Chuyên đề: Căn thức bậc hai và hằng đẳng thức √A² = |A|

Hướng dẫn giải:

a)

\(\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}\\ =\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\\ =\sqrt{2}+1+2-\sqrt{2}=3.\)

b)

 \(\left(\sqrt{5-2\sqrt{6}}+\sqrt{2}\right)\sqrt{3}\\ =\left(\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{2}\right)\sqrt{3}\\ =\left(\sqrt{3}-\sqrt{2}+\sqrt{2}\right)\sqrt{3}\\ =\sqrt{3}.\sqrt{3}=3.\)

Hướng dẫn giải:

a)

 \(M=\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\\ =\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}\\ =2-\sqrt{2}+3\sqrt{2}-2=2\sqrt{2}.\)

b)

\(N=\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\\ =\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}\\ =3-2\sqrt{2}+2\sqrt{2}+1=4.\)

Hướng dẫn giải:

a)

\(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\\ =\left(\sqrt{3}-\sqrt{2}\right).\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\\ =\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)\\ =\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2=3-2=1.\)

b) 

\(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\\ =\sqrt{\left(2+2\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\\ =2+2\sqrt{5}+\sqrt{5}-2\\ =3\sqrt{5}.\)

Hướng dẫn giải:

a)

\(\sqrt{\sqrt{5}-\sqrt{9-\sqrt{29+12\sqrt{5}}}}\\ =\sqrt{\sqrt{5}-\sqrt{9-\sqrt{\left(2\sqrt{5}+3\right)^2}}}\\ =\sqrt{\sqrt{5}-\sqrt{9-\left(2\sqrt{5}+3\right)}}\\ =\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\\ =\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\\ =\sqrt{1}=1.\)

b)

 \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\\ =\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\\ =\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\\ =\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\\ =\sqrt{13+30.\left(\sqrt{2}+1\right)}\\ =\sqrt{43+30\sqrt{2}}\\ =\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5.\)

Hướng dẫn giải:

a)

\(A=x+3+\sqrt{x^2-6x+9}=x+3+\sqrt{\left(x-3\right)^2}\\ =x+3+\left|x-3\right|.\)

Do  \(x\le3\)nên  \(\left|x-3\right|=3-x\). Khi đó  \(A=x+3+3-x=6.\)

b)

\(B=\dfrac{\sqrt{x^2-2x+1}}{x-1}=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)

Do  \(x>1\) nên \(\left|x-1\right|=x-1\). Khi đó \(B=1.\)

c)

\(C=\sqrt{x^2+4x+4}-\sqrt{x^2}=\sqrt{\left(x+2\right)^2}-\sqrt{x^2}=\left|x+2\right|-\left|x\right|\)

Do \(-2\le x\le0\) nên  \(\left|x+2\right|=x+2,\left|x\right|=-x\). Khi đó \(C=x+2-\left(-x\right)=2x+2.\)

Hướng dẫn giải:

a)

\(A=\sqrt{1-4a+4a^2}-2a=\sqrt{\left(1-2a\right)^2}-2a=\left|1-2a\right|-2a.\)

+ Nếu \(1-2a\ge0\Leftrightarrow a\le\dfrac{1}{2}\)thì \(\left|1-2a\right|=1-2a\).

Khi đó \(A=1-2a-2a=1-4a\).

+ Nếu \(1-2a< 0\Leftrightarrow a>\dfrac{1}{2}\) thì \(\left|1-2a\right|=2a-1\).

Khi đó \(A=2a-1-2a=-1\).

b)

\(x-2y-\sqrt{x^2-4xy+4y^2}=x-2y-\sqrt{\left(x-2y\right)^2}=x-2y-\left|x-2y\right|\)+ Nếu \(x-2y\ge0\Leftrightarrow x\ge2y\) thì \(\left|x-2y\right|=x-2y\)

Khi đó \(B=x-2y-\left(x-2y\right)=0.\)

+ Nếu \(x-2y< 0\Leftrightarrow x< 2y\)thì \(\left|x-2y\right|=2y-x\)

Khi đó \(B=x-2y-\left(2y-x\right)=x-2y-2y+x=2x.\)

c)

\(C=2x-1-\dfrac{\sqrt{x^2-10x+25}}{x-5}=2x-1-\dfrac{\sqrt{\left(x-5\right)^2}}{x-5}=2x-1-\dfrac{\left|x-5\right|}{x-5}\)Điều kiện xác định \(x\ne5\)

Với \(x-5>0\Leftrightarrow x>5\). Khi đó \(C=2x-1-\dfrac{x-5}{x-5}=2x-1-1=2x-2.\)

Với \(x-5< 0\Leftrightarrow x< 5\). Khi đó \(C=2x-1-\dfrac{-\left(x-5\right)}{x-5}=2x-1+1=2x.\)

Hướng dẫn giải:

a)

\(M=\sqrt{x+2\sqrt{x-1}}=\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}=\sqrt{\left(\sqrt{x-1}+1\right)^2}=\left|\sqrt{x-1}+1\right|\)Với \(x\ge1\) thì \(\sqrt{x-1}\) được xác định. 

Vì \(\sqrt{x-1}\ge0\Rightarrow\sqrt{x-1}+1>0.\)

Khi đó \(M=\sqrt{x-1}+1\).

b) 

\(N=\sqrt{2x+3+4\sqrt{2x-1}}=\sqrt{\left(2x-1\right)+2.\sqrt{2x-1}+4}=\sqrt{\left(\sqrt{2x-1}+2\right)^2}=\left|\sqrt{2x-1}+2\right|\)

Với \(x\ge\dfrac{1}{2}\) thì \(\sqrt{2x-1}\) được xác định.

Vì \(\sqrt{2x-1}\ge0\Rightarrow\sqrt{2x-1}+2>0.\)

Khi đó \(N=\sqrt{2x-1}+2\).