tách ra

\(1\)) \(5-\left(10-x\right)=7\)
\(10-x=5-7\)
\(10-x=-2\)
\(x=10-\left(-2\right)\)
\(x=12\)

\(2\)) \(-32-\left(x-5\right)=0\)
\(x-5=-32-0\)
\(x-5=-32\)
\(x=-32+5\)
\(x=-27\)

1)10-x=5-7

x=10-(-2)

x=12

2)x+5=0+32

x=32-5

x=27

3) -12+2x-9+x=0

-12+(2x+x-9)=0

3x-9=0+12

3x=12+9

x=21:3

x=7

4) 15-x=1-11

x=15-(-10)

x=25

5) 4-(27+3)=x-9

4-30=x-9

-26+9=x

x=-17

c: -62+x=-14+46

=>\(x-62=46-14=32\)

=>\(x=32+62=94\)

d: \(25+\left(x-5\right)=-123-\left(15-123\right)\)

=>\(x-5+25=-123-15+123\)

=>\(x+20=-15\)

=>\(x=-15-20=-35\)

`c) -62 + x = -14 + 46`

`-62 + x = 32`

`x = 32 - (-62)`

`x = 94`

Vậy `x = 94`

`d) 25 + ( x - 5 ) = -123 - ( 15 - 123 )`

`25 + ( x - 5 ) = -123 - 15 + 123`

`25 + ( x - 5 ) = - 15`

`x - 5 = -15 - 25`

`x - 5 = -40`

`x = -40 + 5`

`x = -35`

Vậy `x = -35`

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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 \(x\in\left\{5;-5;-10\right\}.\)

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

25+(x-5)=123-(15-123)

=>x-5+25=123-15+123

=>x+20=246-15=231

=>x=231-20=211

a) `123 - 27 + 27 + 13 - 123`

`= (123 - 123) + (27 - 27) + 13`

`= 0 + 0 + 13`

`= 13`

`---`

b) `175 + 25 + 13 + 15 - 175 - 25`

`= (175-175)+(25-25)+25+13`

`=0+0+25+13`

`=28`

`---`

c) \(53\times39+47\times39+53\times21+47\times21\)

\(=\text{[}\left(53+47\right)\times39]+\left[\left(53+47\right)\times21\right]\)

\(=\left(100\times39\right)+\left(100\times21\right)\)

\(=3900+2100\)

`=6000`

`---`

d) \(49\times35+49\times16-49\times61\)

`=` \(49\times\left(35+16-61\right)\)

`=` \(49\times\left(-10\right)\)

`=-490`

\(25:x^5-16=3^{15}:3^{13}\)

=>\(25:x^5-16=3^2=9\)

=>\(25:x^5=25\)

=>\(x^5=1\)

=>x=1

Câu 1: 

a:=17+5=22

b: =40-25=15