Tìm x, biêt:
a) \(\left(\frac{-3}{4}\right)^{3x-1}=\frac{256}{81}\)
b) \(172.x^2-7^9:98^3=2^{-3}\)
TÌm x
a)\(\left(\frac{-3}{4}\right)^{3x-1}=\frac{256}{81}\)
b) 172 . x2 - 79 : 98 3 = 2 -3
tìm x biết
a .\(\left(\frac{-3}{4}\right)^{3x-1}=\frac{256}{81}\)
b, \(172.x^2-7^9:98^3=2^{-3}\)
giúp mk đi tiên tài ơi mình cho 1000000000000000000 tick
Tìm x:
a)\(5^x.\left(5^3\right)^2=625\)
b)\(\left(\frac{12}{25}\right)^x=\left(\frac{5}{3}\right)^{-2}-\left(-\frac{3}{5}\right)^4\)
c)\(\left(-\frac{3}{4}\right)^{3x-1}=\frac{256}{81}\)
d)\(172.x^2-7^9:98^3=2^{-3}\)
Tìm x biết :
\(172x^2-7^9:98^3=2^{-3}\)
\(\left(\frac{-3}{4}\right)^{3x-1}=\frac{256}{81}\)
\(a.172x^2-7^9=2^{-3}.98^3=117649\)
\(172x^2=117649+7^9=40471256\)
\(x^2=40471256:172=235298\)
\(x=\sqrt{235298}=485.07......\)
cái gì ? Mk đã học đến \(\sqrt{ }\) cái này đâu
Devil Girl lớp 7 đâu đã học \(\sqrt{ }\)
tìm x:
\(a,5^x.\left(5^2\right)^3=625\)
\(b,\left(\dfrac{12}{15}\right)^x=\left(\dfrac{5}{4}\right)^{-2}-\left(\dfrac{-3}{5}\right)^4\)
\(c,\left(\dfrac{-3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
\(d,172x^2-7^9:98^3=2^{-3}\)
tìm x biết:
a) \(5^x.\left(5^3\right)^2=625\)
b)\(\left(\dfrac{12}{15}\right)^x=\left(\dfrac{5}{3}\right)^{-5}-\left(-\dfrac{3}{5}\right)^4\)
c)\(\left(-\dfrac{3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
d)\(172x^2-7^9:98^3=2^{-3}\)
Tìm x:
a/\(\left(\dfrac{12}{25}\right)^x\)=\(\left(\dfrac{3}{5}\right)^2-\left(-\dfrac{3}{5}\right)^4\)
b/\(\left(-\dfrac{3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
c/172\(x^2-7^9:98^3=2^{-3}\)
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: Tìm x,y biết rằng:
\(x+(-\frac{31}{12})^2=\left(\frac{49}{12}\right)^2-x=y^2\)
bài 2: tìm x biết:
a.\(5^x.\left(5^3\right)^2=625\) b.\(\left(\frac{12}{25}\right)^x=\left(\frac{5}{3}\right)^{-2}-\left(-\frac{3}{5}\right)^4\) c.\(\left(-\frac{3}{4}\right)^{3x-1}=\frac{256}{81}\)
d.\(172x^2-7^9:98^3=2^{-3}\)
Bài 3: Tìm x \(\varepsilon\)N biết:
a.\(8< 2^x\le2^9\times2^{-5}\) b.\(27< 81^3:3^x< 243\) \(\left(\frac{2}{5}\right)^x>\left(\frac{5}{2}\right)^{-3}\times\left(-\frac{2}{5}\right)^2\)c.
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: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
Tìm x, biết
a, \(\left(\frac{12}{21}\right)^x=\left(\frac{3}{5}\right)^2-\left(-\frac{3}{5}\right)^4\)
b, \(\left(-\frac{3}{4}\right)^{3x-1}=\frac{256}{81}\)
Help me!!!