mình không hiểu

c) \(3^2+2^4-\left(6^8:6^6-6^2\right)< 5^x< 125\)

\(=9+16-\left(6^{8-6}-36\right)< 5^x< 5^3\)

\(=25-\left(6^2-36\right)< 5^x< 5^3\)

\(=25-\left(36-36\right)< 5^x< 5^3\)

\(=25-0< 5^x< 5^3\)

\(=25< 5^x< 5^3\)

\(=5^2< 5^x< 5^3\)

Vì \(5^2=25\) và \(5^3=125\) nên \(x\) không thể thỏa mãn đề bài

⇒ \(x\) không thỏa mãn đề bài

a,5^x=125

=>5^x=5^3

=>x=3

b,3^2x=81

=>3^2x=3^4

=>2x=4

=>x=4:2=2

c,5^2x-3-2*5^2=5^2+3

5^2x-3-50=75

5^2x-3=75+50=125

5^2x-3=5^3

=>2x-3=3

=>2x=3+3=6

=>6:2=3

k cho mk nhé

\(a,125=5\cdot5\cdot5=5^3\Leftrightarrow x=3\)

\(b,81=3\cdot3\cdot3\cdot3=3^4\Leftrightarrow2x=4\Leftrightarrow x=4:2\Leftrightarrow x=2\)

\(c,5^{2x-3}-2\cdot5^2=5^2\cdot3\)

\(\Leftrightarrow5^{2x-3}=2\cdot5^2+5^2\cdot3\)

\(\Leftrightarrow5^{2x-3}=5^2\cdot\left(2+3\right)\)

\(\Leftrightarrow5^{2x-3}=5^2\cdot5\Leftrightarrow5^{2x-3}=5^3\)

\(\Leftrightarrow2x-3=3\Leftrightarrow2x=3+3\Leftrightarrow2x=6\Leftrightarrow x=6:2\Leftrightarrow x=3\)

3⁸:3⁴+2²:2³=81,5

3.4²-2.3²=30

(4⁶.3⁴.9³):6¹²=1/9

3⁸:3⁴+2²:2³=3^8-4+2^2-3=3⁴+2^-1=81+0,5=81,5

3.4²-2.3²=3.4²-2.3.3=3.4²-6.3=3.(4²-6)=3.10=30

(4⁶.3⁴.9³):6¹²=(4⁶.3⁴.3⁶):6¹²=(4⁶.3⁴⁺⁶):6¹²=(2¹².3¹⁰):6¹²=\(\frac{2^{12}.3^{10}}{6^{12}}\)=\(\frac{1}{9}\)

\(A=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)

\(A=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)

\(A=\frac{2^2\left(5.2^{28}.3^{18}\right)-2^{29}.3^{18}}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)

\(A=\frac{2^2-1}{1-7}\)

\(A=\frac{3}{-6}=\frac{1}{-2}=-\frac{1}{2}\)

bài này ư hơi dễ đó nhìn sơ qua là biết

\(\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)

= \(\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{\left(5-7.2\right).2^{28}.3^{18}}\)

= \(\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{-9.2^{28}.3^{18}}\)

= \(\frac{\left(5.2^2-2\right).2^{28}.3^{18}}{-9.2^{28}.3^{18}}\)

= \(\frac{18.2^{28}.3^{18}}{-9.2^{28}.3^{18}}\)

= 18/-9 = -2

\(\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)

\(=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)

\(=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)

\(=\frac{2^{28}.3^{18}.\left(5.2^2-2\right)}{2^{28}.3^{18}.\left(5-7.2\right)}\)

\(=\frac{5.4-2}{5-14}\)\(=\frac{18}{-9}=-2\)

T**k mik nha!

\(A=\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^{14}}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)

\(A=\frac{5.\left(2^2\right)^9.\left(3^2\right)^9.\left(2^2\right)^6-2.\left(2^2\right)^{14}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)

\(A=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{18}}{5.2^{28}.3^{18}-7.2.2^{28}.3^{18}}\)

\(A=5.\frac{\left(2^{18}.2^{12}\right).3^{18}.2^{29}.3^{18}}{2^{28}.3^{18}.\left(5-7-2\right)}\)

\(A=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}.\left(-9\right)}\)

\(A=\frac{5.2.2^{29}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}.\left(-9\right)}\)

\(A=\frac{2^{20}.2^{18}.\left(5.2-1\right)}{2^{28}.3^{18}.\left(-9\right)}\)

\(A=\frac{2^{29}.3^{18}.9}{2^{28}.3^{18}.\left(-9\right)}\)

\(A=\frac{2.11}{11.\left(-2\right)}\)

\(A=-2\)

Vậy : A = -2

Số số hạng là :

      (2x - 2) : 2 + 1 = x - 1 + 1 = x (số)

Tổng là : 

       (2x + 2).x : 2 = 210

=> (2x² + 2x) : 2 = 210

=> x² + x = 210

=> x(x + 1) = 210

=> x(x + 1) = 20.21

=> x = 20

Vậy x = 20 

Ta có : \(\frac{x}{2}=\frac{10}{x+1}\)

=> x(x + 1) = 10.2

=> x(x + 1) = 20

=> sai đề 

Ta có: \(\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6}\)

\(=\frac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6}=\frac{2^{12}\cdot3^4\left(3-1\right)}{2^{12}\cdot3^6}\)

\(=\frac{3^4\cdot2}{3^6}=\frac{2}{3^2}=\frac29\)

Ta có: \(\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)

\(=\frac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\frac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\frac{-6}{9}=5\cdot\frac{-2}{3}=-\frac{10}{3}\)

Ta có: \(A=\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)

\(=\frac29+\frac{10}{3}=\frac29+\frac{30}{9}=\frac{32}{9}\)