Bài 2: Căn thức bậc hai và hằng đẳng thức căn bậc hai của bình phương

cho hỏi là lớp mấy vậy

a, \(5\sqrt{\left(-2\right)^4}=5\sqrt{2^4}=5.2^2=5.4=20\)

b, \(-4\sqrt{\left(-3\right)^6}=-4\sqrt{3^6}=-4.3^3=-4.27=-108\)

c,\(\sqrt{\sqrt{\left(-5\right)^8}}=\sqrt{\sqrt{5^8}}=\sqrt{5^4}=5^2=25\)

d ,\(2\sqrt{\left(-5\right)^6}+3\sqrt{\left(-2\right)^8}\)

\(=2\sqrt{5^6}+3\sqrt{2^8}\)

=\(2.5^3+3.2^4=2.125+3.16=298\)

a)\(\sqrt{\left(4+\sqrt{2}\right)^2}=\left|4+\sqrt{2}\right|=4+\sqrt{2}\)

b)\(\sqrt{\left(3-\sqrt{3}\right)^2}=\left|3-\sqrt{3}\right|=3-\sqrt{3}\)

c)\(\sqrt{\left(4-\sqrt{17}\right)^2}=\left|4-\sqrt{17}\right|=\sqrt{17}-4\)

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

a) \(9+4\sqrt{5}=4+4\sqrt{5}+5=2^2+2\cdot2\sqrt{5}+\left(\sqrt{5}\right)^2=\left(\sqrt{5}+2\right)^2\left(ĐPCM\right)\)

a, Để \(\sqrt{\left(x-1\right)\left(x-3\right)}\) xác định thì (x-1)(x-3)\(\ge\)0

TH1: \(\left\{{}\begin{matrix}x-1\ge0\\x-3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\ge3\end{matrix}\right.\Leftrightarrow}x\ge3}\)TH2:\(\left\{{}\begin{matrix}x-1\le0\\x-3\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x\le3\end{matrix}\right.\Leftrightarrow}x\le1}\) Vậy nếu \(x\ge3\) hoặc \(x\le1\) thì biểu thức có nghĩa

b, Để \(\sqrt{x^2-4}=\sqrt{\left(x-2\right)\left(x+2\right)}\)có nghĩa thì (x-2)(x+2)\(\ge0\)

TH1: \(\left\{{}\begin{matrix}x-2\ge0\\x+2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ge-2\end{matrix}\right.\Leftrightarrow x\ge}2}\)TH2:\(\left\{{}\begin{matrix}x-2\le0\\x+2\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le2\\x\le-2\end{matrix}\right.\Leftrightarrow}x\le-2}\)Vậy nếu \(x\ge2\) hoặc \(x\le-2\) thì biểu thức có nghĩa

a) \(\sqrt{9x^2}=2x+1\) (1)

\(\Leftrightarrow3\cdot\left|x\right|=2x+1\)

\(\Leftrightarrow3\cdot\left|x\right|-2x=1\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2x=1\left(đk:x\ge0\right)\\3\cdot\left(-x\right)-2x=1\left(đk:x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(đk:x\ge0\right)\\x=-\dfrac{1}{5}\left(đk:x< 0\right)\end{matrix}\right.\)

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

Vậy tập nghiệm phương trình (1) là \(S=\left\{-\dfrac{1}{5};1\right\}\)

b) \(\sqrt{x^2+6x+9}=3x-1\) (2)

\(\Leftrightarrow x^2+6x+9=\left(3x-1\right)^2\)

\(\Leftrightarrow x^2+6x+9=9x^2-6x+1\)

\(\Leftrightarrow x^2+6x+9-9x^2+6x-1=0\)

\(\Leftrightarrow-8x^2+12x+8=0\)

\(\Leftrightarrow2x^2-3x-4=0\)

\(\Leftrightarrow x=\dfrac{-\left(-3\right)\pm\sqrt{\left(-3\right)^2-4\cdot2\cdot\left(-2\right)}}{2\cdot2}\)

\(\Leftrightarrow x=\dfrac{3\pm\sqrt{9+16}}{4}\)

\(\Leftrightarrow x=\dfrac{3\pm5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+5}{4}\\x=\dfrac{3-5}{4}\end{matrix}\right.\)

