câu cuối là 999,7650928....

rồi mà, câu 3 là 100 

khó quá tui chưa học

    4.[-2(8:4) + 15(-3) - (-12)]

= 4.[-2.2 - 45 + 12]

= 4.[-4 - 45 + 12]

= 4.[-49 + 12]

= 4.[-37]

= - 148

    3.(25 : 5 - 14 : 2) - 5.(6:2)

= 3.(5 - 7) - 5.3

= 3.(-2) - 15

= - 6 - 15

= -21

   [-15:(-3)] - 3.[2.(5 - 9):3]

= 5 - 3.[(2.-4):3]

= 5 - 3.[-8:3]

= 5 + 24:3

= 5 + 8

= 13

`1)1/2:2/3 .... 2/3 : 1/2`

`=>1/2xx3/2 .... 2/3xx2`

`=>3/4 .... 4/3`

Vì `3/4 < 1` và `4/3>1` 

`=>3/4<4/3`

__

`4/7:2/5 ... 4/7 : 3/5`

`=>4/7xx5/2....4/7xx5/3`

`=>20/14...20/21`

`=>10/7...20/21`

Vì `10/7>1` và `20/21<1` 

`=>10/7>20/21`

__

`4/15:4/7....2/5xx10/3`

`=>4/15xx7/4...20/15`

`=>7/15...20/15`

Vì `7<20` nên `7/15<20/15`

__

`5/6...15/18-11/18`

`=>5/6...4/18`

Ta có : MSC : `18`

`5/6 = 15/18`

Vì `15>4` nên `5/6 > 4/18`

`2)2/3=(2xx6)/(3xx6)=12/18`

`7/9=(7xx7)/(9xx7)=49/63`

`6/5=(6xx3)/(5xx3)=18/15`

`2/3=(2xx5)/(3xx5)=10/15`

`5/9=(5xx5)/(9xx5)=25/45`

`49/56=(49:7)/(56:7)=7/8`

`6/8=(6xx7)/(8xx7)=42/56`

`2/9=(2xx7)/(9xx7)=14/63`

`49/56=(49:7)/(56:7)=7/8`

Ta tính nhẩm như sau:

3x4=12     2x5=10    5x3=15     4x2=8

12:3=4      10:2=5    15:3=5      8:2=4

12:4=3      10:5=2     15:5=3     8:4=2

3x4 = 12               2x5 = 10        5x3 = 15                  4x2 =8

12 : 3 = 4              10 : 2 = 5       15 : 3 = 5                8 : 2 = 4

12 : 4 =3               10 :5 = 2        15 : 5 = 3                8 : 4 = 2

a: \(\dfrac{16}{15}\cdot\dfrac{-5}{14}\cdot\dfrac{54}{24}\cdot\dfrac{56}{21}\)

\(=\dfrac{16}{24}\cdot\dfrac{-5}{15}\cdot\dfrac{54}{21}\cdot\dfrac{56}{14}\)

\(=\dfrac{2}{3}\cdot\dfrac{-1}{3}\cdot\dfrac{18}{7}\cdot4\)

\(=\dfrac{-2\cdot2\cdot4}{7}=-\dfrac{16}{7}\)

b: \(\dfrac{7}{3}\cdot\dfrac{-5}{2}\cdot\dfrac{15}{21}\cdot\dfrac{4}{-5}\)

\(=\dfrac{7}{3}\cdot\dfrac{5}{7}\cdot\dfrac{5}{2}\cdot\dfrac{4}{5}\)

\(=\dfrac{5}{3}\cdot\dfrac{4}{2}=\dfrac{5}{3}\cdot2=\dfrac{10}{3}\)

c: \(\left(-\dfrac{2}{5}+\dfrac{1}{3}\right)\left(\dfrac{3}{2}-\dfrac{3}{7}\right)\)

\(=\dfrac{-2\cdot5+3}{15}\cdot\dfrac{3\cdot7-3\cdot2}{14}\)

