a) \(\left(x+1\right)\left(x^2+1\right)=0\)
Vì \(\left(x^2+1\right)>0\forall x\)
\(\Rightarrow x=-1\)
b) \(5y^2-20=0\)
\(y^2-4=0\)
\(\left(y-2\right)\left(y+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\)
a, Ta có : \(\left(x+1\right)\left(x^2+1>0\right)=0\Leftrightarrow x=-1\)
b, \(5y^2=20\Leftrightarrow y^2=4\Leftrightarrow\left[{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\)
c, \(\left|x-2\right|-1=0\Leftrightarrow\left|x-2\right|=1\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
d, \(\left|y-2\right|+5=0\)( vô lí )
Vậy ko có gtr y để bth bằng 0
c) \(\left|x-2\right|-1=0\)
\(\left|x-2\right|=1\)
\(\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
d) \(\left|y-2\right|+5=0\)
\(\left|y-2\right|=-5\)
Vì \(\left|y-2\right|\ge0\forall y\)
⇒ pt vô nghiệm
a, Ta có : (x+1)(x2+1>0)=0⇔x=−1(x+1)(x2+1>0)=0⇔x=−1
b, 5y2=20⇔y2=4⇔[y=2y=−25y2=20⇔y2=4⇔[y=2y=−2
c, |x−2|−1=0⇔|x−2|=1⇔[x−2=1x−2=−1⇔[x=3x=1|x−2|−1=0⇔|x−2|=1⇔[x−2=1x−2=−1⇔[x=3x=1
d, |y−2|+5=0|y−2|+5=0( vô lí )
Vậy ko có gtr y để bth bằng 0
