\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

   =  \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

    = \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)

     = \(\frac{1}{4}+\frac{1}{2}\)

      =  \(\frac{3}{4}\)

b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)

    =\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)

   = \(-\frac{35}{27}+\frac{47}{21}\)

   =        \(\frac{178}{189}\)

c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)

  = \(\frac{117}{13}-\frac{311}{65}\)

 =       \(\frac{274}{65}\)

d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)

= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)

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

=     \(\frac{1}{3}+\frac{5}{2}\)

=         \(\frac{17}{6}\)

a.-1,75-(-\(\dfrac{1}{9}\)-2\(\dfrac{1}{8}\))
-1,75-\(\dfrac{1}{9}+\dfrac{17}{8}\)
\(-\dfrac{7}{4}-\dfrac{1}{9}+\dfrac{17}{8}\)
\(\dfrac{-126}{72}-\dfrac{8}{72}+\dfrac{153}{72}\)
=\(\dfrac{19}{72}\)

b.\(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\dfrac{21}{8}+\dfrac{1}{3}\)
\(\dfrac{-2}{24}-\dfrac{63}{24}+\dfrac{64}{24}\)
=\(\dfrac{-1}{24}\)

\(\dfrac{-5}{6}-\left(\dfrac{-3}{8}+\dfrac{1}{10}\right)\)
\(\dfrac{-5}{6}-\dfrac{-3}{8}-\dfrac{1}{10}\)
 

4:

a: =4/15-2,9+11/15=1-2,9=-1,9

b: \(=-36,75+3,7-63,25+6,3=10-100=-90\)

c: \(=6,5+3,5-\dfrac{10}{17}-\dfrac{7}{17}=10-1=9\)

d: \(=\dfrac{13}{25}\left(-39,1-60,9\right)=\dfrac{13}{25}\left(-100\right)=-52\)

e: =-5/12-7/12-3,7-6,3=-1-10=-11

f: =2,8(-6/13-7/13)-7,2=-2,8-7,2=-10

cậu tách ra từng bài thôi nha.

1) 2,75 - 5/6 × 2/5 = 2,75 - (5/6) × (2/5) = 2,75 - 1/3 = 2,75 - 0,33 = 2,42

2) 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 3/4) - 3/5 = 1,25 - (5/6 - 9/12) - 3/5 = 1,25 - (10/12 - 9/12) - 3/5 = 1,25 - 1/12 - 3/5 = 1,25 - 0,08 - 0,6 = 1,25 - 0,68 = 0,57

3) 4/9 × 0,75 + 8/5 + 3,125 = (4/9) × 0,75 + 8/5 + 3,125 = 0,44 + 8/5 + 3,125 = 0,44 + 1,6 + 3,125 = 0,44 + 4,725 = 5,165

4) 1,125 - 4/7 - 0,12 = 1,125 - (4/7) - 0,12 = 1,125 - 0,57 - 0,12 = 0,435 - 0,12 = 0,315

5) (1/3 + 0,4) × 3,5 + (1/6 + 0,75) × 6/5

3: 

\(=\dfrac{1}{7}\cdot\dfrac{3}{5}\cdot\dfrac{5}{6}\cdot\dfrac{5}{8}=\dfrac{1}{7}\cdot\dfrac{1}{2}\cdot\dfrac{5}{8}=\dfrac{5}{112}\)

4:

=>2/3:x=-2-1/3=-7/3

=>x=-2/3:7/3=-2/7

5:

AC=CB=12/2=6cm

IB=6/2=3cm

Bài 4 là:

\(\dfrac{1}{2.5}\) + \(\dfrac{1}{5.8}\) +....+\(\dfrac{1}{92.95}\) + \(\dfrac{1}{95.98}\)

                  Đúng không bạn

Bạn chỉ nên đưa câu khó lên vì làm cả đề thì ko làm kịp đc

`@` `\text {Ans}`

`\downarrow`

`a.`

`A=(1/2-7/13-1/3)+(-6/13+1/2+1 1/3)`

`= 1/2 - 7/13 - 1/3 - 6/13 + 1/2 + 1 1/3`

`= (1/2 + 1/2) + (-7/13 - 6/13) + (-1/3 + 1 1/3) `

`= 1 - 1 + 1`

`= 1`

`b.`

`B=0,75+2/5+(1/9-1 1/2+5/4)`

`= 3/4 + 2/5 + 1/9 - 3/2 + 5/4`

`= (3/4+5/4)+ 1/9 + 2/5 - 3/2`

`= 2 + 1/9 - 11/10`

`= 19/9 - 11/10`

`= 91/90`

`c.`

`(-5/9).3/11+(-13/18).3/11`

`= 3/11*[(-5/9) + (-13/18)]`

`= 3/11*(-23/18)`

`= -23/66`

`d.`

`(-2/3).3/11+(-16/9).3/11`

`= 3/11* [(-2/3) + (-16/9)]`

`= 3/11*(-22/9)`

`= -2/3`

`e.`

`(-1/4).(-2/13)-7/24.(-2/13)`

`= (-2/13)*(-1/4-7/24)`

`= (-2/13)*(-13/24)`

`= 1/12`

`f.`

`(-1/27).3/7+(5/9).(-3/7)`

`= 3/7*(-1/27 - 5/9)`

`= 3/7*(-16/27)`

`= -16/63`

`g.`

`(-1/5+3/7):2/11+(-4/5+4/7):2/11`

`=[(-1/5+3/7)+(-4/5+4/7)] \div 2/11`

`= (-1/5+3/7 - 4/5 + 4/7) \div 2/11`

`= [(-1/5-4/5)+(3/7+4/7)] \div 2/11`

`= (-1+1) \div 2/11`

`= 0 \div 2/11 = 0`

a ) 18^2 : 9^2 .2^2 +12-12:3

= 4^2 :3 = 16/3

b) 5! +2^10 :2^8

= 120 + 4 = 124

c)28 :[ 117 - ( 117- 4) ] . 7 + 5^3 :125

= 28 : 4 * 7 + 1 = 50

d)3^3. 3^5.3 +2^7 :64

= 3^9 + 2^7 : 2^6 = 3^9 + 1 (tính nhé)

bài 2: tìm x thuôc N biết 

a) (2x -2^5). 5^3=2^3.5^3

2x - 2^5 = 2^3 

2x = 40 

x=20

b) 2^2x -4= 72:3^2

2^2 * x = 12 

x= 3

c)3^x + 3^x+1 =108

\(3^x+3^{x+1}=108\) (là thế này ak??)

3^x (1+3) = 108

3^x = 27

x=3 

d)(x -3).(3x+5) =0

\(\Rightarrow\orbr{\begin{cases}x-3=0\\3x+5\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=-\frac{5}{3}\end{cases}}}\)