\(=\dfrac{-7}{15}\cdot\dfrac{15}{14}=\dfrac{-7}{14}=-\dfrac{1}{2}\)

d: \(\left(\dfrac{1}{2}-\dfrac{1}{3}\right)\left(5-\dfrac{1}{4}\right)\)

\(=\dfrac{3-2}{6}\cdot\dfrac{20-1}{4}\)

\(=\dfrac{1}{6}\cdot\dfrac{19}{4}=\dfrac{19}{24}\)

e: \(\left(\dfrac{4}{5}\right)^2=\dfrac{4^2}{5^2}=\dfrac{16}{25}\)

\(a,\dfrac{3}{4}-\dfrac{2}{5}+\dfrac{5}{12}=\dfrac{45}{60}-\dfrac{24}{60}+\dfrac{25}{60}=\dfrac{46}{60}=\dfrac{23}{30}\\ b,\dfrac{15}{7}-\dfrac{4}{3}-\dfrac{2}{9}=\dfrac{135}{63}-\dfrac{84}{63}-\dfrac{14}{63}=\dfrac{37}{63}\\ c,\dfrac{3}{5}+\dfrac{2}{5}\times\dfrac{4}{5}=\dfrac{15}{25}+\dfrac{8}{25}=\dfrac{23}{25}\\ d,\dfrac{4}{5}-\left(\dfrac{6}{15}+\dfrac{1}{5}\right)=\dfrac{4}{5}-\dfrac{1}{5}-\dfrac{2}{5}=\dfrac{1}{5}\)

A 23/30

B 37/63

C 23/25

D 22/5

a) 12 + 3 = 15     15 + 4 = 19     8 + 2 = 10     14 + 3 = 17

15 - 3 = 12     19 - 4 = 15     10 - 2 = 8     17 - 3 = 14

b) 11 + 4 + 2 = 17     19 - 5 - 4 = 10     14 + 2 - 5 = 11

1) \(2^3\times x-5^2\times x=2\times\left(5^2+2^2\right)-33\)

\(x\times\left(2^3-5^2\right)=2\times\left(25+4\right)-33\)

\(x\times\left(8-25\right)=2\times29-33\)

\(x\times-17=25\)

\(x=-\dfrac{25}{17}\)

2) \(15\div\left(x+2\right)=\left(3^3+3\right)\div1\)

\(15\div\left(x+2\right)=\left(27+3\right)\div1\)

\(15\div\left(x+2\right)=30\div1\)

\(15\div\left(x+2\right)=30\)

\(x+2=\dfrac{1}{2}\)

\(x=-\dfrac{3}{2}\)

3) \(20\div\left(x+1\right)=\left(5^2+1\right)\div13\)

\(20\div\left(x+1\right)=\left(25+1\right)\div13\)

\(20\div\left(x+1\right)=26\div13\)

\(20\div\left(x+1\right)=2\)

\(x+1=20\div2\)

\(x+1=10\)

\(x=9\)

4) \(320\div\left(x-1\right)=\left(5^3-5^2\right)\div4+15\)

\(320\div\left(x-1\right)=\left(125-25\right)\div4+15\)

\(320\div\left(x-1\right)=100\div4+15\)

\(320\div\left(x-1\right)=25+15\)

\(320\div\left(x-1\right)=40\)

\(x-1=8\)

\(x=9\)

5) \(240\div\left(x-5\right)=2^2\times5^2-20\)

\(240\div\left(x-5\right)=4\times25-20\)

\(240\div\left(x-5\right)=100-20\)

\(240\div\left(x-5\right)=80\)

\(x-5=30\)

\(x=35\)

6) \(70\div\left(x-3\right)=\left(3^4-1\right)\div4-10\)

\(70\div\left(x-3\right)=\left(81-1\right)\div4-10\)

\(70\div\left(x-3\right)=80\div4-10\)

\(70\div\left(x-3\right)=20-10\)

\(70\div\left(x-3\right)=10\)

\(x-3=7\)

\(x=10\)

a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)

b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)

c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)