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\(60=2^2\cdot3\cdot5;20=2^2\cdot5\)
Do đó: ƯCLN(60;80)\(=2^2\cdot5=20\)
VÌ UCLN[A:B]->A=20*A1:B=20*B1
VOI [B1,A1]=1 VA A1>B1
MA A+B=60<=>20*A1+20*B1=60
<=>20*[B1+A1]=60
<=>B1+A1=3
VI [B1:A1]=1 NEN TA CO CAC TRUONG HOP SAU:
+TH1:A=1 1=2->A=2*20=40;B1=1->B=1*20=20
+TH2:A1=1->A=1*20=20;B1=2->B=2*20=40
VAY TAT CA CAC CAP [A,B] LA :[20;40],[40;20]
30 và 45
Ta có: \(30=2\cdot3\cdot5;45=3^2\cdot5\)
Do đó: ƯCLN(30;45)\(=3\cdot5=15\)
=>ƯC(30;45)=Ư(15)={1;-1;3;-3;5;-5;15;-15}
42 và 70
TA có: \(42=2\cdot3\cdot7;70=2\cdot7\cdot5\)
Do đó: ƯCLN(42;70)\(=2\cdot7=14\)
=>ƯC(42;70)=Ư(14)={1;-1;2;-2;7;-7;14;-14}
50 và 60
Ta có: \(50=2\cdot5^2;60=2^2\cdot3\cdot5\)
Do đó: ƯCLN(50;60)\(=2\cdot5=10\)
=>ƯC(50;60)=Ư(10)={1;-1;2;-2;5;-5;10;-10}
Ta có:
a) \(A=2000\cdot2009=2000\cdot\left(2005+4\right)=2000\cdot2005+2000\cdot4\)
\(B=2004\cdot2005=\left(2000+4\right)\cdot2005=2000\cdot2005+2005\cdot4\)
Do \(2000< 2005\)
\(\Rightarrow2000\cdot4< 2005\cdot4\)
\(\Rightarrow2000\cdot2005+2000\cdot4< 2000\cdot2005+2005\cdot4\)
Vậy A < B
b) \(A=3004^2=\left(3000+4\right)\cdot3004=3000\cdot3004+3004\cdot4\)
\(B=3000\cdot3008=3000\cdot\left(3004+4\right)=3000\cdot3004+3000\cdot4\)
Do \(3004>3000\)
\(\Rightarrow3004\cdot4>3000\cdot4\)
\(\Rightarrow3000\cdot3004+3004\cdot4>3000\cdot3004+3000\cdot4\)
Vậy A > B
a)
\(\frac{8}{11}.\frac{14}{23}+\frac{9}{23}:\frac{11}{8}-\frac{8}{11} \)
\(=\frac{8}{11}.\frac{14}{23}+\frac{9}{23}.\frac{8}{11}-\frac{8}{11}\)
\(=\frac{8}{23}.(\frac{14}{23}+\frac{9}{23}-1)\)
\(=\frac{8}{23}.0\)
=0
b)
\(1,8.\frac{-20}{27}\)+(75%\(-\frac{5}{16}):3\frac{1}{2}\)
=\(\frac{9}{5}.\frac{-20}{27}+(\frac{3}{4}-\frac{5}{16}):\frac{7}{2}\)
=\(\frac{-4}{3}+\frac{1}{8}\)
=\(\frac{-29}{24}\)
Trả lời
Bài 1:
-5/80;7/16;-9/20
-5/80=-5/80
7/16=35/80
-9/20=-36/80
Bài 2:
-54/90=-6/10=-3/5=-216/360
-180/288=-20/32=-5/8=-225/360
60/-135=-12/27=-4/9=160/380
a) ƯCLN (16, 24) = 8, ƯC (16, 24) = {1; 2; 4; 8};
b) Ta có 180 = 22 . 32 . 5; 234 = 2 . 32 . 13;
ƯCLN (180, 234) = 2 . 32 = 18, ƯC (180, 234) = {1; 2; 3; 6; 9; 18};
c) Ta có 60 = 22 . 3 . 5; 90 = 2 . 32 . 5; 135 = 33 . 5. Do đó
ƯCLN (60, 90, 135) = 3 . 5 = 15; ƯC (60, 90, 135) = {1; 3; 5; 15}.
Chúc bạn học tốt! 
a) ƯCLN (16, 24) = 8, ƯC (16, 24) = {1; 2; 4; 8};
b) Ta có 180 = 22 . 32 . 5; 234 = 2 . 32 . 13;
ƯCLN (180, 234) = 2 . 32 = 18, ƯC (180, 234) = {1; 2; 3; 6; 9; 18};
c) Ta có 60 = 22 . 3 . 5; 90 = 2 . 32 . 5; 135 = 33 . 5. Do đó
ƯCLN (60, 90, 135) = 3 . 5 = 15; ƯC (60, 90, 135) = {1; 3; 5; 15}.
ƯCLN (60;20) = 20
ƯCLN (45;80)= 5
ƯCLN (24;120) = 24
\(Ư C L N \left(\right. 60 , 20 \left.\right) = 2^{2} \times 5^{1} = 4 \times 5 = 20\).
ƯCLN(24,120)=23×31=8×3=24.
a: \(60=2^2\cdot3\cdot5;20=2^2\cdot5\)
Do đó: ƯCLN(60;80)\(=2^2\cdot5=20\)
b: \(45=5\cdot3^2;80=5\cdot2^4\)
Do đó: ƯCLN(45;80)=5
c: \(24=2^3\cdot3;120=2^3\cdot3\cdot5\)
Do đó: ƯCLN(24;120)\(=2^3\cdot3=24\)
ƯCLN ( 60;20 ) = 20
ƯCLN ( 45;80 ) = 5
ƯCLN ( 24;120 ) = 24