3) \(-12+2x-9+x=0\\ -21+3x=0\\ 3x=21\\ x=7\)

4)
\(11+\left(15-x\right)=1\)
\(15-x=1-11\)
\(15-x=-10\)
\(x=15-\left(-10\right)\)
\(x=25\)

5) 
\(4-\left(27-3\right)=x-\left(13-4\right)\)
\(4-24=x-9\)
\(x-9=-20\)
\(x=-20+9\)
\(x=-11\)
 

\(3.-12+\left(2x-9\right)+x=0.\)

\(\Leftrightarrow-12+2x-9+x=0.\Leftrightarrow3x=21.\Leftrightarrow x=7.\)

Vậy \(x=7.\)

\(4.11+\left(15-x\right)=1.\Leftrightarrow11+15-x=1.\Leftrightarrow26-x=1.\Leftrightarrow x=25.\)

Vậy \(x=25.\)

\(5.4-\left(27-3\right)=x-\left(13-4\right).\Leftrightarrow4-24=x-9.\Leftrightarrow-20=x-9.\Leftrightarrow x=-11.\)

Vậy \(x=-11.\)

\(6.8-\left(x-10\right)=23-\left(-4+12\right).\Leftrightarrow8-x+10=23-8.\Leftrightarrow18-x=15.\Leftrightarrow x=3.\)

Vậy \(x=3.\)

\(7.105-5\left(10-5x\right)=-20.\Leftrightarrow105-50+25x=-20.\Leftrightarrow25x=-75.\Leftrightarrow x=-3.\)

Vậy \(x=-3.\)

\(8.\left(x-1\right)\left(8-2x\right)\left(3x+123\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\8-2x=0.\\3x+123=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=4.\\x=-41.\end{matrix}\right.\)

Vậy \(x\in\left\{1;4;-41\right\}.\)

\(9.\left(x^2-25\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0.\\x+5=0.\\x+10=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5.\\x=-5.\\x=-10.\end{matrix}\right.\)

Vậy \(x\in\left\{5;-5;-10\right\}.\)

\(10.x\left(x^2+5\right)=0.\Leftrightarrow x=0.\)

a, Ta có: \(142-\left(12x+30\right)=10^{5-3}\)

\(=>142-12x-30=10^2\)

\(112-12x=100\)

\(=>12x=112-100=12=>x=1\)

Vậy số cần tìm là 1;

b, Ta có: \(=>\left(5x+3^4\right)=6^9:6^8.3^4\)

\(=>5x+3^4=6.3^4=>5x=6.3^4-3^4\)

\(=>5x=5.3^4=>x=3^4=81\)

Vậy x=81;

CHÚC BẠN HỌC TỐT.......

Ta có: \(71.2-6(2x+5)=10^5:10^3\)

\(\Rightarrow142-12x-30=10^2\)

\(\Rightarrow142-30-10^2=12x\)

\(\Rightarrow142-30-100=12x\)

\(\Rightarrow12=12x\)

\(\Rightarrow x=\dfrac{12}{12}\)

\(\Rightarrow x=1\)

Vậy \(x=1\)

a) 71.2-6(2x+5)=10⁵:10³

<=>  71.2-6(2x+5)=10²

142-6(2x+5)=100

6(2x+5)=142-100

6(2x+5)=42

2x+5=7

2x=7-5

2x=2

x=2:2

x=1

địt mẹ mầy

a,\(4x\left(2x+3\right)-x\left(8x-1\right)=5\left(x+2\right)\)

\(< =>8x^2+12x-8x^2+x=5x+10\)

\(< =>13x=5x+10< =>8x=10\)

\(< =>x=\frac{10}{8}=\frac{5}{4}\)

b, \(\left(3x-5\right)\left(3x+5\right)-x\left(9x-1\right)=4\)

\(< =>9x^2-25-9x^2+x=4\)

\(< =>x=4+29=33\)

c,\(3-4x\left(25-2x\right)=8x^2+x-300\)

\(< =>3-100x+8x^2=8x^2+x-300\)

\(< =>x+100x=3+300\)

\(< =>101x=303< =>x=\frac{303}{101}=3\)

d,\(2\left(1-\frac{3x}{5}\right)-\frac{2+3x}{10}=7-\frac{3\left(2x+1\right)}{4}\)

\(< =>2-\frac{6x}{5}-\frac{2+3x}{10}=7-\frac{6x+3}{4}\)

\(< =>-\frac{24x}{20}-\frac{4+6x}{20}+\frac{30x+15}{20}=5\)

\(< =>\frac{30x-6x-24x+15-4}{20}=5\)

\(< =>\frac{11}{5}=5< =>11=25\)(vo li)

e,  \(\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x+2}{5}-5\)

\(< =>\frac{25x+10}{30}-\frac{80x-10}{30}=\frac{24x+12}{30}-5\)

\(< =>\frac{24x+12+80x-10-25x-10}{30}=5\)

\(< =>79x-8=150\)

\(< =>79x=150+8< =>x=\frac{158}{79}\)