\(a,=\dfrac{5}{6}+\dfrac{6}{7}\\ =\dfrac{35}{42}+\dfrac{36}{42}=\dfrac{71}{42}\\ b,=\dfrac{5}{6}+\dfrac{4}{9}\\ =\dfrac{15}{18}+\dfrac{8}{18}=\dfrac{23}{18}\)

a) \(A=\left(-10\right)+\left(-34\right)+\left(-26\right)\)

         \(=\left(-10\right)+\left[\left(-34\right)+\left(-26\right)\right]\)

         \(=\left(-10\right)+\left(-60\right)\)

         \(=-70\)

a) A = ( -34 + -26 ) + (-10)

      =  -60 - 10

        =-70

b) B = 12+35

     =  47

a) = -70
b) = 23

a) 172 + 10 x 2 = 170 +20 = 192

b) 10 x 2 + 300 = 20 + 300 = 320

c) 69 – 54 : 6 = 69 – 9 = 60

d) 900 + 9 x 10 = 900 +90 = 990

e) 20 x 6 + 70 = 120 + 70 = 190

f) 72 + 300 x 3 = 72 + 900 = 972

a) \(\dfrac{9}{5}+\dfrac{9}{5}:\dfrac{9}{5}\)

\(=\dfrac{9}{5}+\dfrac{9}{5}\times\dfrac{5}{9}\)

\(=\dfrac{9}{5}+1\)

\(=\dfrac{14}{5}\)

b) \(\dfrac{7}{5}-\dfrac{1}{2}\times\dfrac{1}{3}\)

\(=\dfrac{7}{5}-\dfrac{1}{6}\)

\(=\dfrac{42}{30}-\dfrac{5}{30}\)

\(=\dfrac{37}{30}\)

\(a,\dfrac{9}{5}+\dfrac{9}{5}:\dfrac{9}{5}\)

\(=\dfrac{9}{5}+\dfrac{9}{5}\times\dfrac{5}{9}\)

\(=\dfrac{9}{5}+1\)

\(=\dfrac{9}{5}+\dfrac{5}{5}\)

\(=\dfrac{14}{5}\)

\(b,\dfrac{7}{5}-\dfrac{1}{2}\times\dfrac{1}{3}\)

\(=\dfrac{7}{5}-\dfrac{1}{6}\)

\(=\dfrac{42}{30}-\dfrac{5}{30}\)

\(=\dfrac{37}{30}\)

a: A

b: =101-4,68+8,96

=105,28

c: x-38,75=206,99

=>x=245,74

a) Có x = 2020 => x + 1 = 2021. Thay 2021 = x + 1 vào A

\(A=x^6-\left(x+1\right)^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)

\(A=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+x+1\)

\(A=1\)

b) Có x = -19 => x - 1 = -20 => - ( x - 1 ) = 20. Thay 20 = - ( x - 1) vào B

\(B=x^{10}-\left(x-1\right)x^9-\left(x-1\right)x^8-\left(x-1\right)x^7-...-\left(x-1\right)x^2-\left(x-1\right)x-x+1\)

\(B=x^{10}-x^{10}+x^9-x^9+...+x^2-x^2+x-x+1\)

\(B=1\)

Chúc bạn học tốt!!!

\(A=\left(\dfrac{x^2}{x^3-4x}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)ĐK : \(x\ne-2;2\)

\(=\left(\dfrac{x}{x-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\left(\dfrac{x^2-4+10-x^2}{x+2}\right)\)

\(=\left(\dfrac{x}{x-4}+\dfrac{2x+4+2-x}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{6}{x+2}\right)=\left(\dfrac{x}{x-4}+\dfrac{x+6}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{6}{x+2}\right)\)

\(=\left(\dfrac{x\left(x^2-4\right)+\left(x+6\right)\left(x-4\right)}{\left(x-4\right)\left(x-2\right)\left(x+2\right)}\right):\dfrac{6}{x+2}\)

\(=\dfrac{x^3-4x+x^2-2x+24}{\left(x-4\right)\left(x-2\right)\left(x+2\right)}:\dfrac{6}{x+2}=\dfrac{x^3+x^2-6x+24}{\left(x-4\right)\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{6}\)

\(=\dfrac{x^3+x^2-6x+24}{6\left(x-4\right)\left(x-2\right)}=\dfrac{\left(x+4\right)\left(x^2-3x+6\right)}{6\left(x-4\right)\left(x-2\right)}\)

P/s : mình thấy đề này cứ sai sai ở đâu ý ! 

b, Ta có : \(\dfrac{\left(x+4\right)\left(x^2-3x+6\right)}{6\left(x-4\right)\left(x-2\right)}=2\)

\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x^2-3x+6\right)-12\left(x-4\right)\left(x-2\right)}{6\left(x-4\right)\left(x-2\right)}=0\)

\(\Rightarrow x^3-11x^2+66x-72=0\)

\(a,ĐK:x\ne-1;x\ne-2\\ b,A=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\dfrac{x-1}{x+2}\\ x=2020\Leftrightarrow A=\dfrac{2019}{2022}=\dfrac{673}{674}\\ c,A=0\Leftrightarrow x-1=0\Leftrightarrow x=1\left(tm\right)\)

a) Với a = 3,05 thì ta có:

\(a\times2,46\) \(=3,05\times2,46=7,503\)

b) Với \(a=\dfrac{15}{8}\) thì ta có:

\(\left(\dfrac{5}{6}+\dfrac{7}{12}\right):a\) \(=\left(\dfrac{5}{6}+\dfrac{7}{12}\right):\dfrac{15}{8}\) \(=\left(\dfrac{10}{12}+\dfrac{7}{12}\right)\times\dfrac{8}{15}\) \(=\dfrac{17}{12}\times\dfrac{8}{15}\) \(=\dfrac{34}{45}\)

a) Thay a=3,05 vào 2,46a, ta được:

\(2.46\cdot3.05=7.503\)

b) Thay \(a=\dfrac{15}{8}\) vào biểu thức \(\left(\dfrac{5}{6}+\dfrac{7}{12}\right):a\), ta được:

\(\left(\dfrac{5}{6}+\dfrac{7}{12}\right):\dfrac{15}{8}\)

\(=\dfrac{17}{12}\cdot\dfrac{8}{15}=\dfrac{34}{45}\)