\(\frac{x-10}{30}+\frac{x-14}{43}+\frac{x-5}{95}+\frac{x-148}{8}=0\)0

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27 tháng 7 2016

\(\frac{x-10}{30}+\frac{x-14}{43}+\frac{x-5}{95}+\frac{x-148}{8}=0\)

\(\Rightarrow\frac{x-10}{30}-3+\frac{x-14}{43}-2+\frac{x-5}{95}-1+\frac{x-148}{8}+6=0\)

\(\Rightarrow\frac{x-100}{30}+\frac{x-100}{43}+\frac{x-100}{95}+\frac{x-100}{8}=0\)

\(\Rightarrow\left(x-100\right)\left(\frac{1}{30}+\frac{1}{43}+\frac{1}{95}+\frac{1}{8}\right)=0\)

mà \(\frac{1}{30}+\frac{1}{43}+\frac{1}{95}+\frac{1}{8}\ne0\)

=> x - 100 = 0 => x = 100

                                                       Vậy x = 100

22 tháng 7 2018

ko biết

4 tháng 6 2016

Tìm x, biết:

3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x2;5;10;17)

2(x1)(x3) +5(x3)(x8) +12(x8)(x20) 1x20 =34 (x1;3;8;20)

x+110 +2+111 x+112 =x+113 +x+114 

x1030 +x1443 +x595 +x1488 =0

4 tháng 6 2016

Trả lời luôn à bạn

3 tháng 8 2019

Ta có : 10 ^ 28 = 10 ..... 0 ( 28 chữ số 0 ) chia hết cho 8​

8 chia hết cho 8

Nên 10 ^ 28 + 8 chia hết cho 8

Ta có : 10 ^ 28 + 8 = 99....9 ( 28 chữ số 9 ) + 1 + 8

=> 10 ^ 28 + 8 = 99....9 ( 28 chữ số 9 ) + 9 chia hết cho 9

Vì ƯCLN ( 8,9 ) = 1

Nên 10 ^ 28 + 8 chia hết cho 72

11 tháng 3 2018

b, |5x-3| >= 7

=> 5x-3 < = -7 hoặc 5x-3 >= 7

=> x < = -4/5 hoặc x >= 2

Vậy ..........

Tk mk nha

8 tháng 8 2017

1) \(\left(x-1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}\right)=0\)

mà \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}\ne0\)

\(\Rightarrow x-1=0\Leftrightarrow x=1\)

2) \(\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-5}{95}-1=\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)

\(\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{95}=\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)

\(\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\right)=\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)

x - 100 = 1

x = 101

10 tháng 4 2019

d) \(\frac{x}{-9}=\left(\frac{2}{6}\right)^2\)

\(\Rightarrow\frac{x}{-9}=\frac{2}{6}.\frac{2}{6}\)

\(\Rightarrow\frac{x}{-9}=\frac{4}{36}\)

\(\Rightarrow\frac{x}{-9}=\frac{1}{9}\)

\(\Rightarrow\frac{-x}{9}=\frac{1}{9}\)

\(\Rightarrow-x=1\)

\(\Rightarrow x=1\)

e) \(\frac{a}{b}+\frac{3}{6}=0\)

\(\Rightarrow\frac{a}{b}=0-\frac{3}{6}\)

\(\Rightarrow\frac{a}{b}=0-\frac{1}{2}\)

\(\Rightarrow\frac{a}{b}=\frac{-1}{2}\)

\(\Rightarrow a=-1;b=2\)

10 tháng 2 2019

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

\(\Rightarrow\frac{x-2}{4}=\frac{9}{x-2}\)

\(\Rightarrow\left(x-2\right)^2=36\)

\(\Rightarrow\left(x-2\right)^2=\left(\pm6\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=6\\x-2=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-4\end{cases}}}\)

\(\frac{x}{15}=\frac{3}{y}\)

\(\Rightarrow xy=45\)

\(\Rightarrow x;y\inƯ\left(45\right)=\left\{\pm1;\pm3;\pm5;\pm9;\pm15;\pm45\right\}\)

Xét bảng 

x1(loại)-13(loại)-35(loại)-545-45(loại)15-15(loại)9-9(loại)
y45(loại)-4515(loại)-159(loại)-91-1(loại)3-3(loại)5-5(loại)

Vậy.......................................

d;Áp dụng tích chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{4}=\frac{y}{3}=\frac{x+y}{4+3}=\frac{14}{7}=2\)

\(\Rightarrow x=4.2=8\)

     \(y=3.2=6\)

22 tháng 8 2020

1) Ta có : \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)

=> x + 1 = 0

=> x = - 1

b) \(\frac{x+4}{2006}+\frac{x+3}{2007}=\frac{x+2}{2008}+\frac{x+1}{2009}\)

=> \(\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+1}{2009}+1\right)\)

=> \(\frac{x+2010}{2006}+\frac{x+2010}{2007}=\frac{x+2010}{2008}+\frac{x+2010}{2009}\)

=> \(\left(x+2010\right)\left(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)

Vì \(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\ne0\)

=> x + 2010 = 0

=> x = -2010

c) \(\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}\)

\(\Rightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)=\left(\frac{x+1975}{75}-1\right)+\left(\frac{x+1969}{69}-1\right)\)

=> \(\frac{x+1900}{45}+\frac{x+1900}{54}=\frac{x+1900}{75}+\frac{x+1900}{69}\)

=> \(\left(x+1900\right)\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0\)

=> \(x+1900=0\left(\text{Vì }\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\ne0\right)\)

