Bài 1: 

a: x/-2=-18/x

=>x²=36

=>x=6 hoặc x=-6

b: x/2+x/5=17/10

=>7/10x=17/10

hay x=17/7

Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}\)

nên \(\dfrac{x}{6}=\dfrac{y}{9}\left(1\right)\)

Ta có: \(\dfrac{x}{3}=\dfrac{z}{5}\)

nên \(\dfrac{x}{6}=\dfrac{z}{10}\left(2\right)\)

Từ (1) và (2) suy ra \(\dfrac{x}{6}=\dfrac{y}{9}=\dfrac{z}{10}\)

Đặt \(\dfrac{x}{6}=\dfrac{y}{9}=\dfrac{z}{10}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=6k\\y=9k\\z=10k\end{matrix}\right.\)

Ta có: \(x^2+y^2+z^2=21\)

\(\Leftrightarrow k^2=\dfrac{21}{217}\)

Trường hợp 1: \(k=\dfrac{\sqrt{93}}{31}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=6k=\dfrac{6\sqrt{93}}{31}\\y=9k=\dfrac{9\sqrt{93}}{31}\\z=10k=\dfrac{10\sqrt{93}}{31}\end{matrix}\right.\)

Trường hợp 2: \(k=-\dfrac{\sqrt{93}}{31}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=6k=\dfrac{-6\sqrt{93}}{31}\\y=9k=\dfrac{-9\sqrt{93}}{31}\\z=10k=\dfrac{-10\sqrt{93}}{31}\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3};\dfrac{x}{3}=\dfrac{z}{5}\Rightarrow\dfrac{x}{6}=\dfrac{y}{9}=\dfrac{z}{10}=\dfrac{x^2}{36}=\dfrac{y^2}{81}=\dfrac{z^2}{100}\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\dfrac{x}{6}=\dfrac{y}{9}=\dfrac{z}{10}=\dfrac{x^2}{36}=\dfrac{y^2}{81}=\dfrac{z^2}{100}=\dfrac{x^2+y^2+z^2}{217}=\dfrac{21}{217}=\dfrac{3}{31}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{31}\cdot6=\dfrac{18}{31}\\y=\dfrac{3}{31}\cdot9=\dfrac{27}{31}\\z=\dfrac{3}{31}\cdot10=\dfrac{30}{31}\end{matrix}\right.\)

a, |3x-2|=2x

Th1: 3x-2 > hoặc = 0 => x> hoặc = 2/3

=> 3x-2=2x

<=> x=2 (thỏa mãn)

Th2: 3x-2< 0 => x<2/3

=> -3x+2 =2x

<=> -5x=-2

<=> x=2/5 (tm)

Vậy S={2; 2/5}

b, |2x-3| = -x+21

Th1: 2x-3> hoặc = 0 =>x > hoặc =3/2

=> 2x-3=-x+21

<=> 3x=24

<=> x=8 (tm)

Th2: 2x-3<0 => x<3/2

=> -2x+3=-x+21

<=> x=18 (loại)

Vậy S={8}

Chúc bạn thi tốt nhaa <33

<=> 5x + 4y = 21

=> x= 9 , y =12 

\(4x=3y\Rightarrow x=\dfrac{3}{4}y\)

\(x+y=21\Rightarrow\dfrac{3}{4}y+y=21\Rightarrow\dfrac{7}{4}y=21\Rightarrow y=12\)

\(\Rightarrow x=9\)

nhanh lên các bạn ơi

D= 1+4+4²+4³+...+4⁵⁸ +4⁵⁹ ⋮ 21

D= (1+4+4²)+(4³⁺4⁴+4⁵)+...(4⁵⁷+4⁵⁸+4⁵⁹)

D= (1+4+4²)+4³.(1+4+4²)+...+4⁵⁷.(1+4+4²)

D= 21+4³.21+....+4⁵⁷.21 ⋮ 21

=>D= 1+4+4²+4³+...+4⁵⁸ +4⁵⁹ ⋮ 21

\(\dfrac{3}{7}=\dfrac{9}{21}\)

7

Ta có : 3x - 7/3 - 2x - 1/2 = 7 .

=>       x ( 3 - 2 ) - ( 7/3 + 1/2 ) = 7 .

=>       x - ( 14/6 + 3/6 ) = 7 .

=>            x - 17/6         = 7 .

=>                x                = 7 + 17/6 .

=>                x                 = 59/6 .

vậy x = 59/6 .

\(3x-\frac{7}{3}-2x-\frac{1}{2}=7\)

\(\Leftrightarrow\left(3x-2x\right)-\left(\frac{7}{3}+\frac{1}{2}\right)=7\)

\(\Leftrightarrow x-\frac{17}{6}=7\)

\(\Leftrightarrow x=7+\frac{17}{6}\)

\(\Leftrightarrow x=\frac{59}{6}\)

\(\frac{3x-7}{3}\)  -    \(\frac{2x-1}{2}\) \(=\) \(7\)

\(\frac{3x-7}{3}\) = \(7\) +  \(\frac{2x-1}{2}\)

\(\frac{3x-7}{3}\)  = \(\frac{14}{2}\) +   \(\frac{2x-1}{2}\)

\(\frac{3x-7}{3}\)  =   \(\frac{2x+13}{2}\)

<=> \(\left(3x-7\right)\)  x  \(2\) =    \(\left(2x+13\right)\)x   \(3\)

<=> \(6x-14=6x+39\) <=> 6x - 6x = 14 + 39

                                                       <=> 0x = 53 => x ko tốn tại

vậy x ko tồn tại

a: =>2x>-6

hay x>-3

e: =>(5-x)/x<0

=>0<x<5

h: \(\Leftrightarrow\dfrac{x+5-x-3}{x+3}< 0\)

\(\Leftrightarrow x+3< 0\)

hay x<-3

g: \(\Leftrightarrow\dfrac{2x+7}{x+4}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{7}{2}\\x< -4\end{matrix}\right.\)