(2²+3).(x-5)+14=5²+124:2²

(4+3).(x-5)+14=25+124:4

7.(x-5)+14=25+31

7.(x-5)+14=56

7.(x-5)=42

x-5=6

x=11

Ta có: \(\left(2^2+3\right)\left(x-5\right)+14=5^2+124:2^2\)

\(\Leftrightarrow7\left(x-5\right)=25+31-14=42\)

\(\Leftrightarrow\left(x-5\right)=6\)

hay \(x=11\)

a)    \(\left(3^2-2\right).\left(x-12+35\right)\)\(=\)\(5^2+279:5\)

\(7.\left(x-12+35\right)=80,8\)

\(x-12+35=80,8:7\)

\(x-12+35=\frac{404}{35}\)

\(x-12=\frac{404}{35}-35\)

\(x-12=\frac{-821}{35}\)

\(x=\frac{-821}{35}+12\)

\(x=\frac{-401}{35}\)

b)    \(260:\left(x+4\right)\)\(=\)\(5\left(2^3+5\right)-3\left(3^2+2^2\right)\)

     \(260:\left(x+4\right)=26\)

     \(x+4=260:26\)

    \(x+4=10\)

   \(x=10-4\)

   \(x=6\)

c)   \(7^{x-3}=343\)

    \(7^x:7^3=343\)

    \(7^x=343.7^3\)

    \(7^x=117649\)

vì \(117649=7^6\Rightarrow x=6\)

d)    \(\left(x-3\right)^{2017}=\left(x-3\right)^{2016}\)

\(\Rightarrow\hept{\begin{cases}x-3=1\\x-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=4\\x=3\end{cases}}}\)

e)     \(\left(2^3+3\right).\left(x-5\right)+14\)\(=\)\(5^2+124:2^2\)

      \(\left(2^3+3\right).\left(x-5\right)+14=56\)

       \(\left(2^3+3\right).\left(x-5\right)=56-14\)

       \(\left(2^3+3\right).\left(x-5\right)=42\)

        \(x-5=42:\left(2^3+3\right)\)

        \(x-5=\frac{42}{11}\)

      \(x=\frac{42}{11}+5\)

     \(x=\frac{97}{11}\)

a) 7(x-5)+14=56

<=>7(x-5)=42

<=>x-5=6

<=>x=11

b)9(x+1)-3=15

<=>9(x+1)=18

<=>x+1=2

<=>x=1

c)12(x+5)-36=48

<=>12(x+5)=84

<=>x+5=7

<=>x=2

d)5(x+14)=145

<=>x+14=29

<=>x=15

1, \(x\div y\div z=3\div8\div5\)

\(\Rightarrow\frac{x}{3}=\frac{y}{8}=\frac{z}{5}\)

\(\Rightarrow\frac{3x}{9}=\frac{y}{8}=\frac{2z}{10}\)

\(\Rightarrow\frac{3x+y-2z}{9+8-10}=\frac{x}{3}=\frac{y}{8}=\frac{z}{10}=\frac{14}{7}=2\)

\(\Rightarrow\hept{\begin{cases}x=2\cdot3=6\\y=2\cdot8=16\\z=2\cdot5=10\end{cases}}\)

vậy_

các phần sau tương tự

1, \(x:y:z=3:8:5;3x+y-2z=14\)

\(\Rightarrow\frac{x}{3}=\frac{y}{8}=\frac{z}{5}\)

\(\Rightarrow\frac{3x}{9}=\frac{y}{8}=\frac{2z}{10}\)

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

\(\frac{3x}{9}=\frac{y}{8}=\frac{2z}{10}=\frac{3x+y-2z}{9+8-10}=\frac{14}{7}=2\)

\(\Rightarrow\hept{\begin{cases}\frac{3x}{9}=2\Rightarrow3x=18\Rightarrow x=6\\\frac{y}{8}=2\Rightarrow y=16\\\frac{2z}{10}=2\Rightarrow2z=20\Rightarrow z=10\end{cases}}\)

Vậy....

