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Tính hợp lí
a,15.(27+ 18 +6 ) + 15.( 23 +12)
= 15(27+18+6+23+12)
= 15.86
= 1290
b,24.(15 + 49 ) + 12. ( 50 + 42 )
=12.2(15+49)+12.(50+42)
= 12(30+98)+12(50+42)
=12(30+98+50+42)
=12.220
= 2640
c,53.(51+4)+53.(49+96) +53
=53(51+4)+53(49+96)+53.1
=53(51+4+49+96+1)
=53.201
=10653
d,42.( 15 +96) +6.(25 +4).7
=42(15+96)+42(25+4)
=42(15+96+25+4)
=42.140
=5880
tick đúng cho mk nha pikachu
Mn oi!!!!! Giup minh lam 2 bai nay voi!!!!!
Bài 1 :
a) \(\dfrac{42}{43}=1-\dfrac{1}{43}\)
\(\dfrac{58}{59}=1-\dfrac{1}{59}\)
Mà \(\dfrac{1}{43}>\dfrac{1}{59}\Leftrightarrow\dfrac{42}{43}< \dfrac{58}{59}\)
b) \(\dfrac{18}{31}>\dfrac{15}{31}>\dfrac{15}{37}\)
\(\Leftrightarrow\dfrac{18}{31}>\dfrac{15}{37}\)
c) \(\dfrac{53}{57}=1-\dfrac{4}{57}\)
\(\dfrac{531}{517}=1-\dfrac{40}{517}\)
Mà \(\dfrac{4}{57}=\dfrac{40}{570}>\dfrac{40}{517}\)
\(\Leftrightarrow\dfrac{53}{57}< \dfrac{531}{517}\)
Bài 1
a, cm : A = 165 + 215 ⋮ 3
A = 165 + 215
A = (24)5 + 215
A = 220 + 215
A = 215.(25 + 1)
A = 215. 33 ⋮ 3 (đpcm)
b,cm : B = 88 + 220 ⋮ 17
B = (23)8 + 220
B = 216 + 220
B = 216.(1 + 24)
B = 216. 17 ⋮ 17 (đpcm)
c, cm: C = 1 - 2 + 22 - 23 + 24 - 25 + 26 -...-22021 + 22022 : 6 dư 1
C=1+(-2+22-23+24- 25+26)+...+(-22017+22018-22019+22020-22021+22022)
C = 1 + 42 +...+ 22016.(-2 + 22 - 23 + 24 - 25 + 26)
C = 1 + 42+...+ 22016.42
C = 1 + 42.(20+...+22016)
42 ⋮ 6 ⇒ C = 1 + 42.(20+...+22016) : 6 dư 1 đpcm
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{48.49}\)
\(A< 1-\frac{1}{49}=\frac{48}{49}< \frac{48}{48}< \frac{40}{48}=\frac{5}{6}\)
5) 24*(15+49)+12*(50+42)
=24*64+12*92
=24*64+12*2*46
=24*64+24*46
=24*(64+46)
=24*110
=2640
Ta có: \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{25}\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)(đpcm)
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\) (đpcm)
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