Tìm x
a) 5x = 125; b) 32x = 81;
c) 52x-3 – 2.52 = 52.3;
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a: 12:x=144
=>\(x=\dfrac{12}{144}\)
=>\(x=\dfrac{1}{12}\)
b: \(x-17=\left(-2\right)\cdot27\)
=>\(x-17=-54\)
=>\(x=-54+17=-37\)
c: \(3x-125=145\)
=>\(3x=125+145=270\)
=>\(x=\dfrac{270}{3}=90\)
\(a,12:x=144\\ x=\dfrac{12}{144}=\dfrac{1}{12}\\ ---\\ b,x-17=\left(-2\right).27\\ x-17=-54\\ x=-54+17=-37\\ ----\\ 3x-125=145\\ 3x=145+125=270\\ x=\dfrac{270}{3}=90\)
\(a,5^x+5^{x+2}=650\\ \Rightarrow a,5^x+5^x.25=650\\ \Rightarrow26.5^x=650\\ \Rightarrow5^x=25\\ \Rightarrow5^x=5^2\\ \Rightarrow x=2\)
\(b,3^{x.1}+5.3^{x.1}=162\\ \Rightarrow3^x+5.3^x=162\\ \Rightarrow6.3^x=162\\ \Rightarrow3^x=27\\ \Rightarrow3^x=3^3\\ \Rightarrow x=3\)
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) \(x^2\sqrt{5}-\sqrt{125}=0\)
\(x^2\sqrt{5}=5\sqrt{5}\)
\(x^2=5\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)
5x2-120x-125=0
x(5x-120)=0+125
x(5x-120)=125
\(\Rightarrow\orbr{\begin{cases}x=125\\5x-120=125\Rightarrow5x=125-120\Rightarrow5x=5\Rightarrow x=5:5=1\end{cases}}\)
Vậy x=125 hoặc x=1
\(5x^2-120x-125=0\)
\(5\left(x^2-24x-25\right)=0\)
\(5\left(x-25\right)\left(x+1\right)=0\)
\(\Rightarrow5=0\)vô nghiệm
\(\Rightarrow\orbr{\begin{cases}x-25=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=25\\x=-1\end{cases}}}\)
a) x = 4
b) x = 5
c) x = 2
d) x = 2
e, x = 1
f, x = 0 hoặc x = 1
a) x = 4
b) x = 5
c) x = 2
d) x = 2
e) x = 1
f) x = 0 hoặc x = 1.
a) x = 3 .
a) 5x = 125
5x = 53
=> x = 3
b) ( mình bó tay )
c) 52x-3 - 2.52 = 52 .3
52x-3 - 50 = 75
52x-3 = 75 + 50
52x-3 = 125
52x-3 = 53
2x-3 = 3
2x = 3+3
2x = 6
x = 6 : 2
x = 3