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\(\frac{11}{15}+\frac{4}{15}.\frac{3}{17}+\frac{4}{15}.\frac{14}{17}=\frac{11}{15}+\frac{4}{15}.\left(\frac{3}{17}+\frac{14}{17}\right)=\frac{11}{15}+\frac{4}{15}.1\)

\(=\frac{11}{15}+\frac{4}{15}=\frac{15}{15}=1\)

4 tháng 2 2017

Phương trình 1:
\(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)
\(\Rightarrow\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}-10=0\)
\(\Rightarrow\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-4\right)=0\)
\(\Rightarrow\frac{x-85-15}{15}+\frac{x-74-26}{13}+\frac{x-67-33}{11}+\frac{x-64-36}{9}=0\)
\(\Rightarrow\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
Do \(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x = 100.

4 tháng 2 2017

Phương trình 3:
\(\frac{1909-x}{91}+\frac{1907-x}{93}+\frac{1905-x}{95}+\frac{1903-x}{97}+4=0\)
\(\Rightarrow\left(\frac{1909-x}{91}+1\right)+\left(\frac{1907-x}{93}+1\right)+\left(\frac{1905-x}{95}+1\right)+\left(\frac{1903-x}{97}+1\right)=0\)
\(\Rightarrow\frac{1909-x+91}{91}+\frac{1907-x+93}{93}+\frac{1905-x+95}{95}+\frac{1903-x+97}{97}=0\)
\(\Rightarrow\frac{2000-x}{91}+\frac{2000-x}{93}+\frac{2000-x}{95}+\frac{2000-x}{97}=0\)
\(\Rightarrow\left(2000-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
Do \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)
\(\Rightarrow2000-x=0\)
\(\Rightarrow x=2000\)
Vậy x = 2000.

15 tháng 3 2019

\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)

\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)

\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)

\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)

\(\Rightarrow-1\le x< 6\)

\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)

Bài b tương tự

17 tháng 3 2019

bạn ơi bạn giải câu b được ko. mk ko biết làm câu b

4 tháng 10 2021

yutyugubhujyikiu

6 tháng 8 2017

Biết rồi mà đi đố người khác . Chịu chị luôn em lạy 2 tay

16 tháng 5 2020

tìm n  \in  N để \(\frac{n}{n+1}\) + \(\frac{n}{n+2}\) là số tự nhiên

giúp mik với sắp thi r

30 tháng 5 2015

a) \(\frac{-8}{3}+\frac{7}{5}+\frac{-71}{15}\)\(x\) < \(\frac{-13}{7}+\frac{19}{14}+\frac{-7}{2}\)

Ta có: \(\frac{-8}{3}+\frac{7}{5}+\frac{-71}{15}\)

=\(\frac{-40}{15}+\frac{21}{15}+\frac{-71}{15}\)

=\(\frac{-90}{15}\)

=\(-6\)

Ta có: \(\frac{-13}{7}+\frac{19}{14}+\frac{-7}{2}\)

=\(\frac{-26}{14}+\frac{19}{14}+\frac{-49}{14}\)

=\(\frac{-56}{14}\)

=\(-4\)

=> \(-6\)\(x\)<\(-4\)

=> \(x=-5\)

 

b)\(\frac{5}{17}+\frac{-4}{9}+\frac{-20}{31}+\frac{12}{17}+\frac{-11}{31}\)\(\frac{x}{9}\)<\(\frac{-3}{7}+\frac{7}{15}+\frac{4}{-7}+\frac{8}{15}+\frac{2}{3}\)

Ta có: \(\frac{5}{17}+\frac{-4}{9}+\frac{-20}{31}+\frac{12}{17}+\frac{-11}{31}\)

=\(\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}+\frac{-11}{31}\right)+\frac{-4}{9}\)

=\(1+\left(-1\right)+\frac{-4}{9}\)

=\(0+\frac{-4}{9}\)

=\(\frac{-4}{9}\)

Ta có: \(\frac{-3}{7}+\frac{7}{15}+\frac{4}{-7}+\frac{8}{15}+\frac{2}{3}\)

=\(\frac{-3}{7}+\frac{7}{15}+\frac{-4}{7}+\frac{8}{15}+\frac{2}{3}\)

=\(\left(\frac{-3}{7}+\frac{-4}{7}\right)+\left(\frac{7}{15}+\frac{8}{15}\right)+\frac{2}{3}\)

=\(\left(-1\right)+1+\frac{2}{3}\)

=\(0+\frac{2}{3}\)

=\(\frac{2}{3}\)

=> \(\frac{-4}{9}\)\(\frac{x}{9}\)<\(\frac{2}{3}\)

=

=> \(\frac{-4}{9}\)<\(\frac{x}{9}\)<\(\frac{6}{9}\)

=> \(-4\)\(x\)<\(6\)

=>\(x\in\left\{-3;-2;-1;0;1;2;3;4;5\right\}\)

30 tháng 12 2018

\(\Leftrightarrow\left(\frac{x+14}{86}+1\right)+\left(\frac{x+15}{85}+1\right)+\left(\frac{x+16}{84}+1\right)+\left(\frac{x+17}{83}+1\right)+\left(\frac{166}{4}-4\right)=0\)

\(\Leftrightarrow\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)

\(\Leftrightarrow\left(x+100\right).\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)

\(\Leftrightarrow\left(x+100\right)=0\Rightarrow x=-100\left(\text{vì }\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)\ne0\)