Question

[Lucy Maylaza]

a)\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

b)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)

c)\(\left(x+20\right)^{100}+\left|y+4\right|=0\)

d)\(2^{x-1}=16\)

Answer 1

[Nhật Hạ]

a, \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)\(\Rightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)\(\Rightarrow x-\frac{1}{2}=\frac{1}{3}\)\(\Rightarrow x=\frac{5}{6}\)

b, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)

\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+6}=0\)

\(\Rightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^4\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^4=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^4=1\end{cases}}\)

Giải: \(\left(x-1\right)^4=1\)\(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)

c, Vì \(\left(x+20\right)^{100}\ge0\)\(\forall x\inℝ\); \(\left|y+4\right|\ge0\)\(\forall y\inℝ\)

\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\)\(\forall x,y\inℝ\)

Dấu " = " xảy ra <=> \(\hept{\begin{cases}x+20=0\\y+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-20\\y=-4\end{cases}}\)

d, \(2^{x-1}=16\)\(\Rightarrow2^{x-1}=2^4\)=> x - 1 = 4 => x = 5