c )
<=> \(\dfrac{x^2}{2}-\dfrac{9}{2}=0\)
=> \(\dfrac{x^2}{2}=\dfrac{9}{2}\)
= > \(x^2=\dfrac{9}{2}.2=9\) = > x = 3
\(c,\dfrac{1}{2}x^2-4\dfrac{1}{2}=0\\ \Rightarrow\dfrac{1}{2}x^2-\dfrac{9}{2}=0\\ \Rightarrow\dfrac{1}{2}x^2=\dfrac{9}{2}\\ \Rightarrow x^2=9\\ \Rightarrow x^2=\left(\pm3\right)^2\\ \Rightarrow x=\left\{-3;3\right\}\)
\(d,16-4\left(x+\dfrac{1}{2}\right)^2=7\\ \Rightarrow4\left(x+\dfrac{1}{2}\right)^2=9\\ \Rightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{9}{4}\\ \Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\pm\dfrac{3}{2}\right)^2\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{3}{2}\\x+\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
a, (x+1/2021 + 1) + ( x+2/2020 + 1) + ( x+3/2019 + 1) + (x+4/2018 + 1) + (x+5/2017 + 1)= - 5+5=0
=> x+2022/2021 + x+2022/2020 + x+2022/2019 + x+2022/2018 + x+2022/2017=0
=> (x+2022)(1/2021+1/2020 + 1/2019+1/2018+1/2017)=0
mà 1/2021+1/2020 + 1/2019+1/2018+1/2017 > 0
=> x+2022=0
=> x= - 2022
b,7^x + 7^x+2 + 7^x+3=2751
7^x(1+7^2 +7^3)=2751
=> 7^x . 393=2751
=> 7^x=7
=> x=1
c,1/2.x^2 - 9/2=0
=> 1/2.x^2=9/2
=> x^2=9
=>x^2=3^2 hoặc x^2=(-3)^2
=> x thuộc {3;-3}
d,16-4(x + 1/2)^2=7
=>4(x+1/2)^2=9
=> (x+1/2)^2=9/4
=> x+1/2=3/2 hoặc x+1/2= - 3/2
=> x=1 hoặc x=-2
=> x thuộc {1;-2}