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− 3 4 + 2 5 : 3 7 + 3 5 + − 1 4 : 3 7 = − 3 4 + 2 5 . 7 3 + 3 5 + − 1 4 . 7 3 = 7 3 . − 3 4 + 2 5 + 3 5 + − 1 4 = 7 3 . − 1 + 1 = 0
a) Ta có: 2002/2003 < 1 (1)
14/13 > 1 (2)
Từ (1) và (2) => 2002/2003 < 14/13
\(\frac{x+15}{35}+\frac{x+16}{36}=\frac{x+17}{37}+\frac{x+18}{38}\)
\(\frac{x+15}{35}-1+\frac{x+16}{36}-1=\frac{x+17}{37}-1+\frac{x+18}{38}-1\)
\(\frac{x-20}{35}+\frac{x-20}{36}-\frac{x-20}{37}-\frac{x-20}{38}=0\)
\(\left(x-20\right)\left(\frac{1}{35}+\frac{1}{36}-\frac{1}{37}-\frac{1}{38}\right)=0\)
\(\Rightarrow x-20=0\Rightarrow x=20\)
Vậy x=20
\(\frac{x+15}{35}+\frac{x+16}{36}=\frac{x+17}{37}+\frac{x+18}{38}\)
\(\Leftrightarrow\frac{x+15}{35}.885780+\frac{x+16}{36}.885780=\frac{x+17}{37}.885780+\frac{x+18}{38}.885780\)
\(\Leftrightarrow25308\left(x+15\right)+24605\left(x+16\right)=23940\left(x+17\right)+23310\left(x+18\right)\)
\(\Leftrightarrow49913x+773300=47250x+826560\)
\(\Leftrightarrow49913x=47250x+53260\)
\(\Leftrightarrow49913x-47250x=47250x+53260-47250x\)
\(\Leftrightarrow2663x=53260\)
\(\Leftrightarrow x=20\)
\(\Rightarrow x=20\)
a: 297/16=1+281/16
306/25=1+281/25
mà 281/16>281/25
nên 297/16>306/25
c: 2002/2003<1<14/13
d: -33/37=-1155/1295
-34/35=-1258/1295
mà -1155>-1258
nên -33/37>-34/35
25 + 37 - 48 - 25 - 37
= (25 - 25) + (37 - 37) - 48
= 0 + 0 - 48
= -48
2575 + 37 - 2576 - 29
= (2575 - 2576) + (37 - 29)
= -1 + 8 = 7
34 + 35 + 36 + 37 - 14 - 15 - 16 - 17
= (34 - 14) + (35 - 15) + (36 - 16) + (37 - 17)
= 20 + 20 + 20 + 20
= 20 x 4
= 80
25 + 37 - 48 - 25 - 37 = - 48