a) \(\dfrac{-x}{4}=\dfrac{-9}{x}\)

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

\(\Rightarrow x^2=36\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

Vậy: \(x\in\left\{6;-6\right\}\)

b) \(\dfrac{5}{9}+\dfrac{x}{-1}=-\dfrac{1}{3}\)

\(\Rightarrow\dfrac{5}{9}+\dfrac{-9x}{9}=\dfrac{-3}{9}\)

\(\Rightarrow5-9x=-3\)

\(\Rightarrow-9x=-8\)

\(\Rightarrow x=\dfrac{8}{9}\)

Vậy: \(x=\dfrac{8}{9}\)

c) \(x:3\dfrac{1}{5}=1\dfrac{1}{2}\)

\(\Rightarrow x:\dfrac{16}{5}=\dfrac{3}{2}\)

\(\Rightarrow x=\dfrac{3}{2}.\dfrac{16}{5}\)

\(\Rightarrow x=\dfrac{24}{5}\)

Vậy: \(x=\dfrac{24}{5}\)

d) \(\dfrac{3x-1}{-5}=\dfrac{-5}{3x-1}\)

\(\Rightarrow\left(3x-1\right)^2=25\)

\(\Rightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{2;-\dfrac{4}{3}\right\}\)

a) Ta có \(-4\dfrac{3}{5}\cdot2\dfrac{4}{3}=-\dfrac{23}{5}\cdot\dfrac{10}{3}=-\dfrac{46}{3}\) và \(-2\dfrac{3}{5}\div1\dfrac{6}{15}=-\dfrac{13}{5}\div\dfrac{7}{5}=-\dfrac{13}{7}\)

Do đó \(-\dfrac{46}{3}< x< -\dfrac{13}{7}\)

Lại có \(-\dfrac{46}{3}\le-15\) và \(-\dfrac{13}{7}\ge-2\)

Suy ra \(-15\le x\le-2\), x ϵ Z

b) Ta có \(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=-\dfrac{13}{3}\cdot\dfrac{1}{3}=-\dfrac{13}{9}\) và \(-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{2}{3}\cdot\dfrac{-11}{12}=\dfrac{11}{18}\)

Do đó \(-\dfrac{13}{9}< x< \dfrac{11}{18}\)

Lại có \(-\dfrac{13}{9}\le-1\) và \(\dfrac{11}{18}\ge0\)

Suy ra \(-1\le x\le0\), x ϵ Z

b, -4\(\dfrac{1}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{6}\)) < \(x\) < - \(\dfrac{2}{3}\).(\(\dfrac{1}{3}\) - \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\))

   - \(\dfrac{13}{3}\).\(\dfrac{1}{3}\) < \(x\) < - \(\dfrac{2}{3}\).(-\(\dfrac{11}{12}\))

    - \(\dfrac{13}{9}\) < \(x\) < \(\dfrac{11}{18}\)

     \(x\) \(\in\) { -1; 0; 1}

a, -4\(\dfrac{3}{5}\).2\(\dfrac{4}{3}\) < \(x\) < -2\(\dfrac{3}{5}\): 1\(\dfrac{6}{15}\)

  - \(\dfrac{23}{5}\).\(\dfrac{10}{3}\) <   \(x\)   < - \(\dfrac{13}{5}\): \(\dfrac{21}{15}\)

   -  \(\dfrac{46}{3}\)     <  \(x\) < - \(\dfrac{13}{7}\) 

          \(x\) \(\in\) {-15; -14;-13;..; -2}

`a, x-2=7/15`

`=>x=7/15 +2`

`=>x= 7/15+ 30/15`

`=>x= 37/15`

`b, 9/20 +x=2/5 +3/20`

`=> 9/20 +x=8/20 +3/20`

`=> 9/20 +x=11/20`

`=>x=11/20-9/20`

`=>x= 2/20=1/10`

x = \(\dfrac{7}{15}\) + 2 = \(\dfrac{37}{15}\) 

x = \(\dfrac{2}{5}\) + \(\dfrac{3}{20}\) - \(\dfrac{9}{20}\) = \(\dfrac{1}{10}\)

Bài 4:

a) \(\dfrac{2.7.13}{26.35}=\dfrac{2.7.13}{13.2.7.5}=\dfrac{1}{5}\)

b) \(\dfrac{23.5-23}{4-27}=\dfrac{23.\left(5-1\right)}{-23}=\dfrac{23.4}{-23}=-4\)

c) \(\dfrac{2130-15}{3550-25}=\dfrac{2115}{3525}=\dfrac{3}{5}\)

Bài 1

a) \(\dfrac{x}{15}=\dfrac{17}{51}\)

51x=17.15

51x=255

⇒x=5

b) \(\dfrac{-7}{y}=\dfrac{12}{24}\)

-7.24=24y

-168=12y

⇒y=-14

\(\dfrac{15-x}{14}=\dfrac{5}{7}\\ \left(15-x\right)\times7=5\times14\\ \left(15-x\right)\times7=70\\ 15-x=70:7\\ 15-x=10\\ x=15-10\\ x=5\)

__

\(\dfrac{x}{9}=\dfrac{17}{3}\\ x\times3=17\times9\\ x\times3=153\\ x=153:3\\ x=51\)

__

\(15-\dfrac{x}{2}=\dfrac{3}{7}\\ \dfrac{x}{2}=15-\dfrac{3}{7}\\ \dfrac{x}{2}=\dfrac{102}{7}\\ x\times7=102\times2\\ x\times7=204\\ x=204:7\\ x=\dfrac{204}{7}\)

???

