30+{96-3.[54-2.(3²+6)]}.x =79

30+{96-3.[54-2.15]}.x=79

30+{96-3.[54-30]}.x=79

30+{96-3.24}.x=79

30+{96-72}.x=79

30+24.x=79

24.x=79-30

24x=49

x=49/24

à cho mk hỏi là x có thuộc Z hay N ko

30+{96-3.[54-2.(3²+6)]}.x =79

   30+{96-3.[54-2.(9+6)]}.x=79

        30+{96-3.[54-2.15]}.x=79

           30+{96-3.[54-30]}.x=79

                   30+{96-3.24}.x=79

                      30+{96-72}.x=79

                              30+24.x=79

                                     24.x=79-30

                                     24.x=49

                                           x=49:24

                                            x=\(\frac{49}{24}\)

\(30+\left\{96-3.\left[54-2\left(3^2+6\right)\right]\right\}.x=78\)

\(30+\left\{96-3.\left[56-2.15\right]\right\}.x=78\)

\(30+\left\{96-3.\left[56-30\right]\right\}.x=78\)

\(30+\left\{96-3.26\right\}.x=78\)

\(30+\left\{96-78\right\}.x=78\)

\(30+18.x=78\)

\(\Rightarrow18.x=48\)

\(\Rightarrow x=\frac{8}{3}\)

học tốt

a, (-0,2)² \(\times\) 5 - \(\dfrac{2^{13}\times27^3}{4^6\times9^5}\)

= 0,04 \(\times\) 5 - \(\dfrac{2^{13}\times3^9}{2^{12}\times3^{10}}\)

= 0,2 - \(\dfrac{2}{3}\)

= \(\dfrac{2}{10}\) - \(\dfrac{2}{3}\)

=  - \(\dfrac{7}{15}\)

b, \(\dfrac{5^6+2^2.25^3+2^3.125^2}{26.5^6}\)

 = \(\dfrac{5^6+4.5^6+8.5^6}{26.5^6}\)

= \(\dfrac{5^6.\left(1+4+8\right)}{26.5^6}\)

= \(\dfrac{1}{2}\)

a, (-0,2)2 $\times$ 5 - 213×27346×9546×95213×273​

= 0,04 $\times$ 5 - 213×39212×310212×310213×39​

= 0,2 - 2332​

= 210102​ - 2332​

=  - 715157​

b, 56+22.253+23.125226.5626.5656+22.253+23.1252​

 = 56+4.56+8.5626.5626.5656+4.56+8.56​

= 56.(1+4+8)26.5626.5656.(1+4+8)​

= 1221​
 

\(10^{30}vs\)\(2^{100}\)

\(10^{30}=\left(10^3\right)^{10}=1000^{10}\)

\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)

Vì \(1000^{10}< 1024^{10}=>10^{30}< 2^{100}\)

\(3^{54}vs2^{81}\)

\(3^{54}=\left(3^6\right)^9=729^9\)

\(2^{81}=\left(2^9\right)^9=512^9\)

Vì \(729^9>512^9=>3^{54}>2^{81}\)

\(3^{200}=\left(3^2\right)^{100}=9^{100}\)

\(2^{300}=\left(2^3\right)^{100}=8^{100}\)

Vì \(8^{100}< 9^{100}=>2^{300}< 3^{200}\)

12x - 10x + 54 = 86
\(\Rightarrow\)12x - 10x = 86 - 54
\(\Rightarrow\)2x = 32 
\(\Rightarrow\)x = 32 : 2
\(\Rightarrow\) x= 16
 

12x-10x+54=86

=(12-10)x+54=86

(12-10)x      =32

     2x            =32

       x            =32chia2

       x            =16

Ta có:

3/2 - 1/3 + 1/6 = 4/3

1/5 + 1/6 + 79/30 = 3

⇒ 4/3 < 2 < 3

Và 2 là số nguyên tố

Vậy có 1 số nguyên tố là x = 2 thỏa mãn đề bài

ta có:
3/2-1/3+1/6<x<1/5+1/6+79/30
=45/30-10/30+5/30<x<6/30+5/30+79/30
=40/30<x<90/30
=>4/3<x<9/3
=>x có 4 số nguyên tố thỏa mãn
Không chắc lắm nha :((

a) \(x^2-5x+6\)

\(=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)

b) \(x^2-9x+18=x^2-3x-6x+18\)

\(=x\left(x-3\right)-6\left(x-3\right)=\left(x-3\right)\left(x-6\right)\)

c) \(x^2-6x+5=x^2-x-5x+5\)

\(=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)

d) \(3x^2+5x-30=3\left(x^2+\dfrac{5x}{3}-10\right)=3\left(x^2+2.x.\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{5347}{500}\right)\)

Câu này bạn xem lại đề nha

e) \(3x^2-5x-2=3x^2-6x+x-2\)

\(3x\left(x-2\right)+x-2=\left(x-2\right)\left(3x+1\right)\)

\(\Leftrightarrow x:3+367\cdot\left(-2\right)=-60\)

=>x:3=674

hay x=2022

x= 2022