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Câu 1
4 p/s cộng thêm 1,p/s cuối trừ 4 rồi nhóm vs nhau
d/s la x= - 329
Câu 2
NHân vs 7 thành 7S rồi rút gọn là đc
Câu 1 :
a) \(\Leftrightarrow\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Leftrightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Rightarrow\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Dễ thấy \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\ne0\) \(\Rightarrow x+329=0\Rightarrow x=-329\)
1/ \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{10}\)
\(\Rightarrow2017\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2017\cdot\frac{1}{10}\)
\(\Rightarrow\frac{2017}{a+b}+\frac{2017}{b+c}+\frac{2017}{c+a}=201,7\)
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=201,7\) (vì a + b + c = 2017)
\(\Rightarrow\left(\frac{c}{a+b}+1\right)+\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)=201,7\)
\(\Rightarrow M=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+3=201,7\)
\(\Rightarrow M=198,7\)
2/
a, 3n+2 - 2n+2 + 3n + 2n
= 3n.32 + 3n - 2n.22 + 2n
= 3n.10 - 2n.5
= 3n.10 - 2n-1.10
= 10(3n - 2n-1 ) ⋮ 10
Ta có: \(M=\frac{1}{1.2}+\frac{1}{3.4}+.....+\frac{1}{37.38}\)
\(\Rightarrow M=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{37}-\frac{1}{38}\)
\(\Rightarrow M=1-\frac{1}{38}=\frac{37}{38}\)
Tương tự:
=> M/N = ..
Ta có: \(M=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{37.38}\)
\(\Rightarrow M=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{37}-\frac{1}{38}\)
\(\Rightarrow M=1-\frac{1}{38}=\frac{37}{38}\)
Câu tiếp bạn làm tương tự nhé
Và r \(\frac{M}{N}=\)...
Ta có:A= \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{37.38}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{37}-\frac{1}{38}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{37}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{38}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{37}+\frac{1}{38}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{38}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{38}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{19}\right)\)
\(=\frac{1}{20}+\frac{1}{21}+...+\frac{1}{38}\)
B=\(\frac{1}{58}\left(\frac{58}{20.38}+\frac{58}{21.37}+...+\frac{58.}{38.20}\right)\)
=\(\frac{1}{58}\left(\frac{20+38}{20.38}+\frac{21+37}{21+37}+...+\frac{38+20}{38.20}\right)\)
\(=\frac{1}{58}\left(\frac{1}{20}+\frac{1}{38}+\frac{1}{21}+\frac{1}{37}+...+\frac{1}{38}+\frac{1}{20}\right)\)
\(=\frac{1}{58}.2\left(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{38}\right)=\frac{A}{29}\)
=> \(\frac{A}{B}=29\)