a:  =>x-16=74

nên x=90

125 : 5^2 . (x-3) = 3

125 : 25 .(x-3) = 3

5 (x-3) =3

x-3 = 3 : 5

Đề bài sai hay sao ý?

2x -5 = 3

2x = 3+5 =8

x =4

 (19 .x+2.5^2 ): 14 = (13-8)^2 - 4^2

19x + 50) : 14 = 25 - 16

(19x+50) : 14 = 9

19x+50 = 3*14 = 126

 19x = 126-50 = 76

x = 76 / 19 = 4

\(a\)) \(125:5^2.\left(x-3\right)=3\)

\(5^3:5^2.\left(x-3\right)=3\)

\(5.\left(x-3\right)=3\)

\(5x-15=3\)

\(5x=3+15\)

\(5x=18\)

\(x=\frac{18}{5}\)\(.\)Vậy \(x=\frac{18}{5}\)

\(b\)) \(\left(2x-5\right)^3=27\)

\(\left(2x-5\right)^3=3^3\)

\(\Rightarrow2x-5=3\)

\(\Rightarrow2x=8\)

\(\Rightarrow x=4\)\(.\)Vậy \(x=4\)

\(c\))\(\left(19.x+2.5^2\right):14=\left(13-8\right)^2-4^2\)

\(\left(19.x+2.25\right):14=5^2-16\)

\(\left(19.x+2.25\right):14=25-16\)

\(\left(19.x+50\right):14=9\)

\(19.x+50=126\)

\(19.x=126-50\)

\(19.x=76\)

\(x=76:19\)

\(x=4\)\(.\)Vậy \(x=4\)

a) \(\frac{3}{7}-\frac{1}{7}x=\frac{2}{3}\)

=> \(\frac{1}{7}x=\frac{3}{7}-\frac{2}{3}=-\frac{5}{21}\)

=> \(x=-\frac{5}{21}:\frac{1}{7}=-\frac{5}{21}\cdot7=-\frac{5}{3}\)

b) \(3x^2-2=72\)=> 3x² = 74 => x² = 74/3 => x không thỏa mãn

c) \(\left(19x+2\cdot5^2\right):14=\left(13-8\right)^2-4^2\)

=> \(\left(19x+2\cdot25\right):14=5^2-4^2=9\)

=> \(\left(19x+50\right):14=9\)

=> \(19x+50=126\)

=> \(19x=76\)

=> x = 4

d) \(x:\frac{1}{2}+x:\frac{1}{4}+x:\frac{1}{8}+x:\frac{1}{16}+x:\frac{1}{32}=343\)

=> \(x\cdot2+x\cdot4+x\cdot8+x\cdot16+x\cdot32=343\)

=> \(x\left(2+4+8+16+32\right)=343\)

=> x . 62 = 343

=> x = 343/62

b: Ta có: \(2^{x+3}+2^x=144\)

\(\Leftrightarrow2^x\cdot9=144\)

\(\Leftrightarrow2^x=16\)

hay x=4

a) (x ^ 54)^2 = x                                         

         x^108  = x

Để: x^108  = x 

=> x=0 hoặc x=1

Bài 1 :

\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)

\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)

\(\Rightarrow M< N\)

Bài 3 :

a) \(t^2+5t-8\) khi \(t=2\)

\(=5^2+2.5-8\)

\(=25+10-8\)

\(=27\)

b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)

\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)

\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)

c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)

\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)

\(\left(1\right)=1^3=1\)