a) Ta có : \(2.21=3.14\)\(\Rightarrow\)lập được các tỉ lệ thức là : \(\frac{2}{3}=\frac{14}{21};\frac{2}{14}=\frac{3}{21};\frac{3}{2}=\frac{21}{14};\frac{14}{2}=\frac{21}{3}\)

b) Ta có : \(-0,24.1,61=-0,46.0,84\)\(\Rightarrow\)lập được các tỉ lệ thức là : \(\frac{-0,24}{-0,46}=\frac{0,84}{1,61};\frac{-0,24}{0,84}=\frac{-0,46}{1,61};\frac{-0,46}{-0,24}=\frac{1,16}{0,84};\frac{0,84}{-0,24}=\frac{1,16}{-0,46}\)

\(\frac{3^6.45^5-15^{13}.5^{-9}}{27^4.25^3+45^6}=\frac{3^6.\left(3^2.5\right)^5-\left(3.5\right)^{13}.5^{-9}}{\left(3^3\right)^4.\left(5^2\right)^3+\left(3^2.5\right)^6}=\frac{3^{16}.5^5-3^{13}.5^4}{3^{12}.5^6+3^{12}.5^6}\)

\(\frac{3^{13}.5^4\left(3^3.5-1\right)}{2.3^{12}.5^6}=\frac{3.44}{2.5^2}=\frac{3.22}{5^2}=\frac{66}{25}\)

Tìm x

\(\left(152\frac{2}{4}-148\frac{3}{8}\right):0,2=x:0,3\)

\(x:0,3=\left(152\frac{1}{2}-148\frac{3}{8}\right):0,2\)

\(x:0,3=\left(152\frac{4}{8}-148\frac{3}{8}\right):0,2\)

\(x:0,3=4\frac{1}{8}:0,2\)

\(x:\frac{3}{10}=\frac{33}{8}:\frac{1}{5}\)

\(x:\frac{3}{10}=\frac{165}{8}\)

\(x=\frac{165}{8}.\frac{3}{10}\)

\(x=\frac{99}{16}\)

Vậy \(x=\frac{99}{16}\).

\(\frac{x^2}{6}=\frac{24}{25}\)

\(\Rightarrow25x^2=24.6=144\)

\(\Rightarrow x^2=\frac{144}{25}\)

\(\Rightarrow x=\frac{12}{5}\)

Vậy \(x=\frac{12}{5}\).

\(\frac{x-1}{x+5}=\frac{6}{7}\)

\(\Rightarrow7\left(x-1\right)=6\left(x+5\right)\)

\(7x-7=6x+30\)

\(7x-6x=7+30\)

x = 37

Vậy x = 18.

\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)

\(\Rightarrow\left(x-2\right)\left(x+7\right)=\left(x-1\right)\left(x+4\right)\)

x² - 2x + 7x - 14 = x² - x + 4x - 4

(x² - x²) - (2x - 7x - x + 4x) = 14 - 4

- (2x - 7x - x + 4x) = 10

- (-2x) = 10

2x = 10

x = 5

Vậy x = 5.

\(5x=7y\Leftrightarrow\frac{x}{7}=\frac{y}{5}\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có :

\(\frac{y}{5}=\frac{x}{7}=\frac{y-x}{5-7}=\frac{18}{-2}=-9\)

Suy ra :

+) \(\frac{y}{5}=-9\Leftrightarrow y=-45\)

+) \(\frac{x}{7}=-9\Leftrightarrow x=-63\)

Theo đề bài , ta có : 5x=7y  <=> x/7 = y/5 và y-x = 18

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

x/7 = y/5 = y-x/5-7 = 18/ -2 = -9

Suy ra :

x/7 = -9 <=> x = -63

y/5 = -9 <=> y = -45

Vậy , x= - 63 ; y = -45

ĐK \(2018x\ge0\Rightarrow x\ge0\)

Khi đó \(x+\frac{1}{2018}\ge0;x+\frac{2}{2018}\ge0;...;x+\frac{2017}{2018}\ge0\)

Ta có \(\left|x+\frac{1}{2018}\right|+\left|x+\frac{2}{2018}\right|+...+\left|x+\frac{2017}{2018}\right|=2018x\)(Vế trái có 2017 hạng tử)

