b) (2x + 1)3= 125
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a) Ta có: ( 2 x + 1 ) 3 = 3 3 nên 2x + 1 = 3. Do đó x = 1.
b) Ta có: ( 2 x - 1 ) 3 = 5 3 nên 2x - 1 = 5. Do đó x = 3.
(2x-1)\(^3\)=-125
(2x-1)\(^3\)= - 5\(^3\)
2x - 1 = -5
2x = -5+1
2x =-4
x = -4:2
x = -2
Câu 1: 56 - 2.(\(x+3\))\(^3\) = 2
2.(\(x\) + 3)\(^3\) = 56 - 2
2.(\(x+3\))\(^3\) = 54
(\(x+3\))\(^3\) = 54 : 2
(\(x+3\))\(^3\) = 27
(\(x+3\))\(^3\) = 3\(^3\)
\(x+3\) = 3
\(x=0\)
Vậy \(x\) = 0
Câu 2:
4.2\(^{x}\) - 3 = 125
4.2\(^{x}\) = 125 + 3
4.2\(^{x}\) = 128
2\(^{x}\) = 128 : 4
2\(^{x}\) = 32
2\(^{x}\) = 2\(^5\)
\(x\) = 5
Vậy \(x=5\)
Nếu bài dễ thì cố gắng tự làm bạn nhé.
Bài 1:
a) \(2x+3=5\)
\(\Leftrightarrow2x=5-3=2\)
\(\Leftrightarrow x=2:2=1\)
Vậy: x=1
b) \(\left(2x-3\right)^2=9\)\(=3^2=\left(-3\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\left(3+3\right):2=3\\x=\left(-3+3\right):2=0\end{matrix}\right.\)
Vậy: \(x\in\left\{3;0\right\}\)
c) \(3x+4=2x-7\)
\(\Leftrightarrow2x-3x=7+4\)
\(\Leftrightarrow-1x=11\)
\(\Leftrightarrow x=-11\)
Vậy: x=-11
d) \(\left(4x-1\right)^3=27\)\(=3^3\)
\(\Rightarrow4x-1=3\)
\(\Leftrightarrow4x=3+1=4\)
\(\Leftrightarrow x=4:4=1\)
Vậy: x=1
e) \(\left(x-7\right).2=3.\left(2x+4\right)\)
\(\Leftrightarrow2x-14=6x+12\)
\(\Leftrightarrow2x-6x=12+14\)
\(\Leftrightarrow-4x=26\)
\(\Leftrightarrow x=\frac{26}{-4}=\frac{13}{-2}\)
Vậy: \(x=\frac{13}{-2}\)
a) 2x . 4 = 128
<=> 2x = 32
<=> 2x = 25
<=> x = 5
b) x15 = x1
<=> x15 - x = 0
<=> x(x14 - 1) = 0
<=> \(\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{14}=1^{14}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
c) (2x + 1)3 = 125
<=> (2x + 1)3 = 53
<=> 2x + 1 = 5
<=> 2x = 4
<=> x = 2
d) (x - 5)4 = (x - 5)6
<=> (x - 5)6 - (x - 5)4 = 0
<=> (x - 5)4[(x - 5)2 - 1] = 0
<=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)
Khi (x - 5)4 = 0 => x - 5 = 0 => x = 5
Khi (x - 5)2 - 1 = 0 <=> (x - 5)2 = 12 <=> \(\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)
a) \(\left(2x-1\right)^5=243\)
\(\left(2x-1\right)^5=3^5\)
\(\Rightarrow2x-1=3\)
\(2x=3+1\)
\(2x=4\)
\(x=\frac{4}{2}\)
\(x=2\)
Vậy .......
b) \(\left(3x+2\right)^3=125\)
\(\left(3x+2\right)^3=5^3\)
\(\Rightarrow3x+2=5\)
\(3x=5-2\)
\(3x=3\)
\(x=\frac{3}{3}=1\)
Vậy ..........
(2x+1)3 = 125
<=> (2x+1)3 = 53
<=> 2x+1 = 5
<=> 2x = 4
<=> x = 2