a) x+3=12

x=12-3

x=9

b)(x-3):2=5¹⁴:5¹²

=>(x-3):2=5²

=>(x-3):2=25

=>x-3=25.2

=>x-3=50

=>x=50+3

=>x=53

c)4x+3x=30-20:10

=>x(4+3)=30-2

=>7x=28

=>x=28:7

=>x=4

d)2x-138=2³.3²

=>2x-138=8.9

=>2x-138=72

=>2x=72+138

=>2x=210

=>x=210:2

=>x=105

a) x + 3 = 12

x = 12 - 3

x = 9

b) ( x - 3 ) : 2 = 5¹⁴ : 5¹²

( x - 3 ) : 2 = 5^14-12

( x - 3 ) : 2 = 5²

( x - 3 ) : 2 = 25

x - 3 = 50

x = 53

c) 4x + 3x = 30 - 20 : 10

7x = 28

x = 4

d) 2x - 138 = 2³ x 3²

2x - 138 = 8 x 9

2x - 138 = 72

2x = 210

x = 105

1)

a)2x+26-84+7x=-130

(2x+7x)+(-84+26)=-130

9x-58=-130

9x=-130+58

9x=-72

x=-72/9

x=-8

Vậy x=-8

b)-6x+12+12-8x=-130

(-6x-8x)+(12+12)=-130

-14x+24=-130

-14x=-130-24

-14x=-154

x=-154/(-14)

x=11

Vậy x=11

c)4x-26-8=3x-69

4x-3x=-69+26+8

x=-35

Vậy x=-35

\(VT=\left|3x+1\right|+\left|3x-5\right|=\left|3x+1\right|+\left|5-3x\right|\ge\left|3x+1+5-3x\right|=6\)

\(VP=\frac{12}{\left(y+3\right)^2+2}\le\frac{12}{2}=6\)

Như vậy \(VT\ge6;VP\le6\)

Mà \(VT=VP\Leftrightarrow VT=VP=6\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}-\frac{1}{3}\le x\le\frac{5}{3}\\y=-3\end{cases}}\)

g: =>12x+1>=36x+12-24x-3

=>12x+1>=12x+9(loại)

h: =>6(x-1)+4(2-x)<=3(3x-3)

=>6x-6+8-4x<=9x-9

=>2x+2<=9x-9

=>-7x<=-11

=>x>=11/7

i: =>4x^2-12x+9>4x^2-3x

=>-12x+9>-3x

=>-9x>-9

=>x<1

`a)\sqrt{3x}-5\sqrt{12x}+7\sqrt{27x}=12`     `ĐK: x >= 0`

`<=>\sqrt{3x}-10\sqrt{3x}+21\sqrt{3x}=12`

`<=>12\sqrt{3x}=12`

`<=>\sqrt{3x}=1`

`<=>3x=1<=>x=1/3` (t/m)

`b)5\sqrt{9x+9}-2\sqrt{4x+4}+\sqrt{x+1}=36`   `ĐK: x >= -1`

`<=>15\sqrt{x+1}-4\sqrt{x+1}+\sqrt{x+1}=36`

`<=>12\sqrt{x+1}=36`

`<=>\sqrt{x+1}=3`

`<=>x+1=9`

`<=>x=8` (t/m)

9/13 x 7/12 + 9/13 x 5/12 - 9/13

= 9/13 x (7/12 + 5/12 - 1)

= 9/13 x 0

= 0

4/13 x 5/12 + 4/13 x 7/12 - 4/3

= 4/13 x (5/12 + 7/12) - 4/3

= 4/13 x 1 - 4/3

= 4/13 - 4/3

= -40/39

a) 9/13 x 7/12 + 9/13 x 5/12 - 9/13

= 9/13 x ( 5/12 + 7/12 - 1 )

= 9/13 x 0

= 0

b) 4/13 x 5/12 + 4/13 x 7/ 12 - 4/3

= 4/13 x ( 5/12 + 7/12 - 13/3 )

= 4/13 x ( - 10/3 )

= -40/39 

a) Ta có bảng sau:

Vậy: 

b) Ta có bảng sau:

Vậy: ...

`@` `\text {Ans}`

`\downarrow`

`a)`

`(x-1)(y+4) = 5`

`=> (x-1)(y+4) \in \text {Ư(5)} = +-1; +-5`

Ta có bảng sau:

Vậy, ta có các cặp `x,y` thỏa mãn `{2; -9}; {6; -5}; {0; 1}; {-4; -8}`

\(\dfrac{2}{5}+\dfrac{1}{3}=\dfrac{6+5}{15}=\dfrac{11}{15}\)

\(\dfrac{16}{24}+\dfrac{1}{3}=\dfrac{2}{3}+\dfrac{1}{3}=1\)

\(\dfrac{7}{12}+\dfrac{4}{6}+\dfrac{3}{8}=\dfrac{14}{24}+\dfrac{16}{24}+\dfrac{9}{24}=\dfrac{39}{24}=\dfrac{13}{8}\)

\(1+\dfrac{1}{12}=\dfrac{12+1}{12}=\dfrac{13}{12}\)

2/5+1/3= 6/15+5/15=11/15