Ta có :

\(P=-\dfrac{1}{2}.\dfrac{5}{9}.x.\left(-\dfrac{7}{13}\right).\left(-\dfrac{3}{5}\right)\)

\(\Rightarrow P=\left[-\dfrac{1}{2}.\dfrac{5}{9}.\left(-\dfrac{7}{13}\right).\left(-\dfrac{3}{5}\right)\right].x\)

\(\Rightarrow P=\left[\dfrac{-5}{18}.\left(-\dfrac{7}{13}\right).\left(-\dfrac{3}{5}\right)\right].x\)

\(\Rightarrow P=\left[\dfrac{35}{234}.\left(-\dfrac{3}{5}\right)\right].x\)

\(\Rightarrow P=\dfrac{-7}{78}.x\)

Nếu P<0 thì x>0

Nếu P>0 thì x<0

Khoảng cách thực tế từ ổ cắm điện đến vòi nước thể hiện trên bản đồ là: \(2,5\times20=50cm\)

Khoảng cách an toàn là 60 cm trong khi thiết kế chỉ có 50 cm nên bản vẽ như vậy không phù hợp với yêu cầu của kiến trúc sư

( x+ 3/2)\(^4\)  = 16

( x+ 3/2)\(^4\) = 2\(^4\)

x+ 3/2= 2

x= 2- 3/2

x= 1/2

`(x+3/2)^4 =16 `

`<=> (x+3/2)^2 = (+- 2)^4`

`<=> [(x+3/2=2),(x+3/2=-2):}`

`<=> [(x = 2-3/2 = 4/2 -3/2 =1/2),(x=-2 -3/2 = -4/2-3/2 =-7/2):}`

Vậy `x in {1/2;-7/2}`

1/1x2 = 1 - 1/2

A = 1-1/2+1/2-1/3 + ... -1/2023

= 1-1/2023 

=2022/2023

( x - 3 )² - 1 = 24

( x - 3 )² = 24 + 1

( x - 3 )² = 25

( x - 3 )² = 5²

x - 3 = 5

x = 5 + 3

x = 8

\(=\dfrac{\dfrac{25}{3}\times\dfrac{21}{4}+\dfrac{73}{19}\times\dfrac{21}{4}}{\left(\dfrac{48}{17}-\dfrac{69}{34}\right)\times34}=\dfrac{\dfrac{21}{4}\left(\dfrac{25}{3}+\dfrac{73}{19}\right)}{\left(\dfrac{48\times2-69}{34}\right)\times34}=\dfrac{\dfrac{21}{4}\times\dfrac{694}{57}}{\dfrac{27}{34}\times34}=\dfrac{\dfrac{2429}{38}}{27}=\dfrac{2429}{1026}\)

\(=\dfrac{20\times0,02}{0,8}=\dfrac{0,4}{0,8}=\dfrac{1}{2}\)

    \(\dfrac{(11,81+8,19).0,02}{9:11,25}\)

=   \(\dfrac{20.0,02}{9:11,25}\)

= \(\dfrac{0,4}{\dfrac{9}{11,25}}\)

=  0,4x \(\dfrac{11,25}{9}\)

= 0,5

  (0,8.7 + 0,8²).( 1,25.7 - \(\dfrac{4}{5}\).1,25)+31,64

=0,8(7+0,8).[ 1,25.(7-0,8) + 31,64

=0,8.1,25.7,8.6.2  + 31,64

= 48,36 + 31,64

= 80

(6³ + 2. 6² + 2³)/37

= \(\dfrac{6^2.(6+2)+2^3}{37}\)

= \(\dfrac{6^2.8+8}{37}\)

= \(\dfrac{8.(36+1)}{37}\)

= 8

\(B=-\dfrac{1}{6}.\left(-\dfrac{15}{19}\right).\dfrac{38}{45}\\ =\left(-1\right).\left(-1\right).\dfrac{1}{6}.\dfrac{15}{19}.\dfrac{38}{45}\\ =\dfrac{1}{3.2}.\dfrac{5.3}{19}.\dfrac{19.2}{9.5}\\ =\dfrac{1}{9}\)

\(D=\left(2\dfrac{2}{15}.\dfrac{9}{17}.\dfrac{3}{32}\right):\left(-\dfrac{3}{17}\right)\\ =\left(\dfrac{32}{5.3}.\dfrac{3.3}{17}.\dfrac{3}{32}\right).\left(-\dfrac{17}{3}\right)\\ =\dfrac{1}{5}.\dfrac{3.3}{17}.\dfrac{-17}{3}\\ =\dfrac{-3}{5}\)