<=> \(x+\frac{1}{2018}+x+\frac{2}{2018}+...+x+\frac{2017}{2018}=2018x\)

<=> \(\left(x+x+...x\right)+\left(\frac{1}{2018}+\frac{2}{2018}+...+\frac{2017}{2018}\right)=2018x\)

           2017 hạng tử x                   2017 số hạng

=> \(2017x+\frac{1+2+...+2017}{2018}=2018x\)

=> \(x=\frac{2017.\left(2017+1\right):2}{2018}\)

\(\Rightarrow x=\frac{2017}{2}=1008,5\)(tm)

Vậy x = 1008,5

Vì \(\left|x+\frac{1}{2018}\right|\ge0\forall x\)

    \(\left|x+\frac{2}{2018}\right|\ge0\forall x\)

    \(\left|x+\frac{3}{2018}\right|\ge0\forall x\)

     .......................................

    \(\left|x+\frac{2017}{2018}\right|\ge0\forall x\)

\(\Rightarrow\)\(\left|x+\frac{1}{2018}\right|+\left|x+\frac{2}{2018}\right|+\left|x+\frac{3}{2018}\right|+...+\left|x+\frac{2017}{2018}\right|\ge0\forall x\)

mà \(\left|x+\frac{1}{2018}\right|+\left|x+\frac{2}{2018}\right|+\left|x+\frac{3}{2018}\right|+...+\left|x+\frac{2017}{2018}\right|=2018x\)

 \(\Rightarrow\)\(2018x\ge0\forall x\)\(\Rightarrow\)\(x\ge0\)

 \(\Rightarrow\)\(x+\frac{1}{2018}+x+\frac{2}{2018}+x+\frac{3}{2018}+...+x+\frac{2017}{2018}=2018x\)

\(\Leftrightarrow\)\(2017x+\frac{1}{2018}+\frac{2}{2018}+\frac{3}{2018}+...+\frac{2017}{2018}=2018x\)

\(\Leftrightarrow\)\(\frac{1+2+3+...+2017}{2018}=x\)

\(\Leftrightarrow\)\(x=\frac{\left[\left(2017+1\right).2017\right]:2}{2018}\)

\(\Leftrightarrow\)\(x=\frac{2035153}{2018}\)

\(\Leftrightarrow\)\(x=\frac{2017}{2}=1008,5\)

Vậy \(x=1008,5\)

a. 2/3 = 14/21, 3/2 = 21/14, 3/21 = 2/14, 21/3 = 14/2

b. Tương tự với a. nhé

a) Nhận xét: \(x-1< x+4\)

=> \(\hept{\begin{cases}x-1< 0\\x+4>0\end{cases}}\Rightarrow-4< x< 1\)

b) Nếu: \(\hept{\begin{cases}x>0\\4-x>0\end{cases}}\Rightarrow0< x< 4\)

Nếu: \(\hept{\begin{cases}x< 0\\4-x< 0\end{cases}}\Rightarrow∄x\)

c) Nếu: \(\hept{\begin{cases}1-3x>0\\8+x< 0\end{cases}}\Rightarrow x< -8\)

Nếu: \(\hept{\begin{cases}1-3x< 0\\8+x>0\end{cases}\Rightarrow}x>\frac{1}{3}\)

d) Nếu: \(\hept{\begin{cases}2x+6>0\\4-x>0\end{cases}}\Rightarrow-3< x< 4\)

Nếu: \(\hept{\begin{cases}2x+6< 0\\4-x< 0\end{cases}}\Rightarrow∄x\)

a)(x-1).(x+4) < 0 => x-1 và x+4 khác dấu => x-1 < 0 , x+4> 0 ( x-1<x+4) => -1>x>-4
các câu b,c tương tự
d)\(\frac{2x+6}{4-x}=-\frac{-6-2x}{4-x}=-\frac{-14+\left(8-2x\right)}{4-x}=\frac{14}{4-x}-2\)
\(\Rightarrow\frac{14}{4-x}>2\Rightarrow x< 2\)
 

a, \(2-5x< 8\Leftrightarrow-5x< 6\Leftrightarrow x>-\frac{6}{5}\)

b, \(13-6x>x-4\Leftrightarrow7x< 17\Leftrightarrow x< \frac{17}{7}\)