\(\frac{27}{23}+\frac{-4}{23}+\frac{1}{2}+\frac{-4}{8}< x< \frac{7}{3}+\frac{13}{41}+\frac{28}{41}\)

\(\Rightarrow\frac{27}{23}-\frac{4}{23}+\frac{1}{2}-\frac{4}{8}< x< \frac{7}{3}+\frac{13}{41}+\frac{28}{41}\)

\(\Rightarrow1+0< x< \frac{7}{3}+\frac{3}{3}\)

\(\Rightarrow1< x< \frac{10}{3}\)

\(\Rightarrow1< x< 3,333333333\)

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

Vậy : ....

ta co : \(\frac{27}{23}+\frac{-4}{23}+\frac{1}{2}+\frac{-4}{8}< x< \frac{7}{3}+\frac{13}{41}+\frac{28}{41}\)

  =>  \(1< x< \frac{10}{3}\)

vi x la so nguyen => \(1< x\le3\)

con lai ban tu lam

A.-9/2

b.-5/7

C.5/287

D.302/105

nêu cách tính nha bạn 

\(\text{(x + 2)^3 - 41 = 23}\\ \left(x+2\right)^3=23+41\\ \left(x+2\right)^3=64\\ x+2=4\\ x=4-2\\ x=2\)

Lời giải:
$(x+2)^3-41=23$

$\Leftrightarrow (x+2)^3=23+41=64=4^3$

$\Leftrightarrow x+2=4$

$\Leftrightarrow x=2$

( x + 2 )³ = 41 + 23

( x + 2 )^{3 =} 64

( x + 2 )^{3 =} 4³

=> x + 2 = 4

x = 4 - 2

x = 2

Vậy x =2

\(\left(2+x\right)^3-23=41\\ \left(2+x\right)^3=64\\ 2+x=4\\ x=2\)

2

Mik cần lời giải chi tiết nha

\(a,x-36:18=12-15\\ \Rightarrow x-2=-3\\ \Rightarrow x=-1\\ b,92-\left(17+x\right)=72\\ \Rightarrow17+x=20\\ \Rightarrow x=3\\ c,720:\left[41-\left(2x+5\right)\right]=40\\ \Rightarrow41-\left(2x+5\right)=18\\ \Rightarrow2x+5=23\\ \Rightarrow2x=18\\ \Rightarrow x=9\\ d,\left(x+2\right)^3-23=41\\ \Rightarrow\left(x+2\right)^3=64\\ \Rightarrow\left(x+2\right)^3=4^3\\ \Rightarrow x+2=4\\ \Rightarrow x=2\)

b: =>x+17=20

hay x=3

d: =>x+2=4

hay x=2

( 2 + x )³ - 23 = 41

( 2 + x )³ = 64 = 4³

=>  2 + x = 4

x = 2

lam hộ tao đã