\(a.172x^2-7^9=2^{-3}.98^3=117649\)

\(172x^2=117649+7^9=40471256\)

\(x^2=40471256:172=235298\)

\(x=\sqrt{235298}=485.07......\)

cái gì ? Mk đã học đến \(\sqrt{ }\) cái này đâu 

172x² - 7⁹:93³ = 2^-3

172x² - 42,875 = 0,125

172x² = 0,125 + 42,875

172x² = 43

x² = 172 : 43

x² = 4

x = \(\sqrt{4}\) = 2 ( 2² = 4 )

Vậy x = 2

172x^2 - 7^9 : 98^3 = 2^-3

172x^2 - 7^9 : 98^3 = 1/8

172x^2 - 7^9 : ( 7^3 ) ^ 3 = 1/8

172x^2 - 7^ 9 : 7^9 = 1/8

172x^2 - 1 = 1/8

172x² - 7⁹ : 98³ = 2^-3

<=> 172x² - 7⁹ : ( 2 . 7² )³ = 1/8

<=> 172x² - 7⁹ : ( 2³ . 7⁶ ) = 1/8

<=> 172x² - 7⁹ : 2³ : 7⁶ = 1/8

<=> 172x² - 7³ : 2³ = 1/8

<=> 172x² - ( 7 : 2 )³ = 1/8

<=> 172x² - 343/8 = 1/8

<=> 172x² = 43

<=> x² = 43/172 = 1/4

<=> x = ±1/2

a: \(5^{x}\cdot\left(5^3\right)^2=625\)

=>\(5^{x}=\frac{5^4}{5^6}=5^{-2}\)

=>x=-2

b: \(\left(\frac{12}{15}\right)^{x}=\left(\frac53\right)^{-5}-\left(-\frac35\right)^4\)

=>\(\left(\frac45\right)^{x}=\left(\frac35\right)^5-\left(\frac35\right)^4=\left(\frac35\right)^4\cdot\left(\frac35-1\right)=\left(\frac35\right)^4\cdot\frac{-2}{5}=\frac{-2\cdot3^4}{5^5}\)

=>\(x=\log_{0,8}\left(-2\cdot\frac{3^4}{5^5}\right)\)

c: \(\left(-\frac34\right)^{3x-1}=\frac{256}{81}\)

=>\(\left(-\frac34\right)^{3x-1}=\left(-\frac34\right)^{-4}\)

=>3x-1=-4

=>3x=-3

=>x=-1

d: \(172x^2-7^9:98^3=2^{-3}\)

=>\(172x^2=\frac18+\frac{7^9}{7^6\cdot2^3}=\frac18+\frac{7^3}{2^3}=\frac{1+343}{8}=\frac{344}{8}\)

=>\(x^2=\frac{344}{8}:172=\frac{344}{8\cdot172}=\frac28=\frac14\)

=>\(\left[\begin{array}{l}x=\frac12\\ x=-\frac12\end{array}\right.\)

Bài 1 : Ta có:

\(\frac{7+\frac{7}{11}+\frac{7}{23}+\frac{7}{31}}{9+\frac{9}{11}+\frac{9}{23}+\frac{9}{31}}\)

= \(\frac{7.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}{9.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}\)

= \(\frac{7}{9}\)

Bài 2 :

 \(\frac{x}{2}+\frac{3x}{4}+\frac{5x}{6}=\frac{10}{24}\)

=> \(\frac{12x+18x+20x}{24}=\frac{10}{24}\)

=> 50x = 10

=> x = 10 : 50

=> x = 1/5

Bài 3 : Để A nhận giá trị nguyên thì 3 \(⋮\)x + 3

                                         <=> x + 3 \(\in\)Ư(3) = {1; -1; 3; -3}

Lập bảng :

Vậy 

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