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3\(\frac{x}{5}\) = \(\frac{3x+7}{5}\)
Mà 3\(\frac{x}{5}\) =\(\frac{15+x}{5}\)
Suy ra: \(\frac{15+x}{5}\) =\(\frac{3x+7}{5}\)
15+x=3x+7
15+x-7=3x+7-7
8+x=3x
Suy ra: 8=2x
x=8:2
x=4
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Suy ra:(3x-5)*3=4x*2
9x-15=8x
15=9x-8x
15=x
Vậy x=15(tick nha)
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\(A=2-\left(\frac{2^3}{25}+\frac{2^3}{63}+...+\frac{2^3}{255}+\frac{2^3}{323}\right)\)
\(=2-4.\left(\frac{2}{35}+\frac{2}{63}+...+\frac{2}{255}+\frac{2}{323}\right)\)
\(=2-4.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{15.17}+\frac{2}{17.19}\right)\)
\(=2-4.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}\right)\)
\(=2-4.\left(\frac{1}{5}-\frac{1}{19}\right)\)
\(=2-4.\frac{14}{95}=2-\frac{56}{95}=\frac{134}{95}\)
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Ta có :\(A=10\left(\frac{1}{1.2}+\frac{5}{2.3}+...+\frac{89}{9.10}\right)\)
\(\Rightarrow10\left(\frac{1}{2}+\frac{5}{6}+...+\frac{89}{90}\right)\)
\(\Rightarrow10\left[\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\right]\)
\(\Rightarrow10\left[9-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\right]\)
\(\Rightarrow10\left[9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\right]\)
\(\Rightarrow10\left[9-\left(1-\frac{1}{10}\right)\right]\)
\(\Rightarrow\)\(10\left[9-\frac{9}{10}\right]\)
\(\Rightarrow10.\frac{81}{10}\)
\(\Rightarrow A=81\)
Làm rờ đó sai thì thôi nha.
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Ta có:
\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)=\left(-4\right)+1+1+1+1\)
\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}=0\)
\(\Rightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\right)=0\)
Mà \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\ne0\)
\(\Rightarrow x+329=0\)
\(\Rightarrow x=-329\)
\(\Leftrightarrow\dfrac{x+15}{5}=\dfrac{3x+7}{5}\)
=>3x+7=x+15
=>2x=8
hay x=4