Tìm x, biết:
\((-5)^9x^2+1/25=-125\)
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\(\dfrac{\left(-5\right)^{9x^2+1}}{25}=-125\)
\(\Leftrightarrow\left(-5\right)^{9x^2+1}=-5^5\)
\(\Leftrightarrow9x^2=4\)
hay \(x\in\left\{\dfrac{2}{3};-\dfrac{2}{3}\right\}\)
\(\Leftrightarrow\left(-5\right)^{9x^2+1}=-125\cdot25=\left(-5\right)^5\\ \Leftrightarrow9x^2+1=5\Leftrightarrow x^2=\dfrac{4}{9}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
a) \(5\left(x+7\right)-10=2^3\cdot5\)
\(\Rightarrow5\left(x+7\right)-10=40\)
\(\Rightarrow5\left(x+7\right)=40+10\)
\(\Rightarrow x+7=\dfrac{50}{5}\)
\(\Rightarrow x+7=10\)
\(\Rightarrow x=10-7\)
\(\Rightarrow x=3\)
b) \(9x-2\cdot3^2=3^4\)
\(\Rightarrow9x-18=81\)
\(\Rightarrow9x=81+18\)
\(\Rightarrow9x=99\)
\(\Rightarrow x=\dfrac{99}{9}\)
\(\Rightarrow x=11\)
c) \(5^{25}\cdot5^{x-1}=5^{25}\)
\(\Rightarrow5^{x-1}=5^{25}:5^{25}\)
\(\Rightarrow5^{x-1}=1\)
\(\Rightarrow5^{x-1}=5^0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
a) 5(�+7)−10=23⋅55(x+7)−10=23⋅5
⇒5(�+7)−10=40⇒5(x+7)−10=40
⇒5(�+7)=40+10⇒5(x+7)=40+10
⇒�+7=505⇒x+7=550
⇒�+7=10⇒x+7=10
⇒�=10−7⇒x=10−7
⇒�=3⇒x=3
b) 9�−2⋅32=349x−2⋅32=34
⇒9�−18=81⇒9x−18=81
⇒9�=81+18⇒9x=81+18
⇒9�=99⇒9x=99
⇒�=999⇒x=999
⇒�=11⇒x=11
c) 525⋅5�−1=525525⋅5x−1=525
⇒5�−1=525:525⇒5x−1=525:525
⇒5�−1=1⇒5x−1=1
⇒5�−1=50⇒5x−1=50
⇒�−1=0⇒x−1=0
⇒�=1⇒x=1
a) \(\sqrt{2x}=12\left(đk:x\ge0\right)\)
\(2x=144\)
\(x=72\)
b) \(\sqrt{9x^2-6x}+1=10\)\(\left(Đk:x\le0;x\ge\dfrac{2}{3}\right)\)
\(\sqrt{9x^2-6x}=9\)
\(9x^2-6x=81\)
\(\left(3x-1\right)^2=82\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{82}+1}{3}\\x=\dfrac{1-\sqrt{82}}{3}\end{matrix}\right.\)
c) \(x^2\sqrt{5}-\sqrt{125}=0\)
\(x^2\sqrt{5}=5\sqrt{5}\)
\(x^2=5\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)
a.
(2x+1)2 = 25 = 52
=> 2x + 1 = 5
=> 2x = 4
=> x = 2
Vậy x = 2
b.
52x+1 = 125 = 53
=> 2x + 1 = 3
=> 2x = 2
=> x = 1
Vậy x = 1
Chúc em học tốt!!!
Ta có: 25x + 1 . 125x . 625x + 2 = (52)5
=> (52)x + 1 . (53)x . (54)x+ 1 = 510
=> 52x + 2 . 53x . 54x + 8 = 510
=> 2x + 2 + 3x + 4x + 8 = 10
=> 9x + 2 + 8 = 10
=> 9x = 10 - 2 - 8
=> 9x = 0
=> x = 0 : 9
=> x = 0
a: x^3=7^3
=>x^3=343
=>\(x=\sqrt[3]{343}=7\)
b: x^3=27
=>x^3=3^3
=>x=3
c: x^3=125
=>x^3=5^3
=>x=5
d: (x+1)^3=125
=>x+1=5
=>x=4
e: (x-2)^3=2^3
=>x-2=2
=>x=4
f: (x-2)^3=8
=>x-2=2
=>x=4
h: (x+2)^2=64
=>x+2=8 hoặc x+2=-8
=>x=6 hoặc x=-10
j: =>x-3=2 hoặc x-3=-2
=>x=1 hoặc x=5
k:
9x^2=36
=>x^2=36/9
=>x^2=4
=>x=2 hoặc x=-2
l:
(x-1)^4=16
=>(x-1)^2=4(nhận) hoặc (x-1)^2=-4(loại)
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1