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a: =371-271-531+731=100+200=300
b: =-321-28-327+27
=-350-300
=650
c: =-424-371+424+29=-342
d: =-249+248=-1
a)371+(-271)+(-531)+731=300
b)-312+(-327+(-28)+27=650
c)-424+(-371)-(-424)-(-29)=-342
d)-249+197+248+(-197)=-1
e)3+(-5)+(-9)+11+(-13)=-13
g)-8+10+(-12)+14+(-16)+18=6
Ta có:
B=196+197/197+198<196+197/198=196/198+197/198<A
=> B<A
Vậy A>B
= nhau mòa dễ wa chắc cậu ghi đề sai chứ toán này so esay ok
B=\(\frac{196+197}{197+198}\)= \(\frac{196}{197+198}\)+ \(\frac{197}{197+198}\)
ta có \(\frac{196}{197+198}\)< \(\frac{196}{197}\)
\(\frac{197}{197+198}\)< \(\frac{197}{198}\)
=> \(\frac{196}{197+198}\)+ \(\frac{197}{197+198}\)< \(\frac{196}{197}\)+ \(\frac{197}{198}\)
=> B < A
\(B=\frac{196+197}{197+198}=\frac{196}{197+198}+\frac{197}{197+198}\)
\(\frac{196}{197}>\frac{196}{197+198};\frac{197}{198}>\frac{197}{197+198}\)
=>A>B
\(A=\frac{196}{197}+\frac{197}{198}=\left(1-\frac{1}{197}\right)+\left(1-\frac{1}{198}\right)=1-\frac{1}{197}+1-\frac{1}{198}=1-\frac{1}{197}+\frac{197}{197}-\frac{1}{198}\)\(=1-\frac{198}{197}-\frac{1}{198}=\frac{197}{197}-\frac{198}{197}-\frac{1}{198}=\frac{-1}{197}-\frac{1}{198}<\frac{196+197}{197+198}=\frac{393}{395}\)
Dễ thấy \(B=\dfrac{196+197}{197+198}< 1\)
Ta có:
\(A=\dfrac{196}{197}+\dfrac{197}{198}\)
\(A=1-\dfrac{1}{197}+1-\dfrac{1}{198}\)
\(A=\left(1+1\right)-\left(\dfrac{1}{197}+\dfrac{1}{198}\right)\)
\(A=2+\left(\dfrac{1}{197}+\dfrac{1}{198}\right)\)
Mà \(\left(\dfrac{1}{197}+\dfrac{1}{198}\right)< 1\) vì \(\dfrac{1}{197}< 0,5;\dfrac{1}{198}< 0,5\) nên \(\left(\dfrac{1}{197}+\dfrac{1}{198}\right)< 1\)
\(\Rightarrow2-\left(\dfrac{1}{197}+\dfrac{1}{198}\right)>1\)
\(\Rightarrow A>1\)
Vì \(A>1;B< 1\) nên \(A>B\).
-249+197+248+(-197)
=(-249+248)+(-197+197)
=(-1)+0
=-1
= (-249 + 248) + (197 + -197)
= -1 + 0
= -1