1,D

2,B

3,B

4,C

1,D

2,A

3,A

4,D

5,C

\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)

\(=\left(\frac{12}{8}+\frac{3}{8}\right)+\left(\frac{12}{128}+\frac{3}{128}\right)+\frac{3}{512}\)

\(=\frac{15}{8}+\frac{15}{128}+\frac{3}{512}\)

\(=\frac{240}{128}+\frac{15}{128}+\frac{3}{512}\)

\(=\frac{255}{128}+\frac{3}{512}\)

\(=\frac{1020}{512}+\frac{3}{512}\)

\(=\frac{1023}{512}\)

\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}=\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)
                                                        \(=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{8}-\frac{3}{8}+\frac{3}{16}-\frac{3}{16}+\frac{3}{32}\)
                                                         \(=3+\frac{3}{32}=\frac{3.32}{32}+\frac{3}{32}=\frac{96+3}{32}=\frac{99}{32}\)

;-; tách ra đuy bn hìn oi 

a: 7/9=7/9; 2/3=6/9

b: 4/10=8/20

11/20=11/20

c: 4/25=12/75

3/75=3/75

d: 4/25=16/100

5/100=5/100

e: 17/60=17/60

4/5=48/60

g: 5/9=20/36

7/36=7/36

h: 2/10=2/10

9/5=18/10

i: 53/16=53/16

3/4=12/16

a: 

\(MSC:9\\ \dfrac{7}{9};\dfrac{2}{3}=\dfrac{2\times3}{3\times3}=\dfrac{6}{9}\)

b: 

\(MSC:20\\ \dfrac{4}{10}=\dfrac{4\times2}{10\times2}=\dfrac{8}{20}\)

c:

\(MSC:75\\ \dfrac{4}{25}=\dfrac{4\times3}{25\times3}=\dfrac{12}{75}\)

d:

\(MSC:100\\ \dfrac{4}{25}=\dfrac{4\times4}{25\times4}=\dfrac{16}{100}\)

e:

\(MSC:60\\ \dfrac{4}{5}=\dfrac{4\times12}{5\times12}=\dfrac{28}{60}\)

À nhầm : BALLOON

B..LON => BALLON(bóng bay)

Sorry

\(\dfrac{2}{5}-\left|\dfrac{1}{2}-x\right|=6\)

\(\Leftrightarrow\left|\dfrac{1}{2}-x\right|=\dfrac{2}{5}-6\)

\(\Leftrightarrow\left|\dfrac{1}{2}-x\right|=-\dfrac{28}{5}\)( vô lý do \(\left|\dfrac{1}{2}-x\right|\ge0\forall x\))

Vậy \(x\in\left\{\varnothing\right\}\)

\(\Rightarrow\left|\dfrac{1}{2}-x\right|=\dfrac{2}{5}-6=-\dfrac{28}{5}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-x=-\dfrac{28}{5},\forall\dfrac{1}{2}-x\ge0\\\dfrac{1}{2}-x=\dfrac{28}{5},\forall\dfrac{1}{2}-x< 0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{61}{10},\forall x\le\dfrac{1}{2}\left(loại\right)\\x=-\dfrac{51}{10},\forall x>\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\Rightarrow x\in\varnothing\)

bạn qua online Math hỏi xem

e) từ 5 đến 99 có (99-5):2 +1=48 số

99-97+95-93+91-89+..............+7-5+3

=(99-97)+(95-93)+(91-89)+..............+(7-5)+3

=2+2+2+.............+2+3

=2x48+3

=96+3

=99

các câu còn lại tí nữa mình giải sau

Đặt  99-97+95-93+91-89+....+7-5+3 là A

Từ 3 tới 99 có (99-3):3+1= 33 số.Ta nhóm các số như sau:

            A =  (99-97)+(95-93)+(91-89)+....+(7-5)+3  [có 16 cặp và 1 số 3]

               =       2     +    2      +    2     + ....+  2  +  3

               =  2.16+3   =35