a.
\(\sqrt{2x+3}=1\)
\(2x+3=1\)
\(2x=1-3\)
\(2x=-2\)
\(x=-\frac{2}{2}\)
\(x=-1\)
b.
\(\left(3x-1\right)^2-25=0\)
\(\left(3x-1\right)^2=25\)
\(\left(3x-1\right)^2=\left(\pm5\right)^2\)
\(3x-1=\pm5\)
TH1:
\(3x-1=5\)
\(3x=5+1\)
\(3x=6\)
\(x=\frac{6}{3}\)
\(x=2\)
TH2:
\(3x-1=-5\)
\(3x=-5+1\)
\(3x=-4\)
\(x=-\frac{4}{3}\)
Vậy \(x=2\) hoặc \(x=-\frac{4}{3}\)
c.
\(\left(2x+4\right)\left(x^2+1\right)\left(x-2\right)=0\)
TH1:
\(2x+4=0\)
\(2x=-4\)
\(x=-\frac{4}{2}\)
\(x=-2\)
TH2:
\(x^2+1=0\)
\(x^2=-1\)
mà \(x^2\ge0\) với mọi x
=> loại
TH3:
\(x-2=0\)
\(x=2\)
Vậy \(x=2\) hoặc \(x=-2\)
\(a.\)\(=>2x+3=1\)\(=>2x=-2\)\(=>x=-1\)
\(b.\)\(=>\left(3x-1\right)^2=25\)\(=>\left(3x-1\right)^2=5^2=>3x-1=5=>3x=6=>x=2\)
\(c.\)\(=>2x+4=0\)hoac \(x^2+1=0\)hoac \(x-2=0\)
=> * 2x=4 => x= 2
* x^2=-1=> x=-1
* x = 2
\(=>x\in\left(2;-1\right)\)
\(\sqrt{2x+3}=1\)
Mình viết gon lại thành \(\left(2x+3\right)=1^2\)nha
\(\Leftrightarrow2x+3=1\)
\(\Leftrightarrow2x=1-3\)
\(\Leftrightarrow2x=-2\)
\(\Leftrightarrow x=-2:2\)
\(\Leftrightarrow x=-1\)
\(b,\left(3x-1\right)^2-25=0\)
\(\Leftrightarrow\left(3x-1\right)^2=25\)
\(\Leftrightarrow\left(3x-1\right)^2=\left(5,-5\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=-5\\3x-1=5\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x=-4\\3x=6\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-4}{3}\\x=2\end{cases}}\)
Vậy \(x\in\left\{\frac{-4}{3},2\right\}\)
\(c,\left(2x+4\right)\left(x^2+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}2x+4=0\\x^2+1=0\\x-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x=-4\\x^2=-1\\x=2\end{cases}}\)\(\hept{\begin{cases}x=-2\\x^2=\left(-1\right)^2\\x=2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=-2\\x=-1\\x=2\end{cases}}\)
Vậy \(x\in\left\{-2,-1,2\right\}\)
a) suy ra 2x+3=1 <=> 2x=-1<=>x=\(\frac{-1}{2}\)
b)<=> \(\left(3x-1\right)^2\)=25<=> 3x-1=5 hoặc 3x-1=-5
th1 : 3x-1=5<=> 3x=6 <=> x=2
th2 : 3x-1=-5 <=> 3x=-4 <=> x=\(\frac{-4}{3}\)
c) th1 : <=> 2x+4=0 <=> 2x=-4 <=> x=-2
th2 : <=> \(x^2\)+1=0 <=> \(x^2\)=-1 ( vô lý )
th3 : <=> x-2=0 <=> x=2
