72+x-1-(342)=0 

x-271 = 0 

x = 271 

học tốt

ko chắc nhé bn

7^2+x -1=342

7^2+x = 342 + 1

7^2+x = 343 

7^2+x = 7³ 

x = 3 - 2 

x = 1

quên làm bài 2

| x - | 6 - 8 | | = 16

| x - 2 | = 16

\(\Rightarrow\orbr{\begin{cases}x-2=16\\x-2=-16\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=16+2\\x=-16+2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=18\\x=-14\end{cases}}\)

a) 342 - ( |27 - 36| + 342)

= 342 - ( | -9 | + 342 )

= 342 - ( 9 + 342 )

= 342 - 9 - 342

= -9

b) |5 - |7 - ( 40 - |-421|)|| 

= | 5 - | 7 - ( 40 - 421 ) ||

= | 5 - | 7 + 381 | |

= | 5 - 388 |

= 383

c)  | ( 7 - 5 )² . | -2³ | |

= | 2² . 8 |

= 32

d)| -16| - ( 4 - |18 - 28| )

16 - ( 4 - 10 )

= 16 - 4 + 10

= 22

ta tách ra :

( 342 + 45 : x ) : 3 = 1756 - 703

.......................... = 1053

342 : 3 + 45 : x : 3 = 1053

114 + 15 : x = 1053 

15 : x = 1053 - 114 

15 : x = 939

      x = 15 : 939

      x = \(\frac{15}{939}\)= 0,01597444089 bạn nhé

=15/939

a. 312; 318; 324; 330; 336.

b. 165; 180; 195; 210.

bạn ơi 

đề bài là tìm x nhưng sao ở trong bài lại y z

mk ko hiểu

\(x\left(x+1\right)=156\)

\(\Rightarrow x^2+x=156\)

\(\Rightarrow x^2+x-156=0\)

\(\Rightarrow x^2+13x-12x-156=0\)

\(\Rightarrow x\left(x+13\right)-12\left(x+13\right)=0\)

\(\Rightarrow\left(x+13\right)\left(x-12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=12\\x=-13\end{matrix}\right.\)

___________________

\(x\left(x+1\right)=342\)

\(\Rightarrow x^2+x=342\)

\(\Rightarrow x^2+x-342=0\)

\(\Rightarrow x^2+19x-18x-342=0\)

\(\Rightarrow x\left(x+19\right)-18\left(x+19\right)=0\)

\(\Rightarrow\left(x+19\right)\left(x-18\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-19\\x=18\end{matrix}\right.\)

__________________

\(x\left(x+1\right)=650\)

\(\Rightarrow x^2+x=650\)

\(\Rightarrow x^2-x+650=0\)

\(\Rightarrow x^2+26x-25x-650=0\)

\(\Rightarrow x\left(x+26\right)-25\left(x+26\right)=0\)

\(\Rightarrow\left(x+26\right)\left(x-25\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-26\\x=25\end{matrix}\right.\)

______________________

\(x\left(x+1\right)=380\)

\(\Rightarrow x^2+x=380\)

\(\Rightarrow x^2+x-380=0\)

\(\Rightarrow x^2+20x-19x-380=0\)

\(\Rightarrow x\left(x+20\right)-19\left(x+20\right)=0\)

\(\Rightarrow\left(x+20\right)\left(x-19\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-20\\x=19\end{matrix}\right.\)

a, \(x\).(\(x\) + 1) = 156

156 = 2².3.13 = 12.13

Vậy \(x\).(\(x\) + 1) = 12.13

Vậy \(x\) = 12

b, \(x.\)(\(x\) + 1) =  342 

    342 = 2.3².19 = 18.19

    \(x\).(\(x+1\)) = 18.19

    \(x\) =   18

c, \(x\).(\(x\) + 1) = 650 

    650 = 2.5².13 = 25.26

    \(x\).(\(x\) +1) = 25.26

    \(x\) = 25

d, \(x\).(\(x\) +1) = 380 

    380 = 2².5.19 = 19.20

      \(x\).(\(x\) + 1) = 19.20

      \(x\)    = 19

1. 

\(PTK_{CuSO_4}=64+32+16.4=160\left(đvC\right)\)

\(PTK_{5CaCO_3}=5\left(40+12+16.3\right)=500\left(đvC\right)\)

\(PTK_{Ca\left(OH\right)_2}=40+\left(16+1\right).2=74\left(đvC\right)\)

2.

Theo đề, ta có:

\(d_{\dfrac{X}{Mg}}=\dfrac{M_X}{M_{Mg}}=\dfrac{M_X}{24}=\dfrac{4}{3}\left(lần\right)\)

=> M_X = 32(g)

Vậy X là lưu huỳnh (S)

3. 

Ta có: \(PTK_{Al_x\left(SO_4\right)_3}=27.x+\left(32+16.4\right).3=342\left(đvC\right)\)

=> x = 2

Bài 1.Phân tử khối các chất:

    \(CuSO_4\)\(\Rightarrow64+32+4\cdot16=160\left(đvC\right)\)

    \(CaCO_3\Rightarrow40+12+3\cdot16=100\left(đvC\right)\)

    \(Ca\left(OH\right)_2\Rightarrow40+16\cdot2+2=74\left(đvC\right)\)

Bài 2.Theo bài: \(\overline{M_X}=\dfrac{4}{3}\overline{M_{Mg}}=\dfrac{4}{3}\cdot24=32\left(đvC\right)\)

     Vậy X là lưu huỳnh.KHHH: S.

Bài 3. \(Al_x\left(SO_4\right)_3\) \(\Rightarrow27x+3\cdot\left(32+4\cdot16\right)=342\Leftrightarrow x=2\)