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

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

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

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

        \(x-5=42:\left(2^3+3\right)\)

        \(x-5=\frac{42}{11}\)

      \(x=\frac{42}{11}+5\)

     \(x=\frac{97}{11}\)

Lời giải:
$(2^2+3)(x-5)+14-5^2+124=2^2$

$7(x-5)+113=4$

$7(x-5)=4-113=-109$

$x-5=-109:7=\frac{-109}{7}$

$x=5+\frac{-109}{7}=\frac{-74}{7}$

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 .....

\(2^5.x+3^3.x=6.100-10.205\)

\(x.\left(2^5+3^3\right)=600-2050\)

\(x.59=-1450\)

\(x=-1450:59\)

\(x=-\frac{1450}{59}\)

|x| - 7 = 11

<=> |x| = 18

<=> x = 18

hoặc x = -18

Vậy...