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\(C=\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\)
\(C=5.\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{99}}\right)\)
đặt \(A=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{99}}\)
\(4A=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{98}}\)
\(4A-A=\left(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{98}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{99}}\right)\)
\(3A=1-\frac{1}{4^{99}}\)< 1
\(\Rightarrow A=\frac{1-\frac{1}{4^{99}}}{3}< \frac{1}{3}\)
suy ra \(C=5.\left(\frac{1-\frac{1}{4^{99}}}{3}\right)< 5.\frac{1}{3}=\frac{5}{3}\)
Ta có :
\(C=\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\)
\(\Rightarrow4.C=4\left(\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\right)\)
\(\Rightarrow4C=5+\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{98}}\)
\(\Rightarrow4C-C=5+\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{98}}-\frac{5}{4}-\frac{5}{4^2}-...-\frac{5}{4^{99}}\)
\(\Rightarrow3C=\)\(5-\frac{5}{4^{99}}=5\left(1-\frac{1}{4^{99}}\right)\)
\(\Rightarrow C=\frac{5}{3}.\left(1-\frac{1}{4^{99}}\right)< \frac{5}{3}\left(đpcm\right)\)
Kb vs k cho mình nhé!
A = \(\frac{5}{4}.\left(5-\frac{4}{3}\right).\frac{1}{11}\)
A = \(\frac{5}{44}\left(5-\frac{4}{3}\right)\)
A = \(\frac{25}{44}-\frac{5}{33}\)
A = \(\frac{25.3}{4.11.3}-\frac{5.4}{3.11.4}\)
A = \(\frac{75}{132}-\frac{20}{132}\)
A = \(\frac{55}{132}\)
B = \(\frac{3}{4}:\left(-12\right).\left(-\frac{2}{3}\right)\)
B = \(\frac{-1}{16}.\left(-\frac{2}{3}\right)\)
B = \(\frac{1}{24}\)
C = \(\frac{5}{4}:\left(-15\right).\left(-\frac{2}{5}\right)\)
C = \(\frac{-1}{12}.\left(-\frac{2}{5}\right)\)
C = \(\frac{1}{30}\)
D = \(\left(-3\right).\left(\frac{2}{3}-\frac{5}{4}\right):\left(-7\right)\)
D = \(\frac{3}{7}.\left(\frac{8}{12}-\frac{15}{12}\right)\)
D = \(\frac{3}{7}.\left(-\frac{7}{12}\right)\)
D = \(\frac{-3}{12}\)
D = \(\frac{-1}{4}\)
Câu b) tạm thời ko bít làm =.=
Bài 1 :
\(d)\) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2x\)
\(\Leftrightarrow\)\(\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=2x\)
\(\Leftrightarrow\)\(\frac{4^6}{3^6}.\frac{6^6}{2^6}=2x\)
\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{2^6.3^6}{2^6}=2x\)
\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{3^6}{1}=2x\)
\(\Leftrightarrow\)\(2^{12}=2x\)
\(\Leftrightarrow\)\(x=\frac{2^{12}}{2}\)
\(\Leftrightarrow\)\(x=2^{11}\)
\(\Leftrightarrow\)\(x=2048\)
Vậy \(x=2048\)
Chúc bạn học tốt ~
Bài 1 :
\(a)\) Ta có :
\(4+\frac{x}{7+y}=\frac{4}{7}\)
\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{4}{7}-4\)
\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{-24}{7}\)
\(\Leftrightarrow\)\(\frac{x}{-24}=\frac{7+y}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{-24}=\frac{7+y}{7}=\frac{x+7+y}{-24+7}=\frac{22+7}{-17}=\frac{29}{-17}=\frac{-29}{17}\)
Do đó :
\(\frac{x}{-24}=\frac{-29}{17}\)\(\Rightarrow\)\(x=\frac{-29}{17}.\left(-24\right)=\frac{696}{17}\)
\(\frac{7+y}{7}=\frac{-29}{17}\)\(\Rightarrow\)\(y=\frac{-29}{17}.7-7=\frac{-322}{17}\)
Vậy \(x=\frac{696}{17}\) và \(y=\frac{-322}{17}\)
Chúc bạn học tốt ~
Nguyễn Huy Hải, papa Dung và mama Triều hả?=.=