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Vì \(\frac{9^{1006}-1}{4}\) là số chẵn nên x là số lẻ
\(\Rightarrow\left(-3\right)^x=-3^x\)
Đặt A=1-3+32-33+...-3x
3A=3-32+33-34+...+3x+1
3A+A=[3-32+33-34+...+3x+1] -[1-3+32-33+...-3x]
4A=3x+1-1
\(A=\frac{3^{x+1}-1}{4}=\frac{9^{1006}-1}{4}=\frac{\left(3^2\right)^{1006}-1}{4}=\frac{3^{1012}-1}{4}\)
=>x+1=2012
=>x=2012-1=2011
vậy x=2011
\(a.\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right).\left(2x+1\right)}=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x-1\right).\left(2x+1\right)}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x-1}-\frac{1}{2x+1}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{2x+1}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{x}{2x+1}=\frac{49}{99}\)
\(\Rightarrow99x=49.\left(2x+1\right)\)
\(\Rightarrow99x=98x+49\)
\(\Rightarrow x=49\)
Vậy : \(x=49\)
\(b.\)
\(1-3+3^2-3^3+...+\left(-3^x\right)=\frac{1-9^{1006}}{4}\)
Đặt \(A=1-3+3^2-3^3+...+\left(-3^x\right)\)
\(\Rightarrow3A=3-3^2+3^3-3^4+...+\left(-3^{x+1}\right)\)
\(\Rightarrow3A+A=1+\left(-3^{x+1}\right)\)
\(\Rightarrow4A=1+\left(-3^{x+1}\right)\)
\(\Rightarrow A=\frac{1+\left(-3^{x+1}\right)}{4}\)
\(\Rightarrow\frac{1+\left(-3^{x+1}\right)}{4}=\frac{1-9^{1006}}{4}\)
\(\Rightarrow-3^{x+1}=-9^{1006}\)
\(\Rightarrow-3^{x+1}=-3^{2012}\)
\(\Rightarrow x+1=2012\)
\(\Rightarrow x=2012-1\)
\(\Rightarrow x=2011\)
Vậy : \(x=2011\)
Bn tham khảo ở đây nhé, mk lm r`: Câu hỏi của Su su - Toán lớp 7 | Học trực tuyến
Sửa đề: \(1-3+3^2-3^3+...+\left(-3\right)^x=\dfrac{1-9^{1006}}{4}\)
Đặt \(A=\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+\left(-3\right)^3+...+\left(-3\right)^x\)
\(\Leftrightarrow-3A=\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{x+1}\)
\(\Leftrightarrow-4A=\left(-3\right)^{x+1}-1\)
hay \(A=\dfrac{-\left(-3\right)^{x+1}+1}{4}\)
Theo đề, ta có: \(-\left(-3\right)^{x+1}+1=1-9^{1006}\)
=>x+1=2012
hay x=2011
a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)
\(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)
\(\frac{2}{5}-x=-3\)
\(x=\frac{2}{5}-\left(-3\right)\)
\(x=\frac{2}{5}+3\)
\(x=\frac{3}{5}-\frac{15}{5}\)
\(x=-\frac{12}{5}\)
Vay \(x=-\frac{12}{5}\)
b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)
\(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)
\(-3+\frac{3}{x}=\frac{-25}{12}\)
\(\frac{3}{x}=\frac{-25}{12}+3\)
\(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)
\(\frac{3}{x}=\frac{5}{6}\)
\(\frac{18}{6x}=\frac{5x}{6x}\)
Đèn dây , bạn tự làm tiếp nhé , de rồi chứ