tìm x
3x=9
4x=64
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a, \(=-91x-y+5z\)
b, \(=4x^2+x^2y-5y^2-\dfrac{5}{3}x^3+6xy^2+x^2y\)
\(=4x^2+2x^2y-5y^2-\dfrac{5}{3}x^3+6xy^2\)
\(64-x(3x-27)\div2=40\)
\(\Rightarrow x(3x-27)\div2=64-40\)
\(\Rightarrow x(3x-27)\div2=24\)
\(\Rightarrow x(3x-27)=24\div2\)
\(\Rightarrow x(3x-27)=12\)
\(\Rightarrow4x+27x=12\)
\(\Rightarrow x(27+4)=12\)
\(\Rightarrow31x=12\)
\(\Rightarrow x\in\varnothing\)
\(\frac{2^{7x}}{2^{3x}}=64\)
\(\Rightarrow\frac{\left(2^7\right)^x}{\left(2^3\right)^x}=2^6\)
\(\Rightarrow\left(\frac{128}{8}\right)^x=2^6\)
\(\Rightarrow16^x=2^6\)
\(\Rightarrow2^{4x}=2^6\)
\(\Rightarrow4x=6\)
\(\Rightarrow x=\frac{6}{4}=\frac{3}{2}\)
Ta co 27x / 23x = 64
===> 27x-3x = 64
===> 24x = 64 = 26
=====> 4x = 6
===> x = 1,5
\(\left(3x-1\right)^2=64\)
\(\Rightarrow\left(3x-1\right)^2=8^2\)
\(\Rightarrow3x-1=8\)
\(\Rightarrow3x=9\)
\(\Rightarrow x=3\)
Vậy x = 3
\(b,\left(5x-1\right)^2:2=8\\ \Leftrightarrow\left(5x-1\right)^2=16\\ \Leftrightarrow\left[{}\begin{matrix}5x-1=4\\5x-1=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{3}{5}\end{matrix}\right.\\ c,\left(1-3x\right)^3=-64\\ \Leftrightarrow1-3x=-4\\ \Leftrightarrow3x=5\\ \Leftrightarrow x=\dfrac{5}{3}\)
\(3x+36=-7x-64\)
\(10x=-100\)
\(x=-10\)
\(-5x-178=14x+145\)
\(-19x=323\)
\(x=-17\)
\(\left(3x-4\right).2^3=64\)
\(\Rightarrow3x-4=64:8=8\)
\(\Rightarrow3x=4+8=12\)
\(\Rightarrow x=4\)
(3x-4).2^3 = 64
=> (3x-4).8 = 64
=) 3x-4 = 8
=> 3x = 12
=> x = 12:3
=> x = 4 (tmđk)
Vậy, x = 4
3^2=9
4^3=64
tích mk nha
x=2
x=3