a: Trên tia Ox, ta có: OA<OB

nên A nằm giữa O và B

b: ta có: A nằm giữa O và B

=>OA+AB=OB

=>AB+3=6

=>AB=3(cm)

ta có: A nằm giữa O và B

mà OA=AB(=3cm)

nên A là trung điểm của OB

c: Trên tia đối của tia Ax, ta có: AO<AM

nên O nằm giữa A và M

=>AO+OM=AM

=>OM+3=6

=>OM=3(cm)

=>OM=OA(=3cm)

có cần đoạn thẳng không bạn ?

giải giúp mình đang cần

\(x+10\%\cdot x+20\%\cdot x+30\%\cdot x=4,8\)

=>\(x\left(1+0,1+0,2+0,3\right)=4,8\)

=>\(x\cdot1,6=4,8\)

=>\(x=4.8:1.6=3\)

Số tiền Mai tiết kiệm trong 1 ngày là:

\(20000-12000-5000=3000\left(đồng\right)\)

Số tiền Mai tiết kiệm sau 15 ngày là \(3000\cdot15=45000\left(đồng\right)\)

=>Chọn B

42x(13+78)+9x(87+22)x5

= 42 x 92 + 9 x 109 x 5

= 3864 + 981 x 5

= 3864 + 4905

= 8769

\(42\times\left(13+78\right)+9\times\left(87+22\right)\times5\)

\(=42\times\left(13+87\right)+9\times\left(78+22\right)\times5\)

\(=42\times100+9\times100\times5\)

\(=42\times100+45\times100\)

\(= \left(42+45\right)\times100\)

\(=87\times100\)

\(=8700\)

a: \(\left(-25,5\right):5=\dfrac{-25,5}{5}=-5,1\)

b: \(1\dfrac{3}{2}-0,25+\dfrac{10}{9}\cdot\dfrac{3}{5}\)

\(=\dfrac{5}{2}-0,25+\dfrac{30}{45}\)

\(=\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{2}{3}=\dfrac{9}{4}+\dfrac{2}{3}=\dfrac{27}{12}+\dfrac{8}{12}=\dfrac{35}{12}\)

\(a,\left(-25,5\right):5\)

\(=\left(-5,1\right)\)

\(b,1\dfrac{3}{2}-0,25+\dfrac{10}{9}.\dfrac{3}{5}\)

\(=\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{10}{9}.\dfrac{3}{5}\)

\(=\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{2}{3}\)

\(=\dfrac{10}{4}-\dfrac{1}{4}+\dfrac{2}{3}\)

\(=\dfrac{9}{4}+\dfrac{2}{3}\)

\(=\dfrac{35}{12}\)

50000-10-12-...-200-201

= 50000- ( 10+12+...+200) -201

= 50000- 10080- 201

= 50000- ( 10080 +201)

= 50000-10281

= 39719 

a) \(\dfrac{-3}{7}+\dfrac{4}{21}=\dfrac{-3\cdot3}{21}+\dfrac{4}{21}=\dfrac{-9}{21}+\dfrac{4}{21}=\dfrac{-9+4}{21}=-\dfrac{5}{21}\)

b) \(\left(-0,346\right)+\left(-1,2\right)=-1,546\)

a: \(\dfrac{-3}{7}+\dfrac{4}{21}=-\dfrac{9}{21}+\dfrac{4}{21}=-\dfrac{5}{21}\)

b: \(\left(-0,346\right)+\left(-1,2\right)=-\left(1,2+0,346\right)=-1,546\)

a: \(1,2756\simeq1\)

b: \(\dfrac{x-2}{7}=\dfrac{6}{21}\)

=>\(\dfrac{x-2}{7}=\dfrac{2}{7}\)

=>x-2=2

=>x=2+2=4

a) 1,2756 ≃ 1 

b) 

\(\dfrac{x-2}{7}=\dfrac{6}{21}\\ =>x-2=\dfrac{6}{21}\cdot7=\dfrac{6}{3}=2\\ =>x=2+2=4\)

\(\dfrac{3}{5}:1,5+20\%-3\dfrac{1}{4}\)

\(=\dfrac{3}{5}:\dfrac{3}{2}+\dfrac{1}{5}-\dfrac{13}{4}\)

\(=\dfrac{2}{5}+\dfrac{1}{5}-\dfrac{13}{4}=\dfrac{3}{5}-\dfrac{13}{4}=\dfrac{12}{20}-\dfrac{65}{20}=\dfrac{-53}{20}\)

\(\dfrac{3}{5}:1,5+20\%-3\dfrac{1}{4}\\ =\dfrac{3}{5}:\dfrac{3}{2}+\dfrac{1}{5}-\dfrac{13}{4}\\ =\dfrac{3}{5}\cdot\dfrac{2}{3}+\dfrac{1}{5}-\dfrac{13}{4}\\ =\dfrac{2}{5}+\dfrac{1}{5}-\dfrac{13}{4}\\ =\dfrac{3}{5}-\dfrac{13}{4}\\ =\dfrac{12}{20}-\dfrac{65}{20}\\ =\dfrac{-53}{20}\)

\(a.8,2+\left(-3,5\right)+50+3,5+\left(-8,2\right)\\ =\left(8,2-8,2\right)+\left(3,5-3,5\right)+50\\ =0+0+50\\ =50\\ b.\dfrac{8}{15}\cdot\dfrac{-45}{4}\\ =\dfrac{4\cdot2}{15}\cdot\dfrac{15\cdot-3}{4}\\ =2\cdot-3\\ =-6\)

a: \(8,2+\left(-3,5\right)+50+3,5+\left(-8,2\right)\)

\(=\left(8,2-8,2\right)+\left(3,5-3,5\right)+50\)

=0+0+50

=50

b: \(\dfrac{8}{15}\cdot\dfrac{-45}{4}=\dfrac{8}{4}\cdot\dfrac{-45}{15}=2\cdot\left(-3\right)=-6\)