Tìm x như thường -_-

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

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

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

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

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

 dề bài đâu mà tìm?

ủa mù à

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

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

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

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

a: \(\Leftrightarrow\left(\dfrac{12}{25}\right)^x=\dfrac{9}{25}-\dfrac{81}{625}=\dfrac{144}{625}\)

=>x=2

b: =>3x-1=-4

=>3x=-3

hay x=-1

Bài 1:

Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)

\(\Leftrightarrow2x=\frac{1440}{144}=10\)

\(\Rightarrow x=5\)

Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)

=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)