\(\frac{x-7}{36}=-\frac{4}{7-x}\)

\(\frac{x-7}{36}=\frac{4}{x-7}\)

\(\Rightarrow\left(x-7\right)^2=36.4=144\)

\(\left(x-7\right)^2=12^2\)

\(\Rightarrow\orbr{\begin{cases}x-7=12\Rightarrow x=12+7=19\\x-7=-12\Rightarrow x=-12+7=-5\end{cases}}\)

(x-7)^2=36*(-4)

(x-7)^2=-144

(x-7)^2=(+-12)^2

x-7=12

x=19

4/7x - 2/3 = 1/5

=> 4/7x = 13/15

=> x = 91/60

vậy_

\(\frac{4}{7}x-\frac{2}{3}=\frac{1}{5}\)

\(\frac{4}{7}x=\frac{13}{15}\)

\(x=\frac{91}{60}\)

\(\frac{4}{5}+\frac{5}{7}:x=\frac{1}{6}\)

\(\frac{5}{7}:x=-\frac{19}{30}\)

\(x=-\frac{150}{133}\)

=.= hk tốt!!

Bài 1:

a; \(\dfrac{x}{3}\) = \(\dfrac{4}{y}\)

     \(xy\) = 12

12 = 2².3; Ư(12) = {-12; -6; -4; -3; -2; -1; 1; 2; 3; 4; 6;12}

Lập bảng ta có:

Theo bảng trên ta có các cặp \(x;y\) nguyên thỏa mãn đề bài là:

(\(x\)\(;y\)) =(-12; -1);(-6; -2);(-4; -3);(-2; -6);(-1; 12);(1; 12);(2;6);(3;4);(4;3);(6;2);(12;1)

b; \(\dfrac{x}{y}\) = \(\dfrac{2}{7}\)

    \(x\) = \(\dfrac{2}{7}\).y 

     \(x\) \(\in\)z ⇔ y ⋮ 7

   y = 7k;

   \(x\) = 2k 

Vậy \(\left\{{}\begin{matrix}x=2k\\y=7k;k\in z\end{matrix}\right.\) 

|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|

\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)

\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)

\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)

Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}

|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0

|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|

\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)

Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)

\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)

\(\Rightarrow x=165;y=20;z=25\)

c) pt <=> \(x-\frac{21}{5}=\frac{23}{7}< =>x=\frac{23}{7}+\frac{21}{5}=\frac{262}{35}\)

vậy x = \(\frac{262}{35}\) 

d) \(x-\frac{3}{4}=\frac{51}{8}< =>x=\frac{51}{8}+\frac{3}{4}=\frac{57}{8}\) 

vậy x = \(\frac{57}{8}\) 

e) pt <=> \(\frac{7}{8}:x=\frac{7}{2}< =>\frac{7}{8}.\frac{1}{x}=\frac{7}{2}< =>\frac{7}{8x}=\frac{7}{2}< =>56x=14< =>x=\frac{14}{56}=\frac{1}{4}\)

vậy x = \(\frac{1}{4}\)

a) pt <=> \(x+\frac{11}{4}=\frac{17}{3}< =>x=\frac{17}{3}-\frac{11}{4}=\frac{35}{12}\)

vậy x = \(\frac{35}{12}\)

b) pt <=> \(\frac{x.7}{2}=\frac{19}{4}< =>x=\frac{19.2}{4.7}=\frac{38}{28}=\frac{19}{14}\)

vậy x = \(\frac{19}{14}\) 

\(\frac{4}{7}\times x=\frac{1}{5}+\frac{2}{3}\)

\(\frac{4}{7}x=\frac{13}{15}\)

\(\Rightarrow x=\frac{91}{60}\)

các bài còn lại tương tự nha 

mấy cái này dễ mà toán tìm x này là cơ bản!!

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