góc yOt = góc xOy - góc xOt = 130 độ - 70 độ = 60 độ

chu vi của mảnh vườn : 

(5,5 + 3,75 ) x 2 = 18,5 (m)

số khóm hoa cần trồng: 

18,5 : \(\dfrac{1}{4}\)   = 74 (cây)

đs... 

x. (x+y+z) = -1/3 ⇔ x + y + z = -\(\dfrac{1}{3x}\) (1)

y.(x+y+z) = - 2/3 ⇔ x + y + z  = \(\dfrac{-2}{3y}\) (2)

z. (x+y+z) = 2 ⇔ x + y + z  = \(\dfrac{2}{z}\) (3)

từ (1) ; (2); (3) ta có: 

\(\dfrac{-1}{3x}\) = \(\dfrac{-2}{3y}\) = \(\dfrac{2}{z}\) ⇒ \(\dfrac{-2}{6x}\) = \(\dfrac{-2}{3y}\) = \(\dfrac{-2}{-z}\)

⇒ 6x = 3y = -z

⇒ \(\dfrac{x}{3}\) = \(\dfrac{y}{6}\) =  \(\dfrac{-z}{18}\) ⇒ \(\dfrac{x+y-(-z)}{3+6-18}\) = \(\dfrac{x+y+z}{-9}\)

⇒ x + y + z  = -3x ⇒ -3x. x = -1/3 ⇒ x = +_ 1/3

với x = 1/3 thì y = \(\dfrac{6}{3}\) . x = 2. 1/3 = 2/3; 

                  z = -6x = -6.1/3 = -2

vậy ( x,y,z) =( 1/3; 2/3; -2)

với x = -1/3 thì y = \(\dfrac{6}{3}\) . x = \(\dfrac{6}{3}\) . (-1/3 ) = -2/3

                        z = -(-1/3 )x6 = 2 

vậy (x,y,z) =(-1/3; -2/3 ; 2)

kết luận (x,y,z) = ( 1/3; 2/3; -2) ; (-1/3; -2/3; 2)

`1/5.(-5/5-10)+5x=x-2/3`

`1/5.(-1-10)+5x-x=-2/3`

`1/5 .(-11)+4x=-2/3`

`4x=-2/3+11/5=23/15`

`x=23/15:4=23/60`

cm      A = 3^{n + 3} - 2ⁿ⁺³ + 3ⁿ⁺¹ - 2ⁿ⁺¹ ⋮ 10

ta có : A = 3^{n+ 1}( 3² + 1) - 2ⁿ. ( 2³ + 2)

         ⇔A = 3ⁿ⁺¹.10 - 2ⁿ.10

         ⇔ A = 10.( 3ⁿ⁺¹- 2ⁿ)

            10 ⋮ 10 ⇔ 10. ( 3ⁿ⁺¹ - 2ⁿ)⋮ 10

       ⇔  A = 10 . (3ⁿ⁺¹ - 2ⁿ) ⋮10 (đpcm)

\(\dfrac{9^5.8^2}{27^3.16}\) = \(\dfrac{3^{10}.2^6}{3^9.2^4}\) = 12 

9⁵/27³ • 8²/16

= 3¹⁰/3⁹ • 2⁶/2⁴

=13

Cách giải:

A=1/2.4+1/4.6+........+1/100.102

A=1/2-1/4+1/4-1/6+.......+1/100-1/102

A=1/2-1/102

A=51/102-1/102

A=50/102

A=25/51

Đặt A = \(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{100\cdot102}\)

2A = \(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{100\cdot102}\)

2A = \(\dfrac{2}{2}-\dfrac{2}{4}+\dfrac{2}{4}-\dfrac{2}{6}+...+\dfrac{2}{100}-\dfrac{2}{102}\)

2A = \(1-\dfrac{1}{51}=\dfrac{50}{51}\)

A = \(\dfrac{50}{51}:2=\dfrac{25}{51}\)

a, (x- \(\dfrac{2}{7}\))(x+\(\dfrac{3}{4}\))

⇔ \(\left[{}\begin{matrix}x-\dfrac{2}{7}=0\\x+\dfrac{3}{4}=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=\dfrac{2}{7}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

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

⇔ \(\dfrac{2}{3}\) x - \(\dfrac{1}{2}\) x = \(\dfrac{2}{5}\) -  \(\dfrac{1}{3}\) 

⇔ \(\dfrac{x}{6}\) = \(\dfrac{1}{15}\)

⇔ x = \(\dfrac{1}{15}\) x 6 

     x = \(\dfrac{2}{5}\)

c, \(\dfrac{1}{3}\) x + \(\dfrac{2}{5}\) (x+1)= 0

⇔ \(\dfrac{5x+6x+6}{15}\) = 0

⇔ 11x + 6 = 0

⇔ x = - \(\dfrac{6}{11}\)

tìm x

a, (x - 2/7 ) ( x + 3/4 )=0

<=> x - 2/7= 0 hoặc x+3/4 = 0
=> X = 2/7 hoặc x = -3/4

c,1/3x +2/5 (x+1 )=0

=> 1/3x + 2/5x + 2/5.1 = 0

=> x.(1/3 + 2/5 ) +2/5 = 0

=> x.11/15 + 2/5 = 0

=> x.11/15 = 0 - 2/5

=> x.11/15 = -2/5

=> x = -2/5 : 11/15

=> x = -6/11

Vậy x = -6/11

\(\dfrac{1}{6}\) - 0,4 . \(\dfrac{5}{8}\) + \(\dfrac{1}{2}\)

= \(\dfrac{1}{6}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{2}\)

= \(\dfrac{2}{12}\) - \(\dfrac{3}{12}\) + \(\dfrac{6}{12}\)

= \(\dfrac{5}{12}\)

1/6-0,4.5/8+1/2

=1/6-1/4+1/2

=2/12-3/12+6/12

=5/12