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đặt B=1/3 + 1/3^2 + 1/3^3 +.....+ 1/3^99
=>1/3B=1/3^2 + 1/3^3 +.....+ 1/3^100
=>1/3B-B=1/3^2 + 1/3^3 +.....+ 1/3^100-1/3-1/3^2-1/3^3-...-1/3^100
=>-2/3=1/3^100-1/3
=>B=(1/3^100-1/3):(-2/3)<1/2 (vì kết quả ra số âm )
làm bừa thui,ai tích mình mình tích lại
Số số hạng là :
Có số cặp là :
50 : 2 = 25 ( cặp )
Mỗi cặp có giá trị là :
99 - 97 = 2
Tổng dãy trên là :
25 x 2 = 50
Đáp số : 50
a) \(\frac{x}{4}=\frac{16}{x^2}\)\(=>x^3=16.4\)\(=>x^3=64\)\(=>x=4\)
b) \(\frac{4}{3}:\frac{4}{5}=\frac{2}{3}.\left(\frac{1}{10}.x\right)\)\(=>\frac{4}{3}.\frac{5}{4}=\frac{2}{3}\left(\frac{1}{10}x\right)\)\(=>\frac{5}{3}=\frac{2}{3}\left(\frac{1}{10}x\right)\)\(=>\frac{5}{3}:\frac{2}{3}=\frac{1}{10}x\)\(=>\frac{5}{3}.\frac{3}{2}=\frac{1}{10}x\)\(=>\frac{5}{2}=\frac{1}{10}x\)\(=>x=\frac{5}{2}:\frac{1}{10}\)\(=>x=\frac{5}{2}.10\)\(=>x=25\)
vậy x=25
1.
a) \(\frac{x}{4}=\frac{16}{x^2}\)
\(\Rightarrow x^3=64\)
\(\Rightarrow x^3=4^3\)
\(\Rightarrow x=4\)
b) \(1\frac{1}{3}:0,8=\frac{2}{3}.\left(0,1.x\right)\)
\(\frac{5}{3}=\frac{2}{3}.\frac{x}{10}\)
\(\frac{x}{10}=\frac{5}{2}\)
\(\Rightarrow x=\frac{5.10}{2}=25\)
2.
\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\)
\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)\)
\(2A=1-\frac{1}{3^{99}}< 1\)
\(\Rightarrow A=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)
\(-\frac{13}{3}\cdot\left(\frac{1}{2}-\frac{1}{6}\right)< x< -\frac{2}{3}\cdot\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Leftrightarrow-\frac{13}{3}\cdot\left(\frac{3}{6}-\frac{1}{6}\right)< x< -\frac{2}{3}\cdot\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)
\(\Leftrightarrow-\frac{13}{3}\cdot\frac{1}{3}< x< -\frac{2}{3}\cdot\left(-\frac{11}{12}\right)\)
\(\Leftrightarrow-\frac{13}{9}< x< \frac{11}{18}\)
\(\Leftrightarrow-\frac{26}{18}< x< \frac{11}{18}\)
\(\Leftrightarrow x\in\left\{-25;-24;-23;...;8;9;10\right\}\)
Vậy ...
\(3\frac{1}{3}\div2\frac{2}{5}-1< x< 7\frac{2}{3}\cdot\frac{3}{7}+\frac{5}{7}\)
\(\frac{25}{18}-1< x< \frac{23}{7}+\frac{5}{7}\)
\(\frac{7}{18}< x< \frac{28}{7}\)
\(\frac{49}{126}< x< \frac{504}{126}\)
\(\Rightarrow x=\left(\frac{50}{126};\frac{51}{126};\frac{52}{126};......;\frac{503}{126}\right)\)
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1 +\frac{x+349}{5}-4=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
\(\Leftrightarrow\)\(x+329=0\) (vì 1/327 + 1/326 + 1/325 + 1/324 + 1/5 khác 0 )
\(\Leftrightarrow\)\(x=-329\)
Bài 1 :
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Vì \(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)
\(\Rightarrow\)\(x+329=0\)
\(\Rightarrow\)\(x=-329\)
Vậy \(x=-329\)