tập trung sẽ bt làm

a)  \(\Leftrightarrow\frac{x+7}{2003}+1+\frac{x+4}{2006}+1-\frac{x-1}{2011}-1-\frac{x-5}{2015}-1=0\)

     \(\Leftrightarrow\frac{x+2010}{2003}+\frac{x+2010}{2006}-\frac{x+2010}{2011}-\frac{x+2010}{2015}=0\)

     \(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2003}+\frac{1}{2006}-\frac{1}{2011}-\frac{1}{2015}\right)=0\)

     \(\Leftrightarrow x+2010=0\) ( vì 1/2003  +  1/2006  --  1/2011  -- 1/2015   \(\ne\)0)

    \(\Leftrightarrow x=-2010\)

câu b làm tương tự (có gì không hiểu hỏi mk nha) >v<

a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\) 

\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)

\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)

Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)

Nên x + 1 = 0

=> x = -1

còn b vs c thì sao ạ

b) Sai đề à bạn đề \(\frac{x+2}{42}+\frac{x+4}{22}=\frac{x+5}{23}+\frac{x+3}{43}\)  hả đề này mk làm đc 

a)

\(\begin{array}{l}x + \left( { - \frac{1}{5}} \right) = \frac{{ - 4}}{{15}}\\x = \frac{{ - 4}}{{15}} + \frac{1}{5}\\x = \frac{{ - 4}}{{15}} + \frac{3}{{15}}\\x = \frac{{ - 1}}{{15}}\end{array}\)                 

Vậy \(x = \frac{{ - 1}}{{15}}\).

b)

\(\begin{array}{l}3,7 - x = \frac{7}{{10}}\\x = 3,7 - \frac{7}{{10}}\\x = \frac{{37}}{{10}} - \frac{7}{{10}}\\x=\frac{30}{10}\\x = 3\end{array}\)

Vậy \(x = 3\).

c)

\(\begin{array}{l}x.\frac{3}{2} = 2,4\\x.\frac{3}{2} = \frac{{12}}{5}\\x = \frac{{12}}{5}:\frac{3}{2}\\x = \frac{{12}}{5}.\frac{2}{3}\\x = \frac{8}{5}\end{array}\)

Vậy \(x = \frac{8}{5}\)       

d)

\(\begin{array}{l}3,2:x =  - \frac{6}{{11}}\\\frac{{16}}{5}:x =  - \frac{6}{{11}}\\x = \frac{{16}}{5}:\left( { - \frac{6}{{11}}} \right)\\x = \frac{{16}}{5}.\frac{{ - 11}}{6}\\x = \frac{{ - 88}}{{15}}\end{array}\)

Vậy \(x = \frac{{ - 88}}{{15}}\).

\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot....\cdot\frac{30}{62}\cdot\frac{31}{64}=2^x\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.....\cdot\frac{30}{31}\cdot\frac{31}{32}\right)=2^x\)

\(\Leftrightarrow\frac{1}{32}=2^{x+1}\)

Làm nốt.

ko làm được câu này hay câu b ib với tớ nha.khẳng định tối giải.

\(dat:\frac{x}{2}=\frac{y}{5}=k\)

x=2k   ;  y=5k

x.y=10k²

10 = 10k²

k² = 1

k  = +-1

Voi : k=1 = > x=1.2=2 ; y=5.1=5

voi : k=-1 => x=-1.2=-2 ; y=-1.5=-5

\(\frac{x}{2}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{x}{2}=\frac{4y}{12};\frac{3y}{12}=\frac{z}{5}\Rightarrow\frac{x}{8}=\frac{y}{12};\frac{y}{12}=\frac{z}{15}\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)

Ap dung tinh chat day ti so bang nhau ta co : 

\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)

Suy ra  : \(\frac{x}{8}=2\Rightarrow x=16;\frac{y}{12}=2\Rightarrow y=2.12=24;\frac{z}{15}=2\Rightarrow z=2.15=30\)

nhieu qua lam ko het

a, -3/4-2x= 4

 2x=-19/4

x=-19/8

(-2/3x-3/5).(-29/6)=2/5

-2/3x-3/5=-12/145

-2/3x=15/29

x=-45/58

c, x/2-3x/5+13/5=-7/5-7/10x

x/2-3x/5+7/10x=-7/5-13/5=-4

5x/10-6x/10+7x/10=-4

6x/10=-4

6x=-40

x=-20/3

2). Ta có: x/2=y/3 => x/8 = y/12

                y/4=z/5 => y/12 = z/15

=> x/2=y/12=z/15 và x+y-z=10

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{x}{2}\)=\(\frac{y}{12}\)=\(\frac{z}{15}\)=\(\frac{x+y-z}{2+12-15}\)=\(\frac{10}{-1}\)= -10

=> x=2.(-10)=-20

     y=12.(-10)=-120

     z=15.(-10)=-150

Vậy x=-20; y=-120;z=-150

3). Đặt \(\frac{x}{2}\)=\(\frac{y}{5}\)= k

=> x=2k

     y=5k

Ta có xy = 10

       2k.5k =10

       10. k²=10

       k²      = 10 :10=1

=> k =1; k=-1

+) k = 1

=> x=2.1=2

     y=5.1=5

+) k = -1

=> x= 2.(-1) =-2

     y=5.(-1) = -5

Vậy x=2;y=5 hoặc x=-2;y=-5

Câu 2:

Ta có \(\frac{x}{2}=\frac{y}{3}=\frac{x}{8}=\frac{y}{12}\)(1)

           \(\frac{y}{4}=\frac{z}{5}=\frac{y}{12}=\frac{z}{15}\)(2)

    Từ (1) và (2) suy ra:\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)

Áp dụng dãy tỉ số bằng nhau ta có:

    \(\Rightarrow\)\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)

\(\Rightarrow\begin{cases}\frac{x}{8}=2\\\frac{y}{12}=2\\\frac{z}{15}=2\end{cases}\)\(\Rightarrow\begin{cases}x=16\\y=24\\z=30\end{cases}\)

Vậy x=16;y=24;z=30

Câu 3:

Vì xy=10 nên x,y khác 0

    Đặt \(\frac{x}{2}=k\)\(\Rightarrow\)x=2k(1)

           \(\frac{y}{5}=k\)\(\Rightarrow\)y=5k2)

Suy ra x.y=2k.5k=10k²

      Ta có:x.y=10

Do đó k=1;-1. Thay vào (1) và (2) ta có:

x=2k(Suy ra:x=2;-2)

y=5k(Suy ra:y=5;-5)

Vậy cặp (x;y)là:(2;5)(-2;-5)