\(a,\left(7,8+2,2\right)\times1,2\\ ◻C1:\\ =10\times1,2\\ =12\\ ◻C2:=7,8\times1,2+2,2\times1,2\\ =9,36+2,64\\ =12\\ b,\left(6,75+3,25\right)\times4,2\\ ◻C1:=10\times4,2=42\\ ◻C2:=6,75\times4,2+3,25\times4,2\\ =28,35+13,65\\ =42\\ c,\left(9,6-4,2\right)\times3,6\\ ◻C1:=5,4\times3,6\\ =19,44\\ ◻C2:=9,6\times3,6-4,2\times3,6\\ =34,56-15,12\\ =19,44\)

:v phim 3 và 2 quá gần nhau 

3 bao gạo nếp nặng số kg là :

\(3\times30=90\left(kg\right)\)

1 bao gạo tẻ nặng số lg là : 

\(40\left(kg\right)\)

 3 bao gạo nếp và 1 bao gạo te cân nặng  số kg là : \(90+40=130\left(kg\right)\)

Thay \(x=-1\) vào A
\(A=\dfrac{4.\left(-1\right)}{3.\left(-1\right)-8}=\dfrac{4}{11}\)

Vậy tại \(x=-1\) thì A có giá trị là \(\dfrac{4}{11}\)

\(b,\left(5+x\right)^2-36=0\\ =>\left(5+x\right)^2=36\\ =>\left[{}\begin{matrix}5+x=6\\5+x=-6\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)

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

\(d,\dfrac{1}{x-5x^2}-\dfrac{25x-15}{25x^2-1}\\ =\dfrac{1}{x\left(1-5x\right)}-\dfrac{25x-15}{\left(5x-1\right)\left(5x+1\right)}\\ =\dfrac{-1}{x\left(5x-1\right)}-\dfrac{25x-15}{\left(5x-1\right)\left(5x+1\right)}\\ =\dfrac{-\left(5x+1\right)-\left(25x-15\right).x}{x\left(5x-1\right)\left(5x+1\right)}\\ =\dfrac{-5x-5-25x^2+15x}{x\left(5x-1\right)\left(5x+1\right)}\\ =\dfrac{-25x^2+10x-5}{x\left(5x-1\right)\left(5x+1\right)}\\ =\dfrac{-\left(25x^2-10x+5\right)}{x\left(5x-1\right)\left(5x+1\right)}\\ =\dfrac{-\left(5x-1\right)^2}{x\left(5x-1\right)\left(5x+1\right)}\\ =\dfrac{-\left(5x-1\right)}{x\left(5x+1\right)}\)

\(c,\dfrac{7x+6}{2x\left(x+7\right)}-\dfrac{3x+6}{2x^2+14x}\\ =\dfrac{7x+6}{2x\left(x+7\right)}-\dfrac{3x+6}{2x\left(x+7\right)}\\ =\dfrac{7x+6-3x-6}{2x\left(x+7\right)}\\ =\dfrac{4x}{2x\left(x+7\right)}\\ =\dfrac{2}{x+7}\)

\(236\times123-236\times22-236\\ =236\times\left(123-22-1\right)\\ =236\times100\\ =23600\)

\(ĐK:x\ne\pm3\\ \dfrac{10}{x+3}-\dfrac{8x-36}{x^2-9}\\ =\dfrac{10}{x+3}-\dfrac{8x-36}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{10\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{8x-36}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{10x-30-8x+36}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2}{x-3}\)