b) Tìm $x$

$10(8x+9x)+8(2x-1)-2(5-6x)=20$

$\Leftrightarrow 170x+16x-8-10+12x=20$

$\Leftrightarrow 198x-18=20$

$\Leftrightarrow 198x=38$

$\Leftrightarrow x=\dfrac{19}{99}$.

Vậy: $x=\dfrac{19}{99}$.

c) Tìm $x$

$(5x^2)(194+20x)=50$

$\Leftrightarrow x^2(194+20x)=10$

$\Leftrightarrow 10x^3+97x^2-5=0$

Thử $x=\dfrac15$:

$10\left(\dfrac15\right)^3+97\left(\dfrac15\right)^2-5=0$.

Suy ra: $(5x-1)(2x^2+39x+5)=0$.

Do đó: $x=\dfrac15$ hoặc $x=\dfrac{-39\pm\sqrt{39^2-4\cdot2\cdot5}}{4}$

$=\dfrac{-39\pm\sqrt{1481}}{4}$.

Vậy: $x=\dfrac15\ \text{hoặc}\ x=\dfrac{-39+\sqrt{1481}}4\ \text{hoặc}\ x=\dfrac{-39-\sqrt{1481}}4.$

|3x - 1| ≤ 7

- 7 ≤ 3x - 1 ≤ 7

-6 ≤ 3x ≤ 8

- 2 ≤ x ≤ 8/3

Vậy x ∈ {-2; -1; 0; 1; 2}

Câu b:

xy + 3x - y = 20

x(y + 3) - (y + 3) = 20 - 3

(y + 3)(x - 1) = 17

Ư(17) = {-17; -1; 1; 17}

Lập bảng ta có:

Theo bảng trên ta có:

(x; y) = (-16; -4); (0; -20); (2; 14); (18; -2)

Vậy: (x; y) = (-16; -4); (0; -20); (2; 14); (18; -2)

\(a.97-7\left(x-8\right)=3\cdot2^4\)

\(97-7\left(x-8\right)=48\)

\(7\left(x-8\right)=97-48\)

\(7\left(x-8\right)=49\)

\(x-8=49:7=7\)

\(x=7+8=15\)

\(b.5x^2-74=51\)

\(5x^2=51+74=125\)

\(x^2=125:5=25\)

\(\Rightarrow x=\pm5\)

\(c.3x+2-3x=216\)

\(0x=214\)

⇒ \(x\in\) rỗng

\(5\left(3x-4\right)^3=40\)

\(\left(3x-4\right)^3=40:5=8\)

⇒ 3x - 4 = 2

⇒ 3x = 2 + 4 = 6

⇒ x = 6 : 3 = 2

a.97−7(x−8)=3⋅24

\(97 - 7 \left(\right. x - 8 \left.\right) = 48\)

\(7 \left(\right. x - 8 \left.\right) = 97 - 48\)

\(7 \left(\right. x - 8 \left.\right) = 49\)

\(x - 8 = 49 : 7 = 7\)

\(x = 7 + 8 = 15\)

\(b . 5 x^{2} - 74 = 51\)

\(5 x^{2} = 51 + 74 = 125\)

\(x^{2} = 125 : 5 = 25\)

\(\Rightarrow x = \pm 5\)

\(c . 3 x + 2 - 3 x = 216\)

\(0 x = 214\)

⇒ \(x \in\) rỗng

\(5 \left(\left(\right. 3 x - 4 \left.\right)\right)^{3} = 40\)

\(\left(\left(\right. 3 x - 4 \left.\right)\right)^{3} = 40 : 5 = 8\)

⇒ 3x - 4 = 2

⇒ 3x = 2 + 4 = 6

⇒ x = 6 : 3 = 2

a) \(-3+x=-5\)

   \(x=-5-\left(-3\right)\)

   \(x=-2\)

b) \(\frac{x}{4}=\frac{20}{8}\)

   \(8x=80\)

    \(x=10\)

THANK YOU

Câu 1:

Câu 1: x.(x+ 7) = 0

x = 0 hoặc x + 7 = 0

x + 7 = 0 suy ra x = - 7

Vậy x ∈ {0; - 7}

Câu 2: (x + 12).(x - 3) = 0

x + 12 = 0 hoặc x - 3 = 0

x + 12 = 0 suy ra x = - 12

x - 3 = 0 suy ra x = 3

Vậy x ∈ {-12; 3}

Câu 3:

(-x + 5).(3 - x) = 0

- x + 5 = 0; hoặc 3 - x = 0

- x + 5 = 0 suy ra x = 5

3 - x = 0 suy ra x = 3

Vậy x ∈ {0; 3}

Câu 4:

x(x + 2).(7 - x) = 0

x = 0 hoặc x + 2 = 0 hoặc 7 - x = 0

x + 2 = 0 suy ra x = - 2

7 - x = 0 suy ra x = 7

Vậy x ∈ {0; -2; 7}

\(5\frac{8}{17}:x-\frac{4}{17}:x=\frac{17}{20}=\left(5\frac{8}{17}-\frac{4}{17}\right):x=\frac{17}{20}=5\frac{4}{17}:x=\frac{17}{20}=>x=5\frac{4}{17}:\frac{17}{20}=\frac{1780}{289}\)