2, \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3};4x-3y-2z=36\)

\(\Rightarrow\frac{4x}{4}=\frac{3y}{6}=\frac{2z}{6}\)

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

\(\frac{4x}{4}=\frac{3y}{6}=\frac{2z}{6}=\frac{4x-3y-2z}{4-6-6}=\frac{36}{-8}=\frac{-36}{8}=\frac{-9}{4}\)

Làm tương tự để tìm x;y;z

3, \(x:y:z=3:5:\left(-2\right);5x-y+3z=124\)

\(\Rightarrow\frac{x}{3}=\frac{y}{5}=\frac{z}{\left(-2\right)}\)

\(\Rightarrow\frac{5x}{15}=\frac{y}{5}=\frac{3z}{-6}\)

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

\(\frac{5x}{15}=\frac{y}{5}=\frac{3z}{-6}=\frac{5x-y+3z}{15-5+\left(-6\right)}=\frac{124}{4}=31\)

\(\Rightarrow\hept{\begin{cases}\frac{5x}{15}=31\Rightarrow5x=465\Rightarrow x=93\\\frac{y}{5}=31\Rightarrow y=155\\\frac{3z}{-6}=31\Rightarrow3z=-186\Rightarrow z=-62\end{cases}}\)

Vậy .....

a.

Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=3k\\z=4k\end{matrix}\right.\)

Thế vào \(2x+y-z=81\)

\(\Rightarrow2.5k+3k-4k=81\)

\(\Rightarrow9k=81\)

\(\Rightarrow k=9\)

\(\Rightarrow\left\{{}\begin{matrix}x=5k=45\\y=3k=27\\z=4k=36\end{matrix}\right.\)

b.

Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{2}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\\z=2k\end{matrix}\right.\)

Thế vào \(5x-y+3z=124\)

\(\Rightarrow5.3k-5k+3.2k=124\)

\(\Rightarrow16k=124\)

\(\Rightarrow k=\dfrac{31}{4}\) \(\Rightarrow\left\{{}\begin{matrix}x=3k=\dfrac{93}{4}\\y=5k=\dfrac{155}{4}\\z=2k=\dfrac{31}{2}\end{matrix}\right.\)

c.

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

Thế vào \(xyz=810\)

\(\Rightarrow2k.3k.5k=810\)

\(\Rightarrow k^3=27\)

\(\Rightarrow k=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k=6\\y=3k=9\\z=5k=15\end{matrix}\right.\)

d.

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=6k\end{matrix}\right.\)

Thế vào \(x^2y^2z^2=288^2\)

\(\Rightarrow\left(2k\right)^2.\left(3k\right)^2.\left(6k\right)^2=288^2\)

\(\Rightarrow\left(k^2\right)^3=64\)

\(\Rightarrow k^2=4\)

\(\Rightarrow k=\pm2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k=4\\y=3k=6\\z=6k=12\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=2k=-4\\y=3k=-6\\z=6k=-12\end{matrix}\right.\)

|7 - \(\dfrac{3}{4}\)\(x\)| - \(\dfrac{3}{2}\) = \(\dfrac{1}{\dfrac{1}{2}}\)

|7 - \(\dfrac{3}{4}x\)|  - \(\dfrac{3}{2}\) = 2

|7 - \(\dfrac{3}{4}\)\(x\)| = 2 + \(\dfrac{3}{2}\)

|7 - \(\dfrac{3}{4}x\)| = \(\dfrac{7}{2}\)

\(\left[{}\begin{matrix}7-\dfrac{3}{4}x=\dfrac{7}{2}\\7-\dfrac{3}{4}x=-\dfrac{7}{2}\end{matrix}\right.\) 

\(\left[{}\begin{matrix}\dfrac{3}{4}x=7-\dfrac{7}{2}\\\dfrac{3}{4}=7+\dfrac{7}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{7}{2}\\\dfrac{3}{4}x=\dfrac{21}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{14}{3}\\x=14\end{matrix}\right.\)

 5  - |\(x-3\)| = 5

       |\(x-3\)| = 5 - 5

      |\(x-3\)| = 0

      \(x-3\) = 0 

      \(x\) = 3

|\(\dfrac{7}{2}\) - 42| + 14 = 14 ( vô lý xem lại đề bài nhé em)

\(\dfrac{2}{3}\) - |\(x+2\)| = 2

     |\(x+2\)| = \(\dfrac{2}{3}\) - 2

    |\(x+2\)| = - \(\dfrac{4}{3}\)  

   |\(x+2\)| ≥ 0 ∀ \(x\)

Vậy \(x\in\) \(\varnothing\)