Bài 1:

a) \(=\dfrac{8}{15}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)=\dfrac{8}{15}.1=\dfrac{8}{15}\)

b) \(=\dfrac{3.3-7-2.4}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\)

Bài 2:

 \(\dfrac{x}{2,7}=-\dfrac{2}{3,6}\Rightarrow x=\dfrac{\left(-2\right).2,7}{3,6}\Rightarrow x=-\dfrac{3}{2}\)

Bài 3:

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).5=-10\end{matrix}\right.\)

Tham khảo:

a)      \((4{x^2} - 5):(x - 2) = \dfrac{{4{x^2} - 5}}{{x - 2}} = 4x + 8 + \dfrac{{11}}{{x - 2}}\)

Vậy \( (4{x^2} - 5):(x - 2)= 4x + 8 + \dfrac{{11}}{{x - 2}}\)

b)      \((3{x^3} - 7x + 2):(2{x^2} - 3) = \dfrac{{3{x^3} - 7x + 2}}{{2{x^2} - 3}}\)

Vậy \( (3{x^3} - 7x + 2):(2{x^2} - 3)= \dfrac{3}{2}x + \dfrac{{\dfrac{-5}{2}x + 2}}{{2{x^2} - 3}}\)

Tham khảo:

a)      \((45{x^5} - 5{x^4} + 10{x^2}):5{x^2}\)\( = 9{x^3} - {x^2} + 2\)

b)      \((9{t^2} - 3{t^4} + 27{t^5}):3t = (27{t^5} - 3{t^4} + 9{t^2}):3t\\=(27t^5):(3t) - (3t^4):(3t)+(9t^2):(3t) = 9{t^4} - {t^3}+3t\)

Tham khảo:

a) \((8{x^6} - 4{x^5} + 12{x^4} - 20{x^3}):4{x^3}\)

\( = (8{x^6}:4{x^3}) - (4{x^5}:4{x^3}) + (12{x^4}:4{x^3}) - (20{x^3}:4{x^3})\)

\( = 2{x^2} - {x^2} + 3x - 5\)

b) 

Vậy \((2{x^2} - 5x + 3):(2x - 3)= x - 1\)

a.-1,75-(-\(\dfrac{1}{9}\)-2\(\dfrac{1}{8}\))
-1,75-\(\dfrac{1}{9}+\dfrac{17}{8}\)
\(-\dfrac{7}{4}-\dfrac{1}{9}+\dfrac{17}{8}\)
\(\dfrac{-126}{72}-\dfrac{8}{72}+\dfrac{153}{72}\)
=\(\dfrac{19}{72}\)

b.\(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\dfrac{21}{8}+\dfrac{1}{3}\)
\(\dfrac{-2}{24}-\dfrac{63}{24}+\dfrac{64}{24}\)
=\(\dfrac{-1}{24}\)

a: =1/2x^3*x^2-1/2x^3*6x-1/2x^3*10

=1/2x^5-3x^4-5x^3

b: =-3x^2*5x^3+3x^2*4x^2-3x^2*3x+3x^2*3x

=-15x^5+12x^4-9x^3+9x^2

c: \(=3x\cdot5x^2-3x\cdot2x-3x=15x^3-6x^2-3x\)

d: \(=\dfrac{1}{2}x^2y\cdot2x^3-\dfrac{1}{2}x^2y\cdot\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)