a) \(\Leftrightarrow\left(2x+1\right)^3-125=0\)
\(\Leftrightarrow\left(2x+1-5\right)\left(4x^2+4x+1+10x+5+5\right)=0\)
\(=\left(2x-4\right)\left(4x^2+14x+11\right)=0\)
\(4x^2+14x+11>0\Leftrightarrow x=2\)
b) \(\Leftrightarrow\left(x+1\right)^2-64=0\Leftrightarrow\left(x+1+8\right)\left(x+1-8\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=7\end{matrix}\right.\)
c) \(x^2+2x+4^x-2^{x+1}+2=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(4^x-2^{x+1}+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(2^x-1\right)^2=0\)
x=-1;0 suy ra pt vô nghiệm
Lời giải:
a) \((2x+1)^3-25=100\)
\(\Rightarrow (2x+1)^3=125=5^3\)
\(\Rightarrow 2x+1=5\Rightarrow x=2\)
b)
\((x+1)^2-4=60\)
\(\Leftrightarrow (x+1)^2=64\)
\(\Rightarrow \left[\begin{matrix} x+1=\sqrt{64}=8\\ x+1=-\sqrt{64}=-8\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=7\\ x=-9\end{matrix}\right.\)
c)
\(x^2+2x+4^x-2^{x+1}+2=0\)
\(\Leftrightarrow (x^2+2x+1)+(4^x-2^{x+1}+1)=0\)
\(\Leftrightarrow (x+1)^2+(2^x-1)^2=0\)
Vì \((x+1)^2; (2^x-1)^2\geq 0\Rightarrow \) để tổng của chúng bằng 0 thì:
\(\left\{\begin{matrix} (x+1)^2=0\\ (2^x-1)^2=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=-1\\ x=0\end{matrix}\right.\) (vô lý)
Do đó pt vô nghiệm
