(2x+1)^3=729
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a,(5x-1)6=36
5x-1=3
x=4/5
b,(2x+1)3=0,13
2x+1=0,1
x=-0,45
c,(2x-3)4=(2x-3)4(2x-3)2
(2x-3)2=0
2x-3=0
x=3/2
d,(2x+1)5=(2x+1)5(2x+1)2005
(2x+1)2005=0
2x+1=0
x=-1/2
a)\(\orbr{\begin{cases}5x-1=3\\5x-1=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}5x=4\\5x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{5}\\x=\frac{-2}{5}\end{cases}}\)
b) 2x+1=-0,1 <=> 2x=-1,1=>x=-0,55
c) (2x-3)4 .[1-(2x-3)2 ]=0
do (2x-3)4 lớn hơn 0 nên 1-(2x-3)2=0=>(2x-3)2=1=>2x-3=1=>2x=4=>x=2
d) tương tự câu c)
(2x+1)3=729
=>2x+1=\(\sqrt[3]{729}\)
=>2x+1=9
=>2x=8
=>x=4
tick đúng nha
3x+3*32x-1*3x=729
\(\Rightarrow3^{x+3+2x-1+x}=3^6\)
\(\Rightarrow3^{4x+2}=3^6\)
\(\Rightarrow4x+2=6\)
\(\Rightarrow4x=4\)
\(\Rightarrow x=1\)
Ta có: \(A=\left[6.\left(\frac{-1}{3}\right)^2-\left(-\frac{1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)
\(\Rightarrow A=\left[6.\frac{1}{9}+\frac{1}{3}+1\right]:\left(\frac{-1}{3}-\frac{3}{3}\right)\)
\(\Rightarrow A=\left[\frac{2}{3}+\frac{1}{3}+1\right]:\frac{-4}{3}\)
\(\Rightarrow A=\left[1+1\right].\frac{-3}{4}=2.\frac{-3}{4}=\frac{-3}{2}\)
Mà \(B=\left(729-1^3\right)\left(729-2^3\right)\left(729-3^3\right)...\left(729-125^3\right)\)
\(=\left(729-1^3\right)\left(729-2^3\right)...\left(729-9^3\right)...\left(729-125^3\right)\)
\(=\left(729-1^3\right)\left(729-2^3\right)...0...\left(729-125^3\right)=0\)
Vì \(\frac{-3}{2}< 0\)nên A < B
\(3^{2x}=\left(3^2\right)^x\)\(\Rightarrow9^x=729\)\(\Rightarrow x=3\)
\(3^{2x}\) = 729
<=>\(9^x\) = 729
<=>9 . 9 .9 = 729
<=> \(9^3\) = 729
<=> x =3
Vay :x=3
\(\left(2x+1\right)^3=729\)
\(\Leftrightarrow\)\(\left(2x+1\right)^3=9^3\)
\(\Leftrightarrow\)\(2x+1=9\)
\(\Leftrightarrow\)\(2x=8\)
\(\Leftrightarrow\)\(x=4\)
Vậy...
\(\left(2x+1\right)^3=729\)
\(\Rightarrow2x+1=9\)
\(\Rightarrow2x=8\)
\(\Rightarrow x=4\)