C

\(a,\dfrac{3}{5}+\dfrac{1}{5}.\dfrac{-17}{9}=\dfrac{3}{5}-\dfrac{17}{45}=\dfrac{27}{45}-\dfrac{17}{45}=\dfrac{10}{45}=\dfrac{2}{9}\\ b,\left(-\dfrac{4}{15}-\dfrac{18}{19}\right)-\left(\dfrac{20}{19}+\dfrac{11}{15}\right)=-\dfrac{4}{15}-\dfrac{18}{19}-\dfrac{20}{19}-\dfrac{11}{15}=\left(-\dfrac{4}{15}-\dfrac{11}{15}\right)-\left(\dfrac{18}{19}+\dfrac{20}{19}\right)=-1-2=-3\)

\(a,=\dfrac{3}{5}+\left(-\dfrac{17}{45}\right)=\dfrac{2}{9}\)

\(b,=-\dfrac{4}{15}-\dfrac{18}{19}-\dfrac{20}{19}-\dfrac{11}{15}=-3\)

a. 3/5+1/5.-17/9

=3/5+-17/45

=27/45+-17/45

=2/9

b. (-4/15-18/19)-(20/19+11/15)

=-4/15-18/19-20/19-11/15

=(-4/15-11/15)-(18/19-20/19)

=(-1)-(-2/19)

=(-19/19)-(-2/19)

=-17/19  

a.2/9
b.-3

a.2/9 b.(-3)

a.2/9
b.-3

Xét: \(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\)

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

\(=0\)

\(\rightarrow C=0\)

\(a,MSC:180\\ \dfrac{17}{12}=\dfrac{17.15}{12.15}=\dfrac{255}{180};\dfrac{31}{18}=\dfrac{31.10}{18.10}=\dfrac{310}{180};\dfrac{8}{15}=\dfrac{8.12}{15.12}=\dfrac{96}{180}\\ b,MSC:75\\ \dfrac{7}{15}=\dfrac{7.5}{15.5}=\dfrac{35}{75};\dfrac{8}{25}=\dfrac{8.3}{25.3}=\dfrac{24}{75};\dfrac{11}{75}=\dfrac{11}{75}\)

Tham khảo :

\(17A=\dfrac{17^{19}+17}{17^{19}+1}=\dfrac{\left(17^{19}+1\right)+16}{17^{19}+1}=\dfrac{17^{19}+1}{17^{19}+1}+\dfrac{16}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)

\(17B=\dfrac{17^{18}+17}{17^{18}+1}=\dfrac{\left(17^{18}+1\right)+16}{17^{18}+1}=\dfrac{17^{18}+1}{17^{18}+1}+\dfrac{16}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)

Vì \(17^{19}>17^{18}=>17^{19}+1>17^{18}+1\)

\(=>\dfrac{16}{17^{19}+1}< \dfrac{16}{17^{18}+1}\)

\(=>17A< 17B=>A< B\)

a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)

\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)

\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)

\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)

mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)

nên x-17=0

hay x=17

Vậy: x=17

b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)

\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)

\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)

\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)

mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)

nên x+20=0

hay x=-20

Vậy: x=-20

`B17:`

`a)` Với `x \ne +-3` có:

`A=[x+15]/[x^2-9]+2/[x+3]`

`A=[x+15+2(x-3)]/[(x-3)(x+3)]`

`A=[x+15+2x-6]/[(x-3)(x+3)]`

`A=[3x+9]/[(x-3)(x+3)]=3/[x-3]`

`b)A=[-1]/2<=>3/[x-3]=-1/2<=>-x+3=6<=>x=-3` (ko t/m)

   `=>` Ko có gtr nào của `x` t/m

`c)A in ZZ<=>3/[x-3] in ZZ`

   `=>x-3 in Ư_3`

 Mà `Ư_3={+-1;+-3}`

`@x-3=1=>x=4`

`@x-3=-1=>x=2`

`@x-3=3=>x=6`

`@x-3=-3=>x=0`

________________________________

`B18:`

`a)M=1/3`             `ĐK: x  \ne +-4`

`<=>(4/[x-4]-4/[x+4]).[x^2+8x+16]/32=1/3`

`<=>[4(x+4)-4(x-4)]/[(x-4)(x+4)].[(x+4)^2]/32=1/3`

`<=>32/[x-4].[x+4]/32=1/3`

`<=>3x+12=x-4`

`<=>x=-8` (t/m)

Ta có \(\dfrac{6}{15}>\dfrac{6}{16}>...>\dfrac{6}{19}\) nên \(S< \dfrac{6}{15}.5=2\).

Lại có \(S>\dfrac{6}{19}.5>1\) nên \(1< S< 2\)