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9x - 1 = 729
9x - 1 = 93
=> x - 1 = 3
x = 3 + 1
x = 4
Vậy x = 4
=))
thay 11=x+1 ta có:
f(x)= \(x^{10}\)-11\(x^9\)+11\(x^8\)-11\(x^7\)+....+11\(x^2\)-11x+100
=\(x^{10}\)-(x+1)\(x^9\)+(x+1)\(x^8\)-(x+1)\(x^7\)+...+(x+1)\(x^2\)-(x+1)x+100
=\(x^{10}\)-\(x^{10}\)-\(x^9\)+\(x^9\)+\(x^8\)-\(x^8\)-\(x^7\)+......+\(x^3\)+\(x^2\)-\(x^2\)-x+100
=-x+100
=> f(10)=-10+100=90
Thay 11 = x + 1 ta có:
f(x) = \(x^{10}-11x^9+11x^8-11x^7+...+11x^2-11x+100\)
\(=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-\left(x+1\right)x^7+...+\left(x+1\right)x^2-\left(x+1\right)x^2-\left(x+1\right)x+100\)
= \(x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...+x^3+x^2-x^2-x+100\)
= -x+100
=>f(10)= - 10 + 100 = 90
a) \(\left(2x+3\right)^2=\frac{9}{144}\)
\(\Leftrightarrow\left(2x+3\right)^2=\left(\frac{1}{4}\right)^2=\left(-\frac{1}{4}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x+3=\frac{1}{4}\\2x+3=\frac{-1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{-11}{4}\\2x=\frac{-13}{4}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-11}{8}\\x=\frac{-13}{8}\end{cases}}}\)
Vậy ...
b) Ta có: \(\left(3x-1\right)^3=\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)
\(\Leftrightarrow3x-1=\frac{-2}{3}\Leftrightarrow3x=\frac{1}{3}\Leftrightarrow x=\frac{1}{9}\)
Vậy ....
c) \(x^{10}=25x^8\Leftrightarrow x^{10}:x^8=25\Leftrightarrow x^2=25\Leftrightarrow x=\left\{5;-5\right\}\)
Vậy ...
d) \(\frac{x^7}{81}=27\Leftrightarrow x^7=27.81=2187\)
Mà 37 = 2187 => x7 = 37 => x = 3
Vậy ....
e) \(\frac{x^8}{9}=729\Leftrightarrow x^8=729.9=6561\)
Mà 38 = (-3)8 = 6561
=> x8 = 38 = (-3)8
=> x = {-3;3}
Vậy ...
1. \(\frac{125^{10}.8^{30}}{5^{29}.2^{60}}\) = \(\frac{5^{30}.2^{90}}{5^{29}.2^{60}}\) = ............
Kl ............
a/ 72+x + 2.7x-1 =345
73 . 7x-1 + 2.7x-1 = 345
7x-1 (73+2) = 345
7x-1 . 345 = 345
7x-1 = 345 : 345 = 1
7x-1 = 70
x - 1 = 0
x = 0+1 = 1
b/ 81-2x . 27x = 95
(34)-2x . (33)x = (32)5
34.(-2x) . 33.x = 32.5
3-8x . 33x = 310
3-8x+3x = 310
3-5x = 310
-5x = 10
x = 10 : (-5) = -2
Ta có ( x2 )3 = 46656/729
=> x2.3=6 = 46656/729
<=> x6 = 46656/729
=> \(\sqrt[6]{\frac{46656}{729}}\)= x = 2
=> x = 2
\(\frac{x^8}{9}=729\)
\(x^8=729\times9\)
\(x^8=6561\)
\(x^8=3^8\)
\(\Rightarrow x=3\)
Vậy \(x=3\).
\(\frac{x^8}{9}=729\)
\(x^8=729.9\)
\(x^8=6561\)
\(x^8=3^8\)
\(\Rightarrow x=3\)
k imk nha