=> x = -1900

d) \(\frac{x+2008}{10}+\frac{x+2010}{9}=\frac{x+2012}{8}+\frac{x+2014}{7}\)

=> \(\left(\frac{x+2008}{10}+2\right)+\left(\frac{x+2010}{9}+2\right)=\left(\frac{x+2012}{8}+2\right)+\left(\frac{x+2014}{7}+2\right)\)

=> \(\frac{x+2028}{10}+\frac{x+2028}{9}=\frac{x+2028}{8}+\frac{x+2028}{7}\)

=> \(\left(x+2028\right)\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\right)=0\)

=> x + 2028 = 0 \(\left(\text{Vì }\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\ne0\right)\)

=> x = -2028

22 tháng 8 2020

1) Ta có: \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

        \(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

        \(\Leftrightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

  + TH1\(x+1=0\)\(\Leftrightarrow\)\(x=-1\)

  + TH2\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}=0\)

      Vì \(\hept{\begin{cases}\frac{1}{10}>\frac{1}{13}\\\frac{1}{11}>\frac{1}{14}\\\frac{1}{12}>0\end{cases}}\)\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}>\frac{1}{13}+\frac{1}{14}\)

            \(\Rightarrow\)\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0\)

             mà \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}=0\)

             \(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-1\)

2) Ta có: \(\frac{x+4}{2006}+\frac{x+3}{2007}=\frac{x+2}{2008}+\frac{x+1}{2009}\)

        \(\Leftrightarrow\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+3}{2007}+1\right)-\left(\frac{x+2}{2008}+1\right)-\left(\frac{x+1}{2009}+1\right)=0\)

        \(\Leftrightarrow\frac{x+2010}{2006}+\frac{x+2010}{2007}-\frac{x+2010}{2008}-\frac{x+2010}{2009}=0\)

        \(\Leftrightarrow\left(x+2010\right).\left(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)

  + TH1\(x+2010=0\)\(\Leftrightarrow\)\(x=-2010\)

  + TH2\(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}=0\)

      Vì \(\hept{\begin{cases}\frac{1}{2006}>\frac{1}{2008}\\\frac{1}{2007}>\frac{1}{2009}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{2006}+\frac{1}{2007}>\frac{1}{2008}+\frac{1}{2009}\)

              \(\Rightarrow\)\(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}>0\)

               mà \(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}=0\)

               \(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-2010\)

3) Ta có: \(\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}\)

        \(\Leftrightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)-\left(\frac{x+1975}{75}-1\right)-\left(\frac{x+1969}{69}-1\right)=0\)

        \(\Leftrightarrow\frac{x+1900}{45}+\frac{x+1900}{54}-\frac{x+1900}{75}-\frac{x+1900}{69}=0\)

       \(\Leftrightarrow\left(x+1900\right).\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0\)

  

  + TH1\(x+1900=0\)\(\Leftrightarrow\)\(x=-1900\)

  + TH2\(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}=0\)

      Vì \(\hept{\begin{cases}\frac{1}{45}>\frac{1}{75}\\\frac{1}{54}>\frac{1}{69}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{45}+\frac{1}{54}>\frac{1}{75}+\frac{1}{69}\)

              \(\Rightarrow\)\(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}>0\)

               mà \(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}=0\)

               \(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-1900\)

4) Ta có: \(\frac{x-99}{5}+\frac{x-97}{7}=\frac{x-95}{9}+\frac{x-93}{11}\)

         \(\Leftrightarrow\left(\frac{x-99}{5}-1\right)+\left(\frac{x-97}{7}-1\right)-\left(\frac{x-95}{9}-1\right)-\left(\frac{x-93}{11}-1\right)=0\)

         \(\Leftrightarrow\frac{x-104}{5}+\frac{x-104}{7}-\frac{x-104}{9}-\frac{x-104}{11}=0\)

         \(\Leftrightarrow\left(x-104\right).\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)=0\)

  

  + TH1\(x-104=0\)\(\Leftrightarrow\)\(x=104\)

  + TH2\(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}=0\)

      Vì \(\hept{\begin{cases}\frac{1}{5}>\frac{1}{7}\\\frac{1}{9}>\frac{1}{11}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{5}+\frac{1}{7}>\frac{1}{9}+\frac{1}{11}\)

              \(\Rightarrow\)\(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}>0\)

               mà \(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}=0\)

               \(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=104\)

5) Ta có: \(\frac{x+2008}{10}+\frac{x+2010}{9}=\frac{x+2012}{8}+\frac{x+2014}{7}\)

        \(\Leftrightarrow\left(\frac{x+2008}{10}+2\right)+\left(\frac{x+2010}{9}+2\right)-\left(\frac{x+2012}{8}+2\right)-\left(\frac{x+2014}{7}+2\right)=0\)

        \(\Leftrightarrow\frac{x+2028}{10}+\frac{x+2028}{9}-\frac{x+2028}{8}-\frac{x+2028}{7}=0\)

        \(\Leftrightarrow\left(x+2028\right).\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\right)=0\)

    + TH1\(x+2028=0\)\(\Leftrightarrow\)\(x=-2028\)

    + TH2\(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}=0\)

      Vì \(\hept{\begin{cases}\frac{1}{10}< \frac{1}{8}\\\frac{1}{9}< \frac{1}{7}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{9}< \frac{1}{8}+\frac{1}{7}\)

              \(\Rightarrow\)\(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}< 0\)

               mà \(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}=0\)

               \(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-2028\)

Chúc bn hok tốt nha