Question

a, (x-2)^2-1.16=0

b, |x-1|-2=5

Answer 1

b) \(\left|x-1\right|-2=5\)

=> \(\left|x-1\right|=5+2\)

=> \(\left|x-1\right|=7\)

=> \(\left[{}\begin{matrix}x-1=7\\x-1=-7\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=7+1\\x=\left(-7\right)+1\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=8\\x=-6\end{matrix}\right.\)

Vậy \(x\in\left\{8;-6\right\}.\)

Mình chỉ làm câu b) thôi nhé.

Chúc bạn học tốt!

Answer 2

a, \(\left(x-2\right)^2-1,16=0\)

\(\Leftrightarrow x^2-4x+4-1,16=0\)

\(\Leftrightarrow x^2-4x+2,84=0\)

\(\Leftrightarrow x^2-4x+\frac{71}{25}=0\)

\(\Leftrightarrow25x^2-100x+71=0\)

\(\Leftrightarrow x=\frac{-\left(-100\right)\pm\sqrt{\left(-100\right)^2-4.25.71}}{50}\)

\(\Leftrightarrow x=\frac{100\pm\sqrt{10000-7100}}{50}\)

\(\Leftrightarrow x=\frac{100\pm10\sqrt{29}}{50}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{10+\sqrt{29}}{5}\\x=\frac{10-\sqrt{29}}{5}\end{matrix}\right.\)

b,\(|x-1|-2=5\)

\(\Leftrightarrow|x-1|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=7\\x-1=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-6\end{matrix}\right.\)

Answer 3

Vũ Minh Tuấn

a, \(\left(x-2\right)^2-1.16=0\)

\(\Leftrightarrow x^2-4x+4-16=0\)

\(\Leftrightarrow x^2+2x-6x-12=0\)

\(\Leftrightarrow x\left(x+2\right)-6\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=6\end{matrix}\right.\)

Answer 4

a) \(\left(x-2\right)^2-1.16=0\)

\(\Leftrightarrow\left(x-2\right)^2=16=4^2=\left(-4\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)

Vậy : \(x\in\left\{6,-2\right\}\)

b) \(\left|x-1\right|-2=5\)

\(\Leftrightarrow\left|x-1\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=7\\x-1=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-6\end{matrix}\right.\)

Vậy : \(x\in\left\{8,-6\right\}\)

Answer 5

theo lời bạn Vũ Minh Tuấn thì mình xin phép làm lại câu a thôi chứ câu b ez rồi :)

\(\left(x-2\right)^2-1.16=0\)

\(\Leftrightarrow x^2-4x+4-16=0\)

\(\Leftrightarrow x^2-4x-12=0\)

\(\Rightarrow\Delta'=\left(-2\right)^2+12=16\Rightarrow\sqrt{\Delta'}=4\)( đenta cho nhanh :v )

\(\Rightarrow x_1=\frac{-b'+\sqrt{\Delta'}}{a}=\frac{2+4}{1}=6\)

\(\Rightarrow x_2=\frac{-b'+\sqrt{\Delta'}}{a}=\frac{2-4}{1}=-2\)