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<=> x\(-10\left(\frac{1}{11x13}+\frac{1}{13x15}+...+\frac{1}{53x55}\right)\)) =\(\frac{3}{11}\)
x\(-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
X-10\(\left(\frac{1}{11}-\frac{1}{55}\right)\)=\(\frac{3}{11}\)
X-\(\frac{40}{55}\)=\(\frac{3}{11}\)
X=\(\frac{3}{11}+\frac{40}{55}=\frac{15+40}{55}=\frac{55}{55}=1\)
\(\frac{4}{x+1}=\frac{2}{3x+1}\Leftrightarrow4\left(3x+1\right)=2\left(x+1\right)\Leftrightarrow12x+4=2x+2\)
\(\Leftrightarrow12x-2x=2-4\Leftrightarrow10x=-2\Leftrightarrow\frac{-1}{5}\)
Vậy x=-1/5
\(\frac{4}{x+1}=\frac{2}{3x+1}\left(x\ne-1;x\ne-\frac{1}{3}\right)\)
=> \(4\left(3x+1\right)=2\left(x+1\right)\)
=> \(12x+4=2x+2\)
=> \(12x-2x=2-4\)
=> \(10x=-2\)
=> \(5x=-1\)(chia cho 5)
=> \(x=-\frac{1}{5}\left(tm\right)\)
Vậy \(x=-\frac{1}{5}\)
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{97.100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{97}-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\frac{99}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{33}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow x=\frac{0,33\times100}{0,33}=100\)
.........................
= \(\frac{1}{2}\). ( \(\frac{2}{1.3}\) + \(\frac{2}{3.5}\) + \(\frac{2}{5.7}\) ... + \(\frac{2}{x.\left(x+2\right)}\) )
= \(\frac{1}{2}\) . ( 1 - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{5}\) + \(\frac{1}{5}\) - \(\frac{1}{7}\) + ... + \(\frac{1}{x}\)- \(\frac{1}{x+2}\) )
= ................
Bạn tự làm tiếp nhé ! Chúc bạn học tốt :)
a; 3:\(\frac{2x}{5}\)= 1:0.001
3:\(\frac{2x}{5}\)=1000
\(\frac{2x}{5}\)=1000:3
\(\frac{2x}{5}\)=0.003
2x=0.003.5
2x=0.015
x=0.015:2
x=7.5
a) \(\frac{4}{7}=\frac{12}{21}=\frac{28}{49}=\frac{52}{91}\)
b) \(\frac{4}{5}=\frac{12}{15}=\frac{16}{20}=\frac{8\cdot\left(16-15\right)}{10}\)
=> x,y,y phù hợp vs từng vị trí
hok tốt
\(\frac{x}{13}=\frac{-15}{39}=\frac{20}{3y}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{x}{13}=\frac{-15}{39}\\\frac{20}{3y}=\frac{-15}{39}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\left(-\frac{15}{39}\right)\cdot13\\-45y=780\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-5\\y=\frac{780}{-45}=-\frac{52}{3}\end{cases}}\)
Vậy x = -5 và y = \(\frac{-52}{3}\)
\(\frac{\times}{13}=\frac{-15}{39};\frac{-15}{39}=\frac{20}{3y}\)
\(\Rightarrow\frac{3\times}{39}=\frac{-15}{39};\frac{-60}{156}=\frac{-60}{-9y}\)
\(\Rightarrow3\times=-15;-9y=156\)
\(\Rightarrow\times=-5;y=\frac{-52}{3}\)