\(a.\)

\(\dfrac{x}{2}=\dfrac{y}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{35}{7}=5\)

\(\Rightarrow x=5\cdot2=10\\ y=5\cdot5=25\)

\(b.\)

\(\dfrac{x+2}{y+10}=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{x+2}{1}=\dfrac{y+10}{5}\)

\(\Leftrightarrow\dfrac{3x+6}{3}=\dfrac{y+10}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\Leftrightarrow\dfrac{3x+6}{3}=\dfrac{y+10}{5}=\dfrac{y+10-3x-6}{5-3}=\dfrac{2-4}{2}=-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+6=-3\\y+10=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-15\end{matrix}\right.\)

\(c.\)

\(\dfrac{x}{4}=\dfrac{y}{5}\)

\(\Leftrightarrow\dfrac{2x}{8}=\dfrac{y}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\dfrac{2x}{8}=\dfrac{y}{5}=\dfrac{2x-y}{8-5}=\dfrac{15}{3}=5\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=5\cdot8\\y=5\cdot5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=25\end{matrix}\right.\)

a) Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}\)

mà x+y=35

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{35}{7}=5\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{2}=5\\\dfrac{y}{5}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=25\end{matrix}\right.\)

Vậy: (x,y)=(10;25)

b) Ta có: \(\dfrac{x+2}{y+10}=\dfrac{1}{5}\)

nên \(\dfrac{x+2}{1}=\dfrac{y+10}{5}\)

hay \(\dfrac{3x+6}{3}=\dfrac{y+10}{5}\)

mà y-3x=2 

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{3x+6}{3}=\dfrac{y+10}{5}=\dfrac{y-3x+10-6}{5-3}=\dfrac{2+4}{2}=3\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{3x+6}{3}=3\\\dfrac{y+10}{5}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+6=9\\y+10=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=3\\y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)

Vậy: (x,y)=(1;5)

c) Ta có: \(\dfrac{x}{4}=\dfrac{y}{5}\)

nên \(\dfrac{2x}{8}=\dfrac{y}{5}\)

mà 2x-y=15

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{2x}{8}=\dfrac{y}{5}=\dfrac{2x-y}{8-5}=\dfrac{15}{3}=5\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{4}=5\\\dfrac{y}{5}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=25\end{matrix}\right.\)

Vậy: (x,y)=(20;25)

a: Ta có: \(\dfrac{1}{4}:x=3\dfrac{4}{5}:40\dfrac{8}{15}\)

\(\Leftrightarrow x=\dfrac{1}{4}\cdot\dfrac{\dfrac{608}{15}}{3+\dfrac{4}{5}}\)

\(\Leftrightarrow x=\dfrac{152}{15}:\dfrac{19}{5}=\dfrac{8}{3}\)

b: Ta có: \(\left(x+1\right):\dfrac{5}{6}=\dfrac{20}{3}\)

\(\Leftrightarrow x+1=\dfrac{50}{9}\)

hay \(x=\dfrac{41}{9}\)

c: Ta có: \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

\(\Leftrightarrow x^2-1=63\)

\(\Leftrightarrow x^2=64\)

hay \(x\in\left\{8;-8\right\}\)

c. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\) 

    \(7.9=\left(x-1\right).\left(x+1\right)\) 

    \(63=x^2-1\) 

             \(x^2=63+1\) 

             \(x^2=64\) 

             \(x^2=8^2\)

             \(x=8\)          

a. Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{x}{2}=\frac{y}{5}=\frac{x+y}{2+5}=\frac{-21}{7}=-3$

$\Rightarrow x=2(-3)=-6; y=5(-3)=-15$

b. Áp dụng tính chất dãy tỉ số bằng nhau:

$7x=3y=\frac{x}{\frac{1}{7}}=\frac{y}{\frac{1}{3}}=\frac{x-y}{\frac{1}{7}-\frac{1}{3}}=\frac{16}{\frac{-4}{21}}=-84$

$\Rightarrow x=(-84):7=-12; y=-84:3=-28$

c. $\frac{x}{y}=\frac{5}{9}\Rightarrow \frac{x}{5}=\frac{y}{9}$

Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{x}{5}=\frac{y}{9}=\frac{3x}{15}=\frac{2y}{18}=\frac{3x+2y}{15+18}=\frac{66}{33}=2$

$\Rightarrow x=2.5=10; y=9.2=18$

d. Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{x}{15}=\frac{y}{7}=\frac{2y}{14}=\frac{x-2y}{15-14}=\frac{16}{1}=16$

$\Rightarrow x=16.15=240; y=7.16=112$

e.

Đặt $\frac{x}{5}=\frac{y}{2}=k\Rightarrow x=5k ; y=2k$

Khi đó: $xy=5k.2k=10k^2=1000\Rightarrow k^2=100\Rightarrow k=\pm 10$

Với $k=10$ thì $x=5k=50; y=2k=20$

Với $k=-10$ thì $x=5k=-50; y=2k=-20$

Bài 2:

Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{x}{13}-\frac{y}{7}-\frac{z}{5}=\frac{x-y-z}{13-7-5}=\frac{6}{1}=6$

$\Rightarrow x=13.6=78; y=7.6=42; z=5.6=30$