\(a.\left(\dfrac{-1}{2}\right)^2\cdot\left(\dfrac{2}{5}\right)^2 \\ =\left(\dfrac{-1}{2}\cdot\dfrac{2}{5}\right)^2\\ =\left(\dfrac{-1}{5}\right)^2\\ =\dfrac{1}{25}\\ b.\left(\dfrac{1}{9}\right)^2:\left(\dfrac{1}{3}\right)^3\\ =\left[\left(\dfrac{1}{3}\right)^2\right]^2:\left(\dfrac{1}{3}\right)^3\\ =\left(\dfrac{1}{3}\right)^4:\left(\dfrac{1}{3}\right)^3\\ =\dfrac{1}{3}\\ c.\left(\dfrac{-1}{2}\right)^3\cdot\left(\dfrac{3}{2}\right)^3\\ =\left(\dfrac{-1}{2}\cdot\dfrac{3}{2}\right)^3\\ =\left(\dfrac{-3}{4}\right)^3\\ =\dfrac{-27}{64}\)

2h elm phải đi học rồi ạ :((

\(C=\dfrac{6}{1\cdot4}+\dfrac{6}{4\cdot7}+...+\dfrac{6}{301\cdot304}\\ =2\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{301\cdot304}\right)\\ =2\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{301}-\dfrac{1}{304}\right)\\ =2\cdot\left(1-\dfrac{1}{304}\right)\\ =2\cdot\dfrac{303}{304}\\ =\dfrac{303}{152}\) 

\(B=\dfrac{11}{210}-\left(\dfrac{16}{15\cdot31}+\dfrac{13}{31\cdot44}+\dfrac{16}{44\cdot60}\right)\\ =\dfrac{11}{210}-\left(\dfrac{1}{15}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{60}\right)\\ =\dfrac{11}{210}-\left(\dfrac{1}{15}-\dfrac{1}{60}\right)\\ =\dfrac{11}{210}-\dfrac{1}{20}\\ =\dfrac{1}{420}\)

a) Oz là tia đối của Ox

=> \(\widehat{zOy}=180^o-\widehat{xOy}=180^o-30^o=150^o\)

b) Ot là phân giác của góc zOy

=> \(\widehat{yOt}=\widehat{tOz}\)

Mà: \(\widehat{yOt}+\widehat{tOz}=\widehat{zOy}=>2\widehat{yOt}=\widehat{zOy}\)

\(=>\widehat{yOt}=\dfrac{1}{2}\widehat{zOy}=\dfrac{1}{2}\cdot150^o=75^o\\ =>\widehat{tOz}=\widehat{yOt}=75^o\)

Giúp mình với please

63.370 + 63.82 + 37.69 + 41

= (63.370 + 63.82) + 37.69 + 41

= 63.(370 + 82) + 2553 + 41

= 63.452 + 2553 + 41

= 28476 +2553 + 41

= 31029 + 41

= 31070

\(a.\left(x-2\right)\left(x^2+x-1\right)-x\left(x^2-1\right)\\ =\left(x^3+x^2-x-2x^2-2x+2\right)-\left(x^3-x\right)\\ =x^3-x^2-3x+2-x^3+x\\ =-x^2-2x+2\\ b.\left(2x-9\right)\left(2x+9\right)-4x^2\\ =\left[\left(2x\right)^2-9^2\right]-4x^2\\ =4x^2-81-4x^2\\ =-81\\ c.2x^2+3\left(x-1\right)\left(x-1\right)\\ =2x^2+3\left(x-1\right)^2\\ =2x^2+3\left(x^2-2x+1\right)\\ =2x^2+3x^2-6x+3\\ =5x^2-6x+3\)

a; (\(x\) - 2)(\(x^2\) + \(x\) - 1) - \(x\)(\(x^2\) - 1)

 =   \(x^3\) + \(x^2\) - \(x\) - 2\(x^2\) - 2\(x\) + 2 - \(x^3\) + \(x\)

=    (\(x^3\) - \(x^3\)) - ( 2\(x^2\) - \(x^2\)) - (\(x\) + 2\(x\) - \(x\)) + 2

= 0 - \(x^2\) - (3\(x\) - \(x\)) + 2

= - \(x^2\) - 2\(x\) + 2

ΔAEH vuông tại E

mà EI là đường trung tuyến

nên IE=IH

=>ΔIEH cân tại I

ΔBEC vuông tại E

mà EK là đường trung tuyến

nên KE=KB

=>ΔKEB cân tại K

\(\widehat{IEK}=\widehat{IEB}+\widehat{KEB}=\widehat{IHE}+\widehat{KBE}\)

\(=\widehat{BHD}+\widehat{DBH}=90^0\)

=>IE\(\perp\)EK

\(-\dfrac{3}{11}.\dfrac{5}{7}+\dfrac{5}{7}.-\dfrac{8}{11}+\dfrac{19}{7}\)

`=` \(\dfrac{5}{7}.\left(\dfrac{-3}{11}+\dfrac{-8}{11}\right)+\dfrac{19}{7}\)

`=` \(\dfrac{5}{7}.\dfrac{-11}{11}+\dfrac{19}{7}\)

`=` \(\dfrac{5}{7}.\left(-1\right)+\dfrac{19}{7}\)

`=` \(-\dfrac{5}{7}+\dfrac{19}{7}\)

`=` \(\dfrac{14}{7}\)

`=  2`

(-\(\dfrac{1}{2}\))³:(-\(\dfrac{1}{2}\))⁶=(-\(\dfrac{1}{8}\)):\(\dfrac{1}{64}\)=-\(\dfrac{64}{8}\)=-8

`#3107.101107`

\(\left(-\dfrac{1}{2}\right)^3\div\left(-\dfrac{1}{2}\right)^6\\ =\left(-\dfrac{1}{2}\right)^{3-6}\\ =\left(-\dfrac{1}{2}\right)^{-3}\\ =\left(-2\right)^3\\ =-8\)