\(\sqrt{x}=2\)

mà \(x^2=\left(\sqrt{x}\right)^4\)

\(\Rightarrow\left(\sqrt{x}\right)^4=2^4\)

\(\Rightarrow\left(\sqrt{x}\right)^4=16\)hay  \(x^2=16\)

 vậy chọn ý D

hãy giải thích vì sao bạn chọn câu trả lời đó

mk chọn câu D)

Vì mk thích thì mk chọn thôi !

Nhưng mk nghĩ là đúng đó

1:

a: \(\sqrt{25}+\sqrt{49}=5+7=12\)

b: \(\sqrt{121}-\sqrt{81}=11-9=2\)

2: x>-2

=>2x>-4

=>2x+1>-3

=>Với x>-2 thì \(\sqrt{2x+1}\) chưa chắc có nghĩa

3:

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

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

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

b: \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)

\(=\left(3\sqrt{7}-2\sqrt{14}\right)\cdot\sqrt{7}+14\sqrt{2}\)

\(=21-14\sqrt{2}+14\sqrt{2}=21\)

c:

\(\dfrac{\sqrt{27}-\sqrt{108}+\sqrt{12}}{\sqrt{3}}\)

\(=\dfrac{3\sqrt{3}-6\sqrt{3}+2\sqrt{3}}{\sqrt{3}}=3+2-6=-1\)

Ta có:

\(\Rightarrow\)

Vậy đáp án đúng là D. 16

Ta có: \(\sqrt{x}=2\\ \Rightarrow x=4\\ \Rightarrow x^2=4^2\\ \Rightarrow x^2=16\)

Vậy chọn đáp án (D.16)

câu c

\(\sqrt{25x}-\sqrt{16x}=9\) khi \(x\) bằng :

(A) 1 (B) 3 (C) 9 (D) 81

ĐK: x \(\ge\) 0

Ta có: \(\sqrt{25x}-\sqrt{16x}=9\Leftrightarrow5\sqrt{x}-4\sqrt{x}=9\Leftrightarrow\sqrt{x}=9\Leftrightarrow x=81\left(tm\right)\)

Vậy đáp án đúng là D

Bài 1:

a. Ta có \(\sqrt{\dfrac{2}{x^2}}=\dfrac{\sqrt{2}}{\left|x\right|}=\dfrac{\sqrt{2}}{x}\) ,để biểu thức có nghĩa thì \(x>0\)

b. Để biểu thức \(\sqrt{\dfrac{-3}{3x+5}}\) có nghĩa thì \(\dfrac{-3}{3x+5}\ge0\) 

mà \(-3< 0\Rightarrow3x+5< 0\) \(\Rightarrow x< \dfrac{-5}{3}\)

Bài 2:

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

b. \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)

\(=14-14\sqrt{2}+7+14\sqrt{2}\)

\(=21\)

c. \(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}\)

\(=14-6\sqrt{28}+18+6\sqrt{28}\)

\(=32\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left(2x+1\right)^2=6^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)

\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

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

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

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

d) \(pt\Leftrightarrow\left|x-3\right|=9\Leftrightarrow\left[{}\begin{matrix}x-3=-9\\x-3=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=12\end{matrix}\right.\)

Chọn D

`a)\sqrt{9-4sqrt5}-sqrt5`

`=sqrt{5-2.2sqrt5+4}-sqrt5`

`=sqrt{(sqrt5-2)^2}-sqrt5`

`=|\sqrt5-2|-sqrt5`

`=sqrt5-2-sqrt5=-2`

`b)\sqrt{7-4sqrt3}+sqrt{4-2sqrt3}`

`=\sqrt{4-2.2sqrt3+3}+\sqrt{3-2sqrt3+1}`

`=sqrt{(2-sqrt3)^2}+sqrt{(sqrt3-1)^2}`

`=|2-sqrt3|+|sqrt3-1|`

`=2-sqrt3+sqrt3-1=1`

`c)(x-49)/(sqrtx-7)(x>=0,x ne 49)`

`=((sqrtx-7)(sqrtx+7))/(sqrtx-7)`

`=sqrtx+7`

`d)\sqrt{4+2\sqrt3}-\sqrt{13+4sqrt3}`

`=\sqrt{3+2sqrt3+1}-\sqrt{12+2.2sqrt3+1}`

`=sqrt{(sqrt3+1)^2}-\sqrt{(2sqrt3+1)^2}`

`=sqrt3+1-2sqrt3-1=-sqrt3`

`e)2+sqrt{17-4sqrt{9+4sqrt{45}}}`(câu này hơi sai)