Question

giải giúp mình với ạ !!

Answer 1

a. \(\sqrt{1-x^2}=x-1\)

<=> 1 - x² = (x - 1)²

<=> (1 - x)(1 + x) - (x - 1)² = 0

<=> (1 - x)(1 + x) + (1 - x)² = 0

<=> (1 - x)(1 + x + 1 - x) = 0

<=> 2(1 - x) = 0

<=> 1 - x = 0

<=> -x = -1

<=> x = 1

c. (Tương tự a)

b. \(\sqrt{x^2+4x+4}=x-2\)

<=> \(\sqrt{\left(x+2\right)^2}=x-2\)

<=> \(|x+2|=x-2\)

<=> \(\left[{}\begin{matrix}x+2=-\left(x-2\right)\\x+2=x-2\end{matrix}\right.\)

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

<=> \(\left[{}\begin{matrix}x+x-2+2=0\\4=0\left(Vlí\right)\end{matrix}\right.\)

<=> 2x = 0

<=> x = 0

d, f,h. (tương tự b)

e. \(\sqrt{x^2-4}+2-x=0\)

<=> \(\sqrt{x^2-4}=x-2\)

(Tiếp theo như câu a)

g. \(\sqrt{\left(2x+4\right)\left(x-1\right)}=x+1\)

<=> (2x + 4)(x - 1) = (x + 1)²

<=> 2x² - 2x + 4x - 4 = x² + 2x + 1

<=> 2x² - x² - 2x + 4x - 2x - 4 - 1 = 0

<=> x² - 5 = 0

<=> x² = 5

<=> x = \(\sqrt{5}\)