Bài tập tự luận: Rút gọn một số biểu thức chứa căn SVIP

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 $N=\sqrt{2x-1}+2$.