\(a)\sqrt 2  \approx 1,1412... \in I;\,\,\,\,\,b)\sqrt 9  = 3 \notin I;\,\,\,\,c)\,\pi  \approx 3,141... \in I;\,\,\,\,\,d)\sqrt 4  = 2 \in \mathbb{Q}\)

Vậy các phát biểu a,c,d đúng.

Hướng dẫn trả lời:

Chọn C vì:

Mệnh đề I sai vì không có căn bậc hai của số âm

Mệnh đề IV sai vì √100 = 10 (căn bậc hai số học)

Các mệnh đề II và III đúng

mn ơi giải giúp mik bài não cũng đc a

mình cảm ơn mn nhiều ạ =))

bài 1:

a) \(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)

b) \(\dfrac{26}{5-2\sqrt{3}}=\dfrac{26\left(5+2\sqrt{3}\right)}{13}=\dfrac{130+52\sqrt{3}}{13}\)

c)\(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}=\dfrac{\left(9-2\sqrt{3}\right)\left(3\sqrt{6}-2\sqrt{2}\right)}{46}=\dfrac{27\sqrt{6}-18\sqrt{2}-18\sqrt{2}+4\sqrt{6}}{46}=\dfrac{31\sqrt{6}-36\sqrt{2}}{46}\)

a)      \(\sqrt 3  \in \mathbb{Q}\) sai.

Sửa lại: \(\sqrt 3  \notin \mathbb{Q}\)

b)      \(\sqrt 3  \in \mathbb{R}\) đúng.

c)      \(\frac{2}{3} \notin \mathbb{R}\) sai.

Sửa lại: \(\frac{2}{3} \in \mathbb{R}\)

d)      \( - 9 \in \mathbb{R}\) đúng.

\(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)

a, A = \(3\sqrt{3}+4\sqrt{12}-5\sqrt{27}\)

= \(3\sqrt{3}+8\sqrt{3}-15\sqrt{3}\)

= \(-4\sqrt{3}\)

b, B = \(\sqrt{32}-\sqrt{50}+\sqrt{18}\)

= \(4\sqrt{2}-5\sqrt{2}+3\sqrt{2}\)

= \(2\sqrt{2}\)

a) \(\sqrt{16x-8}+\sqrt{36x-18}-\sqrt{64x-32}=\sqrt{10}\)

\(\Leftrightarrow\sqrt{8\left(2x-1\right)}+\sqrt{18\left(2x-1\right)}-\sqrt{32\left(2x-1\right)}=\sqrt{10}\)

\(\Leftrightarrow\sqrt{8}.\sqrt{2x-1}+\sqrt{18}.\sqrt{2x-1}-\sqrt{32}.\sqrt{2x-1}=\sqrt{10}\)

\(\Leftrightarrow\sqrt{2x-1}.\left(\sqrt{8}+\sqrt{18}-\sqrt{32}\right)=\sqrt{10}\)

\(\Leftrightarrow\sqrt{2x-1}.\sqrt{2}=\sqrt{10}\)

\(\Leftrightarrow\sqrt{2x-1}=\sqrt{5}\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow x=3\)

Vậy ...

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

\(\Leftrightarrow\sqrt{x^2-2.x.3+3^2}=x+3\)

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

\(\Leftrightarrow\left|x-3\right|=x+3\)

\(\Leftrightarrow x-3=x+3\) hoặc \(x-3=-x-3\)

\(\Leftrightarrow x=0\)

Vậy ...

bài 2 :

A = \(\left(\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{4\sqrt{ab}}{a-b}\right)\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}-\left(a+b\right)}\right)\)

\(=\left(\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{4\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a+\sqrt{b}}\right)}\right)\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}-\left(a+b\right)}\right)\)

\(=\left(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right)\left(\dfrac{\sqrt{a^3}+\sqrt{b^3}}{\sqrt{ab}-a-b}\right)\)

\(=\left(\dfrac{a+2\sqrt{ab}+b-4\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right)\left(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{-a+\sqrt{ab}-b}\right)\)

\(=\dfrac{a-2\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}.\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{-\left(a-\sqrt{ab}+b\right)}\)

\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}.\left(-\left(\sqrt{a}+\sqrt{b}\right)\right)\)

\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right).\left(-1\right).\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\)

\(=-\left(\sqrt{a}-\sqrt{b}\right)=\sqrt{b}-\sqrt{a}\)

cuối cùng cũng xong, mong bn phù hộ độ trì cho mk

a \(\sqrt{16x-8}+\sqrt{36x-18}-\sqrt{64x-3x}=\sqrt{10}\)

\(\Leftrightarrow\sqrt{8\left(2x-1\right)}+\sqrt{18\left(2x-1\right)}-\sqrt{32\left(2x-1\right)}=\sqrt{10}\)

\(\Leftrightarrow\sqrt{2x-1}\left(\sqrt{8}+\sqrt{18}-\sqrt{32}\right)=\sqrt{10}\)

\(\Leftrightarrow\sqrt{2x-1}\left(2\sqrt{2}+3\sqrt{2}-4\sqrt{2}\right)=\sqrt{10}\)

\(\Leftrightarrow\sqrt{2\left(2x-1\right)}=\sqrt{10}\)

\(\Rightarrow2\left(2x-1\right)=10\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

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

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

\(\Rightarrow x-3=x+3\) ( vô nghiệm )