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 \(x\in\left\{5;-5;-10\right\}.\)

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

\(TH_1:x\ge0\Leftrightarrow x^3\ge0\Leftrightarrow VT>0\left(loại\right)\)

\(TH_2:x< 0\)

Với \(x=-1\Leftrightarrow VT=4\cdot9\cdot14\cdot29>0\left(loại\right)\)

Với \(x=-2\Leftrightarrow VT=-3\cdot2\cdot7\cdot23< 0\left(nhận\right)\)

Với \(x=-3\Leftrightarrow VT=-22\left(-17\right)\left(-12\right)\cdot3< 0\left(nhận\right)\)

Với \(x< -4\Leftrightarrow x^3< -64\Leftrightarrow x^3+5< x^3+10< x^3+15< x^3+30< 0\)

Do đó cả 4 thừa số trong tích đều âm nên tích này luôn dương

Vậy \(x\in\left\{-2;-3\right\}\)

a: =7/12+6/36

=7/12+2/12=9/12=3/4

b: =8/9-2/3=8/9-6/9=2/9

c: \(=\dfrac{7\cdot2+5\cdot3}{18}\cdot\dfrac{3}{5}=\dfrac{29}{18}\cdot\dfrac{3}{5}=\dfrac{87}{90}=\dfrac{29}{30}\)

a) \(\dfrac{7}{12}+\dfrac{3}{4}\times\dfrac{2}{9}=\dfrac{7}{12}+\dfrac{6}{36}=\dfrac{7}{12}+\dfrac{2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

b) \(\dfrac{8}{9}-\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{8}{9}-\dfrac{4}{15}\times\dfrac{5}{2}=\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{8}{9}-\dfrac{6}{9}=\dfrac{2}{9}\)

c) \(\left(\dfrac{7}{9}+\dfrac{5}{6}\right)\times\dfrac{3}{5}=\dfrac{29}{18}\times\dfrac{3}{5}=\dfrac{87}{90}=\dfrac{29}{30}\)

a) x = 4                

b) x = 3     

c) x = 2                

d) x = 1. 

a: =>x-15=20

hay x=35

c: =>12x=360

hay x=30

a: P(x)=6x^3-4x^2+4x-2

Q(x)=-5x^3-10x^2+6x+11

M(x)=x^3-14x^2+10x+9

b: \(C\left(x\right)=7x^4-4x^3-6x+9+3x^4-7x^3-5x^2-9x+12\)

=10x^4-11x^3-5x^2-15x+21

a) (x - 15) × 7 - 270 : 45 = 169

(x - 15) × 7 - 6 = 169

(x - 15) × 7 = 169 + 6

(x - 15) × 7 = 175

x - 15 = 175 : 7

x - 15 = 25

x = 25 + 15

x = 40

b) [(4x + 28) × 3 + 55] : 5 = 35

(4x + 28) × 3 + 55 = 35 × 5

(4x + 28) × 3 + 55 = 175

(4x + 28) × 3 = 175 - 55

(4x + 28) × 3 = 120

4x + 28 = 120 : 3

4x + 28 = 40

4x = 40 - 28

4x = 12

x = 12 : 4

x = 3

c) (455 × x : 2 × 6) : 5 = 31

455 × x : 2 × 6 = 31 × 5

455 × x : 2 × 6 = 155

x × 455 : 2 × 6 = 155

x × 1365 = 155

x = 155 : 1365

x = 31/273

d) 128 × x - 12 × x - 16 × x = 520800

(128 - 12 - 16) × x = 520800

100 × x = 520800

x = 520800 : 100

x = 5208

e) (x × 0,25 + 2022) × 2023 = (50 + 2022) × 2023

(x × 0,25 + 2022) × 2023 = 2072 × 2023

(x × 0,25 + 2022) × 2023 = 4191656

x × 0,25 + 2022 = 4191656 : 2023

x × 0,25 + 2022 = 2072

x × 0,25 = 2072 - 2022

x × 0,25 = 50

x = 50 : 0,25

x = 200

f) 4 × x + 100 = x + 280

4 × x - x = 280 - 100

(4 - 1) × x = 180

3 × x = 180

x = 180 : 3

x = 60

g) (x + 1) + (x + 2) + (x + 3) + ... + (x + 100) = 7450

x + 1 + x + 2 + x + 3 + ... + x + 100 = 7450

100 × x + 100 × 101 : 2 = 7450

100 × x + 5050 = 7450

100 × x = 7450 - 5050

100 × x = 2400

x = 2400 : 100

x = 24