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b: \(\left(2x+1\right)^2=25\)
=>\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
c: \(\left(1-3x\right)^3=64\)
=>\(\left(1-3x\right)^3=4^3\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1
d: \(\left(4-x\right)^3=-27\)
=>\(\left(4-x\right)^3=\left(-3\right)^3\)
=>4-x=-3
=>x=4+3=7
e: \(x^2-5x=0\)
=>\(x\left(x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Bài 3:
a: =>|x+7|=0
=>x+7=0
hay x=-7
b: =>|x-3|=4
=>x-3=4 hoặc x-3=-4
=>x=7 hoặc x=-1
c: =>x-5<=0
hay x<=5
\(C=1+3+3^2+3^3+...+3^{11}\)
\(\Rightarrow C=3^0+3^1+3^2+3^3+...+3^{11}\)
\(\Rightarrow3C=3^1+3^2+3^3+3^4+...+3^{12}\)
\(\Rightarrow3C-C=\left(3^1+3^2+3^3+3^4+...+3^{12}\right)-\left(3^0+3^1+3^2+3^3+...+3^{11}\right)\)
\(\Rightarrow2C=3^{12}-3^0\)
\(\Rightarrow C=\frac{3^{12}-1}{2}\)
-Chào :)
(2.x+1)3 = 27
(2.x+1)3 = 33
2.x = 3 -1
2x = 2
x = 2:2
x = 1
- Học tốt ='>
a) \(5-\left(-3\right)^2-14.\left(-8\right)+\left(-31\right)\)
\(=5-9-\left(-112\right)+\left(-31\right)\)
\(=-4+112-31\)
\(=108-31\)
\(=77\)
b) \(6.\left(-2\right)^3+5.\left(-4\right)-\left(-12\right)\)
\(=6.\left(-8\right)+\left(-20\right)+12\)
\(=-48+\left(-20\right)+12\)
\(=-56\)
Ta có :
\(\left(2x+1\right)^3=27\)
\(\Rightarrow\left(2x+1\right)^3=3^3\)
\(\Rightarrow2x+1=3\)
\(\Rightarrow2x=3-1=2\)
\(\Rightarrow x=2:2=1\)
Vậy \(x=1\)
Ta có : \(\left(2x+1\right)^3=27\)
\(\left(2x+1\right)^3=3^3\)
2x + 1 = 3
2x = 2
x = 1
Vậy x = 1
khó