tìm x bt:
(2x+1)3=125
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\(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=5-1\)
2x=4
x=4/2
x=2
\(\left(4x-1\right)^2=25\cdot9\)
\(\left(4x-1\right)^2=225\)
\(\left(4x-1\right)^2=15^2\)
\(4x-1=15\)
4x=15+1
4x=16
x=16/4
x=4
Bài 3:
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow x-\frac{1}{2}=0\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\Rightarrow x-2=\pm1\)
+) \(x-2=1\Rightarrow x=3\)
+) \(x-2=-1\Rightarrow x=1\)
Vậy \(x=3\) hoặc \(x=1\)
c) \(\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=\frac{-1}{2}\)
Vạy \(x=\frac{-1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\)
\(\Rightarrow x=\frac{-1}{4}\)
Vậy \(x=\frac{-1}{4}\)
suy ra 1/2+2x=0(1)hay2x-3=0(2)
giải(1)1/2+2x=0 giải(2)2x-3=0
2x=0-1/2 2x=0+3
2x=-1/2 2x=3
x=-1/2:2 x=3:2
x=-1/4 x=3/2
vẫy x ϵ {-1/4;3/2}
Sẽ có 2 trường hợp xảy ra
Trường hợp 1:
\(\dfrac{1}{2}\) + 2x = 0
2x = 0 - \(\dfrac{1}{2}\)
2x = -\(\dfrac{1}{2}\)
x = -0,25
Trường hợp 2:
2x - 3 = 0
2x = 0 + 3
2x = 3
x = 3:2
x = 1,5
\(x+\dfrac{1}{x}=3\Leftrightarrow\left(x+\dfrac{1}{x}\right)^3=27\\ \Leftrightarrow x^3+\left(\dfrac{1}{x}\right)^3+3x\cdot\dfrac{1}{x}\left(x+\dfrac{1}{x}\right)=27\\ \Leftrightarrow x^3+\dfrac{1}{x^3}+3\cdot3=27\\ \Leftrightarrow x^3+\dfrac{1}{x^3}=18\)
(2x + 1)3 = 125
(2x + 1) = 125 : 3
(2x + 1) = \(\frac{125}{3}\)
2x = \(\frac{125}{3}\)+1
2x = \(\frac{128}{3}\)
x = \(\frac{128}{3}\): 2
x = \(\frac{128}{6}\)
(2x + 1)3 = 125
(2x + 1) = 125 : 3
2x + 1 = \(\frac{125}{3}\)
2x = \(\frac{125}{3}-1\)
2x = \(\frac{122}{3}\)
x = \(\frac{122}{3}:2\)
x = \(\frac{61}{3}\)