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

sau khi dùng phép thử ta nhận thấy \(x\ne-\dfrac{1}{2}\)

Vậy tập nghiệm phương trình (2) là \(S=\left\{2\right\}\)

c) \(\sqrt{1-4x+4x^2}=5\) (3)

\(\Leftrightarrow1-4x+4x^2=25\)

\(\Leftrightarrow\left(1-2x\right)^2=25\)

\(\Leftrightarrow1-2x=\pm5\)

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

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

Vậy tập nghiệm phương trình (1) là \(S=\left\{-2;3\right\}\)

d) \(\sqrt{x^4}=7\) (4)

\(\Leftrightarrow x^2=7\)

\(\Leftrightarrow x=\pm\sqrt{7}\)

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

Vậy tập nghiệm phương trình (4) là \(S=\left\{-\sqrt{7};\sqrt{7}\right\}\)

a, \(x^2-7=x^2-\left(\sqrt{7}\right)^2=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)

b, \(x^2-2\sqrt{2}x+2=x^2-2\sqrt{2}x+\left(\sqrt{2}\right)^2=\left(x-\sqrt{2}\right)^2\)

c, \(x^2+2\sqrt{13}x+13=x^2+2\sqrt{13}x+\left(\sqrt{13}\right)^2=\left(x+\sqrt{13}\right)^2\)

a) \(\dfrac{x^2-5}{x+\sqrt{5}}\)(với x khác -\(\sqrt{5}\)) =\(\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}\) = x-\(\sqrt{5}\) vậy \(\dfrac{x^2-5}{x+\sqrt{5}}\) = x-\(\sqrt{5}\) với x khác -\(\sqrt{5}\) b) \(\dfrac{x^2+2\sqrt{2}x+2}{x^2-2}\) ( với x khác +-\(\sqrt{2}\) ) = \(\dfrac{\left(x+\sqrt{2}\right)^2}{\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)}\) =\(\dfrac{x+\sqrt{2}}{x-\sqrt{2}}\) vậy \(\dfrac{x^2+2\sqrt{2}x+2}{x^2-2}\) =\(\dfrac{x+\sqrt{2}}{x-\sqrt{2}}\) với x khác +-\(\sqrt{2}\)

a. Ta có : \(\sqrt{8}< \sqrt{9}\) ( vì 8< 9)

hay \(2\sqrt{2}< 3\)

\(\Rightarrow\) \(2\sqrt{2}+6< 3+6\)

hay \(2\sqrt{2}+6< 9\)

b. Ta có : \(\sqrt{6}>\sqrt{4}\) (vì 6 > 4 )

hay \(\sqrt{2.3}>2\)

\(\Rightarrow\) 2\(\sqrt{2.3}\) > 4

\(\Rightarrow\) 2 + \(2\sqrt{2.3}\) + 3 > 9

hay \(\left(\sqrt{2}+\sqrt{3}\right)^2\)> 9

\(\Rightarrow\) \(\sqrt{2}+\sqrt{3}>3\)

c. Ta có: \(\sqrt{80}>\sqrt{49}\) (vì 80>49)

hay \(4\sqrt{5}\) > 7

\(\Rightarrow\) 9 + \(4\sqrt{5}\) > 16

d. Ta có : \(2\sqrt{33}>2\sqrt{25}\) (vì 33> 25 ) hay \(2\sqrt{23}>2.5\)

\(\Rightarrow\) - \(2\sqrt{33}\) < - 2.5

\(\Rightarrow\) 11 - \(2\sqrt{11.3}\) +3 < 11- 2.5 +3

hay \(\left(\sqrt{11}-\sqrt{3}\right)^2\) < 4

\(\Rightarrow\) \(\sqrt{11}-\sqrt{3}< 2\)

a, Ta có : \(4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}\right)^2-2\sqrt{3}\times1+1^2=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\left|\sqrt{3}-1\right|-\sqrt{3}\)

Ta có : \(\sqrt{3}>\sqrt{1}\)(vì 3>1)

\(\Leftrightarrow\sqrt{3}>1\Leftrightarrow\sqrt{3}-1>0\Rightarrow\left|\sqrt{3}-1\right|=\sqrt{3}-1\)

Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\left|\sqrt{3}-1\right|-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)