S = 3 x 3 / 10 + 3 x 3 / 40 + 3 x 3 / 88 + ... + 3 x 3 / 343
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\(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{1}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{\left(x+2\right)\left(x+5\right)}\right)=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+..+\dfrac{3}{\left(x+2\right)\left(x+5\right)}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{x+2}-\dfrac{1}{x+5}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+5}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{2}-\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{20}\)
\(\Rightarrow x+5=20\)
\(\Rightarrow x=20-5\)
\(\Rightarrow x=15\)
làm lại
ta có : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
=>\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
=>\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x+2}-\frac{1}{x+5}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{x+5}=\frac{9}{20}\)
=>\(\frac{1}{x+5}=\frac{1}{2}-\frac{9}{20}\)
=>\(\frac{1}{x+5}=\frac{1}{20}\)
=>\(x+5=20\)
=>\(x=20-5\)
=>\(x=15\)
ta có : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+....+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
=>\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{3}{20}\)
=>\(3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+....+\frac{1}{\left(x+2\right)\left(x+3\right)}\right)=3.\frac{3}{20}\)
=>\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(x+2\right)\left(x+3\right)}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x+2}-\frac{1}{x+3}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{x+3}=\frac{9}{20}\)
=>\(\frac{1}{x+3}=\frac{1}{2}-\frac{9}{20}\)
=>\(\frac{1}{x+3}=\frac{1}{20}\)
=>\(x+3=20\)
=>\(x=20-3\)
=>\(x=17\)
a) \(\frac{14}{15}:\frac{9}{10}=x:\frac{3}{7}\Rightarrow\frac{28}{27}=x:\frac{3}{7}\Rightarrow x=\frac{4}{9}\)
b) \(\left(x-\frac{4}{7}\right)^3=343\Rightarrow\left(x-\frac{4}{7}\right)^3=7^3\Rightarrow x-\frac{4}{7}=7\Rightarrow x=\frac{53}{7}\)
c) \(x^5=x^3\Leftrightarrow\hept{\begin{cases}x=1\\x=0\end{cases}}\)
e) \(\left(x-1\right)^4=16\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^4=2^4\\\left(x-1\right)^4=\left(-2\right)^4\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x-1=2\\x-1=\left(-2\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
\(A=3^x+3^{x+1}+...+3^{x+100}\)
=>\(3A=3^{x+1}+3^{x+2}+...+3^{x+101}\)
=>\(2A=3^{x+101}-3^x\)
=>\(A=\dfrac{3^{x+101}-3^x}{2}\)
=>\(3^{x+101}-3^x=3^{105}-3^4\)
=>x=4
6: =>x=9/10+1/5=9/10+2/10=11/10
7: =>x=3/8-5/12=36/96-40/96=-1/24
8: =>x=7/6-5/4=14/12-15/12=-1/12
9: =>x=1/35+2/7=1/35+10/35=11/35
10: =>x=2/7-7/10=20/70-49/70=-29/70