2+53
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2/6 × 60/53 + 2/6 × 46/53
= 2/6 × (60/53 + 46/53)
= 1/3 × 2
= 2/3
A=532+472+94.53
=> 532+2.47.53+472
=> (53+47)2
=> 1002
=> 10000
\(A=\frac{53^{30}}{53^{30}-6}=\frac{53^{30}-6+6}{53^{30}-6}=1+\frac{6}{53^{30}-6}\)
\(B=\frac{53^{30}-4}{53^{30}-2}=\frac{53^{30}-2+6}{53^{30}-2}=1+\frac{6}{53^{30}-2}\)
Vì 5330-6<5330-2 nên \(\frac{6}{53^{30}-6}>\frac{6}{53^{30}-2}\)
=> A>B
Ta có:
A=\(\frac{53^{30}-6+6}{53^{30}-6}\)=\(1-\frac{6}{5^{30}-6}\)
B=\(\frac{53^{30}-2+6}{53^{30}-2}\)=\(1-\frac{6}{5^{30}-2}\)
Vì 5330-6<5330-2 nên \(\frac{6}{53^{30}-6}\)>\(\frac{6}{53^{30}-2}\)
=> \(1-\frac{6}{53^{30}-6}\)\(< 1-\frac{6}{5^{30}-2}\)
Vậy A<B
a. 53 + 8 x 53 +53
=53 + 424 + 53
=477 + 53= 530
7)
53.(51+4)+53.(49+96)+53
=53.55+53.145+53
=53.(55+145+1)
=53.201
=10653
8)
42.(15+96)+6.(25+4).7
=42.111+(6.7).29
=42.111+42.29
=42.(111+29)
=42.140
=5880
9)
45.(13+78)+9.(87+22).5
=45.91+(9.5).109
=45.91+45.109
=45.(91+109)
=45.200
=9000
10)
16.(27+75)+8.(53+25).2
=16.102+(8.2).78
=16.102+16.78
=16.(102+78)
=16.180
=2880
7)
53.(51+4)+53.(49+96)+53
=53.55+53.145+53
=53.(55+145+1)
=53.201
=10653
8)
42.(15+96)+6.(25+4).7
=42.111+(6.7).29
=42.111+42.29
=42.(111+29)
=42.140
=5880
9)
45.(13+78)+9.(87+22).5
=45.91+(9.5).109
=45.91+45.109
=45.(91+109)
=45.200
=9000
10)
16.(27+75)+8.(53+25).2
=16.102+(8.2).78
=16.102+16.78
=16.(102+78)
=16.180
=2880
55
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