1}{4}.\left(-\frac{1}{2}\right)+2.\frac{1}{4}-5\)
\(\Rightarrow A=-\frac{1}{2}+\frac{1}{8}+\frac{1}{2}-5\)
\(\Leftrightarrow A=\frac{1}{8}-5=\frac{1}{8}-\frac{40}{8}=-\frac{39}{8}\)
Sửa đề \(\frac{3}{2}+\frac{5}{2^2}+\frac{9}{2^3}+...+\frac{2^{100}+1}{2^{100}}=\frac{2+1}{2}+\frac{2^2+1}{2^2}+\frac{2^3+1}{2^3}+...+\frac{2^{100}+1}{2^{100}}\)
= \(\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)(100 hạng tử 1)
= \(100+\left(1-\frac{1}{2^{100}}\right)=101-\frac{1}{2^{100}}< 101\)(1)
Vì \(-\frac{1}{2^{100}}>-1\Rightarrow101-\frac{1}{2^{100}}>101-1\Rightarrow B>100\)(2)
Từ (1) và (2) => 100 < B < 101
Ta có:B = \(\frac{1}{2}+\frac{3}{2^2}+\frac{7}{2^3}+...+\frac{2^{100}-1}{2^{100}}=\frac{2-1}{2}+\frac{2^2-1}{2^2}+\frac{2^3-1}{2^3}+...+1-\frac{1}{2^{100}}\)
\(=1-\frac{1}{2}+1-\frac{1}{2^2}+1-\frac{1}{2^3}+...+1-\frac{1}{2^{100}}=100-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
=> \(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}\)
=> \(B=100-\left(1-\frac{1}{2^{100}}\right)=100-1+\frac{1}{2^{100}}=99+\frac{1}{2^{100}}>99\) (Đpcm)
\(2^{24}=(2^3)^8=8^8\)
\(3^{16}=\left(3^2\right)^8=9^8\)
vì \(8^8< 9^8\Rightarrow2^{24}< 3^{16}\)
\(2^{24}=\left(2^3\right)^8=8^8\)
\(3^{16}=\left(3^2\right)^8=9^8\)
\(8< 9\)
\(\Rightarrow8^9< 9^9\)
\(\Rightarrow2^{24}< 3^{16}\)
\(S=1+2\cdot3+3\cdot3^2+4\cdot3^3+...+101\cdot3^{100}=\left(1+3+3^2+...+3^{100}\right)+\left(3+3^2+...+3^{100}\right)+...+3^{100}\)\(S=\left(1+...+3^{100}\right)+3\left(1+...+3^{99}\right)+3^2\left(1+...+3^{98}\right)+...+3^{100}\)
\(S=1\cdot A_{100}+3\cdot A_{99}+3^2\cdot A_{98}+...+3^{100}\)
\(A_i=1+3+3^2+...+3^i=\frac{3^{i+1}-1}{2}\)
\(S=\frac{3^{101}-1}{2}+\frac{3\left(3^{100}-1\right)}{2}+\frac{3^2\left(3^{99}-1\right)}{2}+...+\frac{3^{100}\left(3-1\right)}{2}\)
\(2S=3^{101}\cdot101-\left(1+2+3+...+3^{100}\right)=101\cdot3^{101}-A_{100}=101\cdot3^{101}-\frac{3^{101}-1}{2}\)
\(2S=\frac{201\cdot3^{101}+1}{2}\Leftrightarrow S=\frac{201\cdot3^{101}+1}{4}\)
a) \(\frac{-2^2}{3}.x=\frac{-2^5}{3}\)
\(x=\frac{-2^5}{3}:\frac{-2^2}{3}\)
\(x=\frac{-2^5}{3}.\frac{-3}{2^2}\)
\(x=\frac{2^3}{1}\)
\(x=2^3\)
\(x=8\)
Vậy x = 8
b)\(\frac{-1^3}{3}.x=\frac{1}{81}\)
\(x=\frac{1}{81}:\frac{-1^3}{3}\)
\(x=\frac{1}{81}.\frac{-3}{1^3}\)
\(x=\frac{1}{3^4}.\frac{-3}{1^3}\)
\(x=\frac{1}{3^3}.\frac{-1}{1^2}\)
\(x=\frac{-1}{3^3}\)
\(x=\frac{-1}{27}\)
Vậy \(x=\frac{-1}{27}\)
Các bạn giúp mik giải bài 1 thôi cũng được vì bài 2 mik lm đc rồi. Hiện tại mik đag cần gấp giúp mik với
\(20^x\div14^x=\left(\frac{20}{14}\right)^x=\left(\frac{10}{7}\right)^x=\frac{10}{7}x\)
Mình biết kết quả là x = 1 nhưng chưa tìm ra cách giải, bạn cố gắng tìm nốt nhé!
\(A=1+2+2^2+...+2^{101}\)
\(2A=2+2^2+...+2^{102}\)
\(2A=\left(2+2^2+...+2^{102}\right)-\left(1+2+2^2+...+2^{101}\right)\)
\(A=2^{102}-1\)
\(B=5.2^{100}>2^{102}\)
Mà \(2^{102}>2^{102}-1\)
Nên B>A