Question

Tìm x

a)\(\sqrt{x^2-1}\) -\(x^2\)+1=0

b)\(\sqrt{x+2\sqrt{x-1}}\) + \(\sqrt{x-2\sqrt{x-1}}\)

c) \(\sqrt{3x^2+12x+16}\) + \(\sqrt{4x^2+6x+25}\)=1-\(x^2\)-4x

2.Rút gọn

a)\(\sqrt{\left(1-\sqrt{2}\right)^2}\) + \(\sqrt{\left(\sqrt{2}-3\right)^2}\)

b) \(\sqrt{4-2\sqrt{3}}\) + \(\sqrt{7}\) - \(\sqrt{48}\)

c) \(\sqrt{3-\sqrt{8}}\) (3+\(\sqrt{5}\)) ( \(\sqrt{16-\sqrt{2}}\))

Answer 1

bài 2 rút gọn :

a) \(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}-3\right)^2}\)

= \(\left|1-\sqrt{2}\right|+\left|\sqrt{2}-3\right|\)

=\(\sqrt{2}-1+3-\sqrt{2}\)

=2

b) \(\sqrt{4-2\sqrt{3}}+\sqrt{7}-\sqrt{48}\)

= \(\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{7}-4\sqrt{3}\)

= \(\sqrt{3}-1+\sqrt{7}-4\sqrt{3}\)

= \(\sqrt{7}-3\sqrt{3}+1\)

c)

Answer 2

Help mee <3

Answer 3

đề bài phần b kiểu j vậy bn ??/

Answer 4

a) \(\sqrt{x^2-1}-x^2+1=0\left(ĐK:x^2\ge1\right)\)

\(\Leftrightarrow\sqrt{x^2-1}=x^2-1\)

\(\Leftrightarrow x^2-1=\left(x^2-1\right)^2\)

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

\(\Leftrightarrow x^4-3x^2+2=0\)

\(\Leftrightarrow x^4-x^3+x^3-x^2-2x^2+2x-2x+2=0\)

\(\Leftrightarrow x^3\left(x-1\right)+x^2\left(x-1\right)-2x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2-2x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+1\right)-2\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)

b. \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\) ( thiếu đề)

2. \(a.\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}-3\right)^2}=1-\sqrt{2}+\sqrt{2}-3=1-3=-2\)

\(b.\sqrt{4-2\sqrt{3}}+\sqrt{7}-\sqrt{48}=\sqrt{3-2\sqrt{3}+1}+\sqrt{7}-\sqrt{48}=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{7}-4\sqrt{3}=\sqrt{3}-1+\sqrt{7}-4\sqrt{3}=-1+\sqrt{7}-3\sqrt{3}\)

Answer 5

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