=-17/15-9/17-2/3+9/17

=-17/15-10/15

=-27/15

=-9/5

Giải:

\(\left(\dfrac{-17}{15}-\dfrac{9}{17}\right)-\left(\dfrac{2}{3}-\dfrac{9}{17}\right)\)

⇔ \(\dfrac{-17}{15}-\dfrac{9}{17}-\dfrac{2}{3}+\dfrac{9}{17}\)

⇔\(\dfrac{-17}{15}-\dfrac{10}{15}\)

\(\Leftrightarrow\dfrac{-27}{15}=\dfrac{-9}{5}\)

=-17/15-9/17-2/3+9/17

=-17/15-10/15

=-27/15

=-9/5

ta có:\(A=\frac{17^{15}+3}{17^{15}-2}=\frac{17^{15}-2+5}{17^{15}-2}=\frac{17^{15}-2}{17^{15}-2}+\frac{5}{17^{15}-2}=1+\frac{5}{17^{15}-2}\)

\(B=\frac{17^{15}}{17^{15}-5}=\frac{17^{15}-5+5}{17^{15}-5}=\frac{17^{15}-5}{17^{15}-5}+\frac{5}{17^{15}-5}=1+\frac{5}{17^{15}-5}\)

vì 17¹⁵-2>17¹⁵-5

=>\(\frac{5}{17^{15}-2}<\frac{5}{17^{15}-5}\)

=>A<B

B>A

17^15-5>17^15-2

(2¹⁷ + 17²).(9¹⁵ - 3¹⁵).(2⁴ - 4²)

= (2¹⁷ + 17²).(9¹⁵ - 3¹⁵).(16 - 16)

= (2¹⁷ + 17²).(9¹⁵ - 3¹⁵).0

= 0

  (217 + 172) (915 - 315) (24 - 42)
= (217 + 172).(915 - 315).(16 - 16)
= (217 + 172).(915 - 315).0 = 0
 

\(\dfrac{2}{3}.\dfrac{15}{17}-\dfrac{5}{3}.\dfrac{15}{17}\)
= \(\dfrac{15}{17}.\left(\dfrac{2}{3}-\dfrac{5}{3}\right)\)
\(=\dfrac{15}{17}.\left(-1\right)=-\dfrac{15}{17}\)
đúng đề ko?

\(\dfrac{2}{3}\).\(\dfrac{15}{17}\)-\(\dfrac{5}{3}\).\(\dfrac{15}{17}\)

= \(\dfrac{15}{17}\).(\(\dfrac{2}{3}\)-\(\dfrac{5}{3}\))

= \(\dfrac{15}{17}\).(-1)

=\(-\dfrac{15}{17}\)

\(\dfrac{2}{3}\cdot\dfrac{15}{17}-\dfrac{5}{3}\cdot\dfrac{15}{17}\)

\(=\dfrac{15}{17}\left(\dfrac{2}{3}-\dfrac{5}{3}\right)\)

\(=-\dfrac{15}{17}\)

_{khum bt}

10 < 10 + 3

11 + 2=2 + 11

9 < 10 + 9

10 = 10 + 0

17 – 4 > 14 - 3

18 – 4 >12

15 > 15 – 1

17 + 1<17 + 2

12+ 5 > 16

16 =19 - 3

15 – 4 =10 + 1

19 – 3 >11