a)\(\frac{-5}{6}\).\(\frac{120}{25}\)<x<\(\frac{-7}{15}\).\(\frac{9}{14}\)

       -4                 <x<\(\frac{-3}{10}\)

\(\frac{-40}{10}\)<      x   <\(\frac{-3}{10}\)=>x E {-39:-38:-37:.....:-4}

b)\(\left(\frac{-5}{3}\right)^3\)<x<\(\frac{-24}{35}.\frac{-5}{6}\)

\(\frac{-875}{189}< x< \frac{108}{189}\)

=> x  E {\(\frac{-874}{189},\frac{-873}{189},......,\frac{107}{189}\)}

1)

a)

\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)

\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)

\(\frac{-20}{5}< x< \frac{-3}{10}\)

\(\frac{-40}{10}< x< \frac{-3}{10}\)

\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)

\(\left(\frac{-5}{3}\right)^3< x< \frac{-24}{35}.\frac{-5}{6}\)

\(\frac{25}{3}< x< \frac{-4}{7}.\frac{1}{1}\)

\(\frac{-25}{3}< x< \frac{-4}{7}\)

\(\frac{-175}{21}< x< \frac{-12}{21}\)

\(\Rightarrow Z\in\left\{-13;-14;-15;-16;...;-174\right\}\)

2)

a)

\(\left(\frac{9}{10}-\frac{15}{16}\right).\left(\frac{5}{12}-\frac{11}{15}-\frac{7}{20}\right)\)

\(=\left(\frac{72}{80}-\frac{75}{80}\right).\left(\frac{25}{60}-\frac{44}{60}-\frac{21}{60}\right)\)

\(=\frac{-3}{80}.\frac{-40}{60}\)

\(=\frac{-1}{-2}.\frac{-1}{-20}\)

\(=\frac{1}{40}\)

Bài 1: Tìm x biết:

1) x +\(\frac{7}{12}\)= \(\frac{17}{18}\)- \(\frac{1}{9}\)                                                                               2) \(\frac{29}{30}\)- (\(\frac{13}{23}\)+ x) = \(\frac{7}{69}\)

 x +\(\frac{7}{12}\)= \(\frac{15}{18}\)                                                                                                                \(\frac{13}{23}\)+ x = \(\frac{29}{30}\)- \(\frac{7}{69}\)

             x = \(\frac{15}{18}\)- \(\frac{7}{12}\)                                                                                                    \(\frac{13}{23}\)+ x = \(\frac{199}{230}\)

             x = \(\frac{1}{4}\)                                                                                                                                   x = \(\frac{3}{10}\)

1 

Ez lắm =)

Bài 1:

Với mọi gt \(x,y\in Q\) ta luôn có: 

\(x\le\left|x\right|\) và \(-x\le\left|x\right|\) 

\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)

Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)

Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)

Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)

Dấu "=" xảy ra khi: \(xy\ge0\)

Bài 3: 

Ta có: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\) (vì a + b + c = 1)

Do đó: \(\left(x+y+z\right)^2=\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\) (vì a² + b² + c² = 1)

Vậy: (x + y + z)² = x² + y² + z²

a) x = 99/20

b) x = 7

c) x = 2

( chỉ lm đc đến đó thui nk )