1)x=1

2)x=0

1, x = 1

2, x = 0

a) Ta có:

1^x = 1 ( 1 lũy thừa lên bao nhiêu vẫn là 1 )

=> x¹⁰ = 1^x

=> x¹⁰ = 1

=> x = 1

Vậy x = 1

b) Ta có:

1^x = 1 ( 1 lũy thừa lên bao nhiêu vẫn bằng 1 )

=> 10^x = 1^x

=> 10^x = 1

=> x = 0

Vậy x = 0

Lời giải:

a. 

$x-\frac{1}{2}=\frac{2}{3}$

$x=\frac{2}{3}+\frac{1}{2}=\frac{7}{6}$

b. 

$x\times \frac{5}{6}=\frac{1}{2}$
$x=\frac{1}{2}: \frac{5}{6}=\frac{3}{5}$

c.

$x:\frac{2}{5}=10$

$x=10\times \frac{2}{5}=4$

d.

$\frac{7}{5}-x=\frac{12}{10}$

$x=\frac{7}{5}-\frac{12}{10}=\frac{1}{5}$

`x+x:0,5+x:10=3,162`

`x+x xx2+x xx0,1=3,162`

`x xx(1+2+0,1)=3,162`

`x xx3,1=3,162`

`x=3,162:3,1`

`x=1,02`

_____________________________________

`2/5xx x+x xx1/5=2/7`

`x xx(2/5+1/5)=2/7`

`x xx3/5=2/7`

`x=2/7:3/5`

`x=10/21`

a: =>(x+10)(x-1)=0

=>x=-10 hoặc x=1

b: \(A=x^3-1-\left(x+5\right)\left(x^2-3\right)-5x^2-10x-5\)

\(=x^3-5x^2-10x-6-x^3+3x-5x^2+15\)

=-7x+9

=110/13

\(\dfrac{x}{10}=\dfrac{1}{2}+\dfrac{2}{5}+\dfrac{3}{10}\)

\(\dfrac{x}{10}=\dfrac{5}{10}+\dfrac{4}{10}+\dfrac{3}{10}\)

\(\dfrac{x}{10}=\dfrac{12}{10}\)

\(=>x=12\)

=>x/10=5/10+4/10+3/10=12/10

=>x=12

\(\Rightarrow\cdot\dfrac{x}{10}=\dfrac{5}{10}+\dfrac{4}{10}+\dfrac{3}{10}=\dfrac{12}{10}\Rightarrow x=12\)

2(x – 1) – 5 (x + 2) = – 10

⇒ 2x – 2 – 5x – 10 = –10

⇒ -3x – 12 = – 10

⇒ – 3x = -10+12

⇒ -3x = 2

⇒ x = (-2)/3

Vậy : x = (-2)/3

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

\(\Leftrightarrow4x^2+12x+9+x^2-2x+1-10=0\)

\(\Leftrightarrow5x^2+10x=0\)

\(\Leftrightarrow5x\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

Vậy \(S=\left\{-2;0\right\}\)