\(a.5x=-20\\ x=-20:5=-4\\ b.\dfrac{5}{6}-x=\dfrac{7}{4}\\ x=\dfrac{5}{6}-\dfrac{7}{4}=\dfrac{-11}{12}\)

\(5x=-20\Leftrightarrow x=-20:5\Leftrightarrow x=-4\)

\(\dfrac{5}{6}-x=\dfrac{7}{4}\Leftrightarrow x=\dfrac{5}{6}-\dfrac{7}{4}\Leftrightarrow x=-\dfrac{11}{12}\)

Bài 5:

a: x(x-4)=0

=>\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

b: Đề thiếu vế phải rồi bạn

Bài 6:

a: \(\left(-5\right)\cdot\left(-6\right)\cdot\left(-4\right)\cdot2\)

\(=-\left(2\cdot5\right)\cdot\left(4\cdot6\right)\)

\(=-24\cdot10=-240\)

b: \(\left(-3\right)\cdot2\cdot\left(-8\right)\cdot5\)

\(=3\cdot2\cdot8\cdot5\)

\(=\left(3\cdot8\right)\cdot\left(2\cdot5\right)\)

\(=24\cdot10=240\)

a)

`x/(-8)=5/4`

`=>x*4=5*(-8)`

`=>4x=-40`

`=>x=-40:4`

`=>x=-10`

b)

`(-13)/x=(-65)/10`

`=>(-13)/x=(-13)/2`

`=>x=2`

c)

`(-6)/5=(-12)/x`

`<=>(-12)/10=(-12)/x`

`<=>x=10`

d)

`(x-2)/2=8/(x-2)`

`=>(x-2)(x-2)=8*2`

`=>(x-2)^2=16`

`=>x-2=4` hoặc`x-2=-4`

`=>x=4+2` hoặc `x=-4+2`

`=>x=6` hoặc `x=-2`

`x/(-8) = 5/4`

`=> 4x=5.(-8)`

`=>4x=-40`

`=>x=-40:4`

`=>x=-10`

__

`-13/x =-65/10`

`=> -65x=-130`

`=>x=-130:(-65)`

`=>x= 2`

__

`-6/5 =-12/x`

`=> -6x=-12.5`

`=> -6x= -60`

`=>x=-60:(-6)`

`=>x=10`

__

`(x-2)/2 = 8/(x-2)`

`=>(x-2)^2=2.8`

`=>(x-2)^2=16`

`=> (x-2)^2=+-4^2`

\(\Rightarrow\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)

Lời giải:

a.

$\frac{2}{3}+\frac{1}{3}:3\times x=20\text{%}$
$\frac{2}{3}+\frac{1}{9}\times x=\frac{1}{5}$

$\frac{1}{9}\times x=\frac{1}{5}-\frac{2}{3}=\frac{-7}{15}$

$x=\frac{-7}{15}: \frac{1}{9}=\frac{-21}{5}$
b.

$\frac{3-x}{5-x}=\frac{6}{11}$
$\Rightarrow 6(5-x)=11(3-x)$

$\Rightarrow 30-6x=33-11x$

$\Rightarrow 5x=3$

$\Rightarrow x=\frac{3}{5}$

a) x² - 4 = 0

x² = 4

x = 2 hoặc x = -2

b) 2x(x + 5) - 3(5 + x) = 0

(x + 5)(2x - 3) = 0

X + 5 = 0 hoặc 2x - 3 = 0

*) x + 5 = 0

x = -5

*) 2x - 3 = 0

2x = 3

x = 3/2

c) x³ - 6x² + 11x - 6 = 0

x³ - x² - 5x² + 5x + 6x - 6 = 0

(x³ - x²) - (5x² - 5x) + (6x - 6) = 0

x²(x - 1) - 5x(x - 1) + 6(x - 1) = 0

(x - 1)(x² - 5x + 6) = 0

(x - 1)(x² - 2x - 3x + 6) = 0

(x - 1)[(x² - 2x) - (3x - 6)] = 0

(x - 1)[x(x - 2) - 3(x - 2)] = 0

(x - 1)(x - 2)(x - 3) = 0

x - 1 = 0 hoặc x - 2 = 0 hoặc x - 3 = 0

*) x - 1 = 0

x = 1

*) x - 2 = 0

x = 2

*) x - 3 = 0

x = 3

Vậy x = 1; x = 2; x = 3

3/5

1 1/6 

a 4/5 - X = 3/15

X = 4/5 - 3/15

X = 3/5

b X : 2/3 = 1/2

X = 1/2 x 2/3

X = 11/6 

3/5

11/6

Tìm x:

\(a.\dfrac{35}{28}-x=\dfrac{5}{14}\\ x=\dfrac{35}{28}-\dfrac{5}{14}\\ x=\dfrac{25}{28}\\ b.\dfrac{6}{7}\times x=\dfrac{2}{3}\\ x=\dfrac{2}{3}:\dfrac{6}{7}\\ x=\dfrac{7}{9}\\ c.x\times7=\dfrac{3}{4}\\ x=\dfrac{3}{4}:7\\ x=\dfrac{3}{28}\\ d.2:x-\dfrac{1}{3}=\dfrac{2}{5}\\ 2:x=\dfrac{2}{5}+\dfrac{1}{3}\\ 2:x=\dfrac{11}{15}\\ x=2:\dfrac{11}{15}\\ x=\dfrac{30}{11}.\)

\(a,\dfrac{35}{28}-x=\dfrac{5}{14}\)

\(x=\dfrac{35}{28}-\dfrac{5}{14}\)

\(x=\dfrac{25}{28}\)

\(b,\dfrac{6}{7}\times x=\dfrac{2}{3}\)

\(x=\dfrac{2}{3}:\dfrac{6}{7}\)

\(x=\dfrac{7}{9}\)

\(c,x\times7=\dfrac{3}{4}\)

\(x=\dfrac{3}{4}:7\)

\(x=\dfrac{3}{28}\)

\(d,2:x-\dfrac{1}{3}=\dfrac{2}{5}\)

\(2:x=\dfrac{2}{5}+\dfrac{1}{3}\)

\(2:x=\dfrac{11}{15}\)

\(x=2:\dfrac{11}{15}\)

\(x=\dfrac{30}{11}\)

#YVA

a) \(\left(x+1\right)^3-\left(x-1\right)^3-6\cdot\left(x-1\right)^2=10\)

\(\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x^2-2x+1\right)=10\)

\(\Rightarrow6x^2+2-6x^2+12x-6=10\)

\(\Rightarrow12x-4=10\)

\(\Rightarrow12x=14\)

\(\Rightarrow x=\dfrac{7}{6}\)

b) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)

\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)

\(\Rightarrow x^3-25x-x^3-8=42\)

\(\Rightarrow-25x-8=42\)

\(\Rightarrow-25x=50\)

\(\Rightarrow x=\dfrac{50}{-25}=-2\)

c) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)

\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)

\(\Rightarrow24x+25=49\)

\(\Rightarrow24x=24\)

\(\Rightarrow x=\dfrac{24}{24}=1\)

\(a,Sửa:2021x-1+2022x\left(1-2021x\right)=0\\ \Leftrightarrow\left(2021x-1\right)\left(1-2022x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2021}\\x=\dfrac{1}{2022}\end{matrix}\right.\)