5/6+ 1/2 x 4/3

= 5/6+ 2/3

= 5/6+4/6

= 11/6

5/6- 3/8

= 20/24- 9/24

= 11/24

2+9+5+52+41+61+1+41+1+5=218

= MÁY TÍNH 

CHO MÌNH 5

= 218

ai k minh min k lai

a, (-41) x 324 - (-41) x 24

= (-41) x (324 - 24)

= (-41) x 300

=  -12300

b, (-1)² + (-1)³ + (-1)⁴ + ... + (-1)²⁰

= 1 - 1 + 1 - 1 + ... + 1 ( có 20 số 1 )

= (1 + 1 +...+ 1) - (1 + 1 +...+ 1)

         10 số 1              10 số 1

= 0

a) (-41) . 324 - (-41) . 24

= (-41) ( 324 - 24 )

= (-41) . 300

= -12 300

=))

a,(-41)*324-(-41)*24

=(-41)*(324-24)

=(-41)*300

=(-12300)

b,\(\left(-1^2\right)+\left(-1^3\right)+...+\left(-1^{20}\right)\)

=\(\left\{\left(-1^2\right)+\left(-1^3\right)\right\}+\left\{\left(-1^4\right)+\left(-1^5\right)\right\}+...+\left\{\left(-1^{19}\right)\right\}+\left\{\left(-1^{20}\right)\right\}\)

=[(-1)+1]+...+[(-1)+1]

=0

c: \(=0.5\cdot10-9=5-9=-4\)

c: 

\(\frac{15}{41}+\frac{-138}{41}\le x< \frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)

\(\Leftrightarrow-3\le x< 1\)

\(\Leftrightarrow x\in\left\{-3;-2;-1;0\right\}\)

\(\frac{15}{41}+\frac{-138}{41}\le x< \frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)

\(\Rightarrow\frac{15+(-138)}{41}\le x< \frac{1\cdot3}{6}+\frac{1\cdot2}{6}+\frac{1}{6}\)

\(\Rightarrow\frac{-123}{41}\le x< \frac{3}{6}+\frac{2}{6}+\frac{1}{6}\)

\(\Rightarrow-3\le x< 1\Leftrightarrow x\in\left\{-3;-2;-1;0\right\}\)

\(\frac{15}{41}+\frac{-138}{41}\le x< \frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)

\(\Rightarrow-3\le x< 1\)

\(\Rightarrow x\in\left\{-3;-2;-1;0\right\}\)

\(B=\frac{23^{41}+1}{23^{42}+1}\)

Vì B < 1

\(\Rightarrow B=\frac{23^{41}+1}{23^{42}+1}< \frac{23^{41}+1+22}{23^{42}+1+22}=\frac{23^{41}+23}{23^{42}+23}=\frac{23(23^{40}+1)}{23\left(23^{41}+1\right)}=\frac{23^{40}+1}{23^{41}+1}=A\)

P/s: Hoq chắc

ta có 

\(B=\frac{23^{41}+1}{23^{42}+1}< \frac{23^{41}+1+22}{23^{42}+1+22}=\frac{23^{41}+23}{23^{42}+23}=\frac{23\left(23^{40}+1\right)}{23\left(23^{41}+1\right)}=\frac{23^{40}+1}{23^{41}+1}=A\)

\(\Rightarrow B< A\)

Cuối cùng kết luận hộ nhé :v

`a,`

`31/23-(7/32+8/23)`

`=31/23-7/32-8/23`

`=(31/23-8/23)-7/32`

`=1-7/32=25/32`

`b,`

`38/45-(8/45-17/51-3/11)`

`=38/45-8/45+17/51+3/11`

`= (38/45-8/45)+17/51+3/11`

`=2/3+17/51+3/11`

`=1+3/11=14/11`

`c,`

`(1/3+12/67+13/41)-(79/67-28/41)`

`= 1/3+12/67+13/41-79/67+28/41`

`= 1/3+(12/67-79/67)+(13/41+28/41)`

`= 1/3+(-1)+1=1/3+(-1+1)=1/3+0=1/3`

`d,`

`1/5+(-1/6)+1/7+(-1/8)+1/9+1/8+(-1/7)+1/6+(-1/5)`

`= (1/5+ -1/5)+(-1/6+1/6)+(1/7+ -1/7)+(-1/8 +1/8)+1/9`

`= 0+0+0+0+1/9=1/9 .`

Với 1/a + 1/b + 1/c = 1/(a + b + c) thì không thể nào có ĐK : a = b = c vì nó sẽ như sau :

1/a + 1/b + 1/c = 1/(a + b + c)

=> 3/a = 3/b = 3/c = 1/(a x 3) = 1/(b x 3) = 1/(c x 3)  (rất vô lý)

Với 1/(a + b + c) thì phần tử rất nhỏ .

=> Dữ liệu không tồn tại.

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