(2x+3)4=625
(3x+1)3=343
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a) 33x+1 . 5 = 10935
=> 33x . 3 . 5 = 10935
=> 27x . 3 . 5 = 10935
=> 27x = 729
=> 27x = 272
=> x = 2
b) 25x + 1 . 3 = 6144
=> 25x . 2 . 3 = 6144
=> 32x . 6 = 6144
=> 32x = 1024
=> 32x = 322
=> x = 2
c) (2x + 3)4 = 625
=> (2x + 3)4 = (\(\pm\)5)4
=> \(\orbr{\begin{cases}2x+3=5\\2x+3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)
d) (3x + 1)3 = 343
=> (3x + 1)3 = 73
=> 3x + 1 = 7 => 3x = 6 => x = 2
a: =>2x+5=4
=>2x=-1
hay x=-1/2
b: \(\Leftrightarrow\left(3x-4\right)^2\cdot\left[\left(3x-4\right)^2-1\right]=0\)
=>(3x-4)(3x-5)(3x-3)=0
hay \(x\in\left\{1;\dfrac{4}{3};\dfrac{5}{3}\right\}\)
c: \(\Leftrightarrow3^{x+1}=3^{2x}\)
=>2x=x+1
=>x=1
d: \(\Leftrightarrow2^{2x+3}=2^{2x-10}\)
=>2x+3=2x-10
=>0x=-13(vô lý)
( 3/7 ) 5 x ( 7/3 ) - 1 x ( 5/3 )6 : ( 343/652) - 2
= ( 3/7 x 7/3 ) + 5 - 1 x ( 5/3 : 343/652 ) - 2
= 1 + 5 - 1 x 2 - 1715/1956
= 6 - 2 - 1715/1956
=4 - 1715/1956
= 6109/1715
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(=>2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)
=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.
\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)
a) 2x + 23x = 625
x (2 + 23) = 625
x. 25 = 625
x = 625 : 25
x = 25
b) (2^2+4^2) x +2^4 x 5 * x = 100 (dấu* là dấu nhân vì mik không muốn trùng nhau nên viết vậy)
( 4 + 16 ) x + 2^4 x 5 * x = 100
20 * x + 16 x 5 * x =100
x ( 20 + 16 x 5 ) = 100
x ( 20 + 80 ) = 100
x * 100 =100
x=100 : 100
x= 1
c) ( 3x +1)^2 = 6^2+8^2
(3x +1)^2 = 36 +64
(3x +1)^2 =100
(3x +1)^2 = 10^2
3x + 1 =10
3x=10 - 1=9
x = 9 :3
x = 3
( d và e làm tương tự )
k mik nha!
\(a)\left(2x+1\right)^2=25\)
\(\Rightarrow\left(2x+1\right)^2=\left(\pm5\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
\(\left(2x-3\right)^2=36\)
\(\Rightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\frac{9}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
\(b)5^x+2=625\)
\(\Rightarrow5^x=623\)
\(\Rightarrow x\in\varnothing\)
Vậy \(x\in\varnothing\)
\(\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\frac{1}{2}\)
\(\left(2x+3\right)^4=625\)
\(\left(2x+3\right)^4=5^4\)
\(\Rightarrow2x+3=5\)
\(2x=5-3\)
\(2x=2\)
\(x=2\div2\)
\(x=1\)
\(\left(3x+1\right)^3=343\)
\(\left(3x+1\right)^3=7^3\)
\(3x+1=7\)
\(3x=7-1\)
\(3x=6\)
\(x=6\div3\)
\(x=2\)
bé nguyên sai rồi câu a ấy . câu đấy có hai trường hợp vì nó mũ chẵn nha bạn