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dễ mà :
\(\left(2x+3\right)^2=25\)
\(\left(2x+3\right)^2=5^2\)
2x + 3 =5
2x=5-3
2x=2
x=2 :2
x=1
1 like nha bạn
\(\left(2x+3\right)^2=25\)
=> \(\left[{}\begin{matrix}\text{(2x+3)}^2=5^2\\\text{(2x+3)}^2=\left(-5\right)^2\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\text{2x+3=5}\\\text{2x+3=-5}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\text{2x=2}\\\text{2x=-}8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
Chúc bạn học tốt!
#Yuii
1: \(=\dfrac{1}{4}:\dfrac{-1}{4}-2\cdot\dfrac{-1}{8}+5-4\)
\(=-1+1+\dfrac{1}{4}=\dfrac{1}{4}\)
2: \(=5^{20}\cdot\dfrac{1}{5^{20}}+\left(\dfrac{3}{8}\cdot\dfrac{4}{3}\right)^8-1=1-1+\dfrac{1}{2}^8=\dfrac{1}{2^8}\)
\(\left(2x-3\right)^2=\dfrac{4}{25}\\ \Rightarrow\left(2x-3\right)^2=\left(\pm\dfrac{2}{5}\right)^2\\ \Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{2}{5}\\2x-3=-\dfrac{2}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=\dfrac{17}{5}\\2x=\dfrac{13}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)
\(\left(2x-3\right)^2=\dfrac{4}{25}\\ \Leftrightarrow\left(2x-3\right)^2=\left(\dfrac{2}{5}\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=\dfrac{2}{5}\\-2x+3=\dfrac{2}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{17}{5}\\-2x=-\dfrac{13}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)
Vậy...
a: \(5^{\left(x-2\right)\left(x+3\right)}=1\)
=>\(\left(x-2\right)\left(x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
c: \(\left|x^2+2x\right|+\left|y^2-9\right|=0\)
mà \(\left\{{}\begin{matrix}\left|x^2+2x\right|>=0\forall x\\\left|y^2-9\right|>=0\forall y\end{matrix}\right.\)
nên \(\left\{{}\begin{matrix}x^2+2x=0\\y^2-9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(x+2\right)=0\\\left(y-3\right)\left(y+3\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\in\left\{0;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)
d: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)
=>\(2^x\left(1+2+2^2+2^3\right)=120\)
=>\(2^x\cdot15=120\)
=>\(2^x=8\)
=>x=3
e: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)
=>\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)
=>\(\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)
a: =>10căn x=30
=>căn x=3
=>x=9
c: =>|1/2x-3/4|=1/2
=>1/2x-3/4=1/2 hoặc 1/2x-3/4=-1/2
=>1/2x=5/4 hoặc 1/2x=1/2
=>x=1 hoặc x=5/2
\(\left(2x-3\right)^2=25\)
\(\left(2x-3\right)^2=5^2\)
\(\Rightarrow2x-3=5\)
\(\Rightarrow2x=8\)
\(\Rightarrow x=4\)
1/3 . (2x - 1/3) - 25/27 = 0
1/3 . (2x - 1/3) = 25/27
2x - 1/3 = 25/27 : 1/3
2x - 1/3 = 25/9
2x = 25/9 + 1/3
2x = 28/9
x = 28/9 : 2
x = 14/9
a, (2x-3)3 = -64
=> (2x-3)3 = -43
=> 2x-3=-4
=> 2x = -1
=> x = -1 : 2
=> x = -1/2
b, (2x-3)2 =25
=> (2x-3)2 =5^2
=> 2x-3 = 5
=> 2x = 8
=> x = 4
c, (3x-4)2 =36
=> (3x-4)2 =62
=> 3x-4 = 6
=> 3x = 10
=> x = 3.(3)
d, 2x+1 = 64
=> 2x+1 = 26
=> x+1 = 6
=> x = 5
Mấy bạn làm thiếu rồi
Ta có
\(\left(2x+3\right)^2=25=5^2=\left(-5\right)^2\)
Suy ra \(\left\{{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\)
Suy ra \(\left\{{}\begin{matrix}x=\dfrac{5-3}{2}=1\\x=\dfrac{-5-3}{2}=-4\end{matrix}\right.\)
Vậy x=1 và x=-4
(2x+3)^2=5^2
Suy ra: 2x+3=5, x=4
^-^