Question

giúp mình nhé!

tìm x biết

phần a: (2x+1)^2= 25 ; (2x-3)^2= 36

phần b: 5^x+2= 625 ; (2x -1)^3= -8

phần c: (x-3)^2+(15x-45)^4= 0 ; |x-3|+(x^2-3x)^2= 0

Answer 1

\(a)\left(2x+1\right)^2=25\)

\(\Rightarrow\left(2x+1\right)^2=\left(\pm5\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

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 \(x\in\varnothing\)

\(\left(2x-1\right)^3=-8\)

\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Rightarrow2x-1=-2\)

\(\Rightarrow2x=-1\)

\(\Rightarrow x=-\frac{1}{2}\)

Answer 2

\(c)\left(x-3\right)^2+\left(15x-45\right)^4=0\)

- Có \(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0,\forall x\\\left(15x-45\right)^4\ge0,\forall x\end{matrix}\right.\Rightarrow\left(x-3\right)^2+\left(15x-45\right)^4\ge0,\forall x\)

Suy ra: Để \(\left(x-3\right)^2+\left(15x-45\right)^4=0\) thì \(\left\{{}\begin{matrix}\left(x-3\right)^2=0\\\left(15x-45\right)^4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-3=0\\15x-45=0\end{matrix}\right.\Rightarrow x=3\)

\(\left|x-3\right|+\left(x^2-3x\right)^2=0\)

- Có \(\left\{{}\begin{matrix}\left|x-3\right|\ge0,\forall x\\\left(x^2-3x\right)^2\ge0,\forall x\end{matrix}\right.\Rightarrow\left|x-3\right|+\left(x^2-3x\right)^2\ge0,\forall x\)

Suy ra: Để \(\left|x-3\right|+\left(x^2-3x\right)^2=0\)thì \(\left\{{}\begin{matrix}\left|x-3\right|=0\\\left(x^2-3x\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-3=0\\x^2-3x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x\left(x-3\right)=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)