Question

Tìm x, biết:

a, \(\sqrt{7+\sqrt{2x}}=3+\sqrt{5}\).

b, \(\sqrt{x^2}-6x+9=\sqrt{4+2\sqrt{3}}\).

Answer 1

Làm lại câu b :v (đề ở dưới thì dễ òi :vv)\(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\\ \Leftrightarrow x^2-6x+9=4+2\sqrt{3}\left(binhphuong\right)\\ \Leftrightarrow x^2-6x+5-2\sqrt{3}=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\)

Vậy.....

Answer 2

b) ĐK:...

\(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)

\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(\Leftrightarrow\left|x-3\right|=\sqrt{3}+1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=\sqrt{3}+1\\3-x=\sqrt{3}+1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}+4\\x=2-\sqrt{3}\end{matrix}\right.\)( thỏa )

Vậy...

Answer 3

Thoi làm câu a :v

ĐK: \(x\ge0\)

Bình phương hai vế ta được

\(7+\sqrt{2x}=\left(3+\sqrt{5}\right)^2=14+6\sqrt{5}\\ \Leftrightarrow\sqrt{2x}=7+6\sqrt{5}\\ \Leftrightarrow2x=\left(7+6\sqrt{5}\right)^2=229+84\sqrt{5}\\ \Leftrightarrow x=\frac{229+84\sqrt{5}}{2}\left(tm\right)\)

Số to dzữ ta OwO

Answer 4

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Answer 5

a) Bình phương hai vế (đk x>=0)

\(7+2x=\left(3+\sqrt{5}\right)^2=14+6\sqrt{5}\Rightarrow2x=7+6\sqrt{5}\)

\(\Rightarrow x=\frac{7+6\sqrt{5}}{2}\left(TM\right)\)

Is it true?

Answer 6

Câu b,

\(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)

Answer 7

b.

\(\sqrt{x^2}-6x+9=\sqrt{4+2\sqrt{3}}\\ \Leftrightarrow\left|x\right|-6x+9=\sqrt{\left(\sqrt{3}+1\right)^2}\\ \Leftrightarrow\left|x\right|-6x+9=\sqrt{3}+1\\ \Leftrightarrow\left|x\right|-6x=\sqrt{3}-8\left(1\right)\)

Chia 2 t/h:

+) T/h 1: Với x ≥ 0 ta có:

\(\left(1\right)\Leftrightarrow x-6x=\sqrt{3}-8\Leftrightarrow-5x=\sqrt{3}-8\Leftrightarrow x=\frac{\sqrt{3}-8}{-5}=\frac{8-\sqrt{3}}{5}\left(tm\right)\)

+) T/h 2: Với x < 0 ta có:

\(\left(1\right)\Leftrightarrow-x-6x=\sqrt{3}-8\Leftrightarrow-7x=\sqrt{3}-8\Leftrightarrow x=\frac{\sqrt{3}-8}{-7}=\frac{8-\sqrt{3}}{7}\left(ktm\right)\)

Vậy.....