a) \(\sqrt{19-6\sqrt{2}}=3\sqrt{2}-1\)

b) \(\sqrt{11-6\sqrt{2}}=3-\sqrt{2}\)

d) \(\sqrt{21+12\sqrt{3}}=2\sqrt{3}+3\)

e) \(\sqrt{57-40\sqrt{2}}=4\sqrt{2}-5\)

\(a.\sqrt{19-6\sqrt{2}}=\sqrt{18-2.3\sqrt{2}+1}=3\sqrt{2}-1\)

\(b.\sqrt{21+12\sqrt{3}}=\sqrt{12+2.2\sqrt{3}.3+9}=2\sqrt{3}+3\)

\(c.\sqrt{57-40\sqrt{2}}=\sqrt{32-2.4\sqrt{2}.5+25}=4\sqrt{2}-5\)

\(d.\sqrt{\left(5-2\sqrt{6}\right)\left(4-2\sqrt{3}\right)}=\sqrt{3-2\sqrt{3}.\sqrt{2}+2}.\sqrt{3-2\sqrt{3}+1}=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-1\right)\) \(e.\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}=\sqrt{18+2.3\sqrt{2}.\sqrt{3}+3}+\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}=6\sqrt{2}\) \(g.\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}=\sqrt{4-2.2\sqrt{3}+3}-\sqrt{4+2.2\sqrt{3}+3}=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3}\)

a)

=\(\sqrt{18-2.3\sqrt{2}.1+1}\)

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

\(=3\sqrt{2}-1\)

b)

=\(\sqrt{12+2.2\sqrt{3}.3+9}\)

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

=\(2\sqrt{3}+3\)

c)

=\(\sqrt{25-2.5.4\sqrt{2}+32}\)

=\(\sqrt{\left(5-4\sqrt{2}\right)^2}\)

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

d)

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

e)

\(=\sqrt{18+2.3\sqrt{2}.\sqrt{3}+3}+\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}\\ =\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\\ =3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}\\ =6\sqrt{2}\)

g)

\(=\sqrt{4-2.2.\sqrt{3}+3}-\sqrt{4+2.2.\sqrt{3}+3}\\ =\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\\ =2-\sqrt{3}-2-\sqrt{3}\\ =-2\sqrt{3}\)

b, \(\sqrt{21+12\sqrt{3}}=\sqrt{21+2.3.2.\sqrt{3}}=\sqrt{9+2.3.\sqrt{12}+12}\)

\(=\sqrt{\left(3+\sqrt{12}\right)^2}=3+\sqrt{12}\)

\(c,\sqrt{57-40\sqrt{2}}=\sqrt{57-2.4.5.\sqrt{2}}=\sqrt{25-2.5.\sqrt{32}}\)

\(=\sqrt{\left(5-\sqrt{32}\right)^2}=\left|5-\sqrt{32}\right|=5-\sqrt{32}\)

\(d,\sqrt{\left(5-2\sqrt{6}\right)\left(4-2\sqrt{3}\right)}=\sqrt{\left(3-2.\sqrt{2}.\sqrt{3}+2\right)\left(3-2\sqrt{3}+1\right)}\) \(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2\left(\sqrt{3}-1\right)^2}=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-1\right)\)

\(e,A=\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}\)

Ta có :

\(21+6\sqrt{6}=\dfrac{42+12\sqrt{6}}{2}=\dfrac{36+2.6.\sqrt{6}+6}{2}=\left(\dfrac{6+\sqrt{6}}{\sqrt{2}}\right)^2\) Tương tự : \(21-6\sqrt{6}=\left(\dfrac{6-\sqrt{6}}{\sqrt{2}}\right)^2\)

Do đó :

\(A=\sqrt{\left(\dfrac{6+\sqrt{6}}{\sqrt{2}}\right)^2}+\sqrt{\left(\dfrac{6-\sqrt{6}}{\sqrt{2}}\right)^2}=\dfrac{6+\sqrt{6}}{\sqrt{2}}+\dfrac{6-\sqrt{6}}{\sqrt{2}}=\dfrac{6+\sqrt{6}+6-\sqrt{6}}{\sqrt{2}}\)\(=\dfrac{12}{\sqrt{2}}=\dfrac{12\sqrt{2}}{2}=6\sqrt{2}\)

Phần g làm tương tự như phần e nha bạn :>

Chúc bạn học tốt :>

tớ ko chép lại đề, kí hiệu nhé

(1) \(=\left(\sqrt{6}-\sqrt{5}\right)^2-\sqrt{\left|\sqrt{6}+\sqrt{5}\right|^2}=\left(\sqrt{6}-\sqrt{5}\right)^2-\left(\sqrt{6}+\sqrt{5}\right)=1-2\sqrt{30}-\sqrt{6}-\sqrt{5}\)

ai ra đề mà để đáp án dài thế này mất thẩm mĩ quá!!!

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

(3) \(=\sqrt{\left|\sqrt{7}+2\right|^2}-\sqrt{\left|3-\sqrt{5}\right|^2}=\sqrt{7}+2-3+\sqrt{5}=\sqrt{7}+\sqrt{5}-1\)

lại thêm 1 phép tính không đẹp....

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

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

kiểm tra lại kết quả nhé ^^! Cảm ơn!

\(a,\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{3}+\sqrt{5}-\left(\sqrt{5}+1\right)=\sqrt{3}-1\\ b,=3-2\sqrt{2}-\left(3\sqrt{2}+1\right)=2-5\sqrt{2}\\ c,=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\\ d,=\sqrt{11}+1-\left(\sqrt{11}-1\right)=2\\ e,=\sqrt{7}-\sqrt{3}-\left(\sqrt{7}-\sqrt{2}\right)=\sqrt{2}-\sqrt{3}\)

Bài 2:

a)\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: \(x\ge2\))

\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+\dfrac{6}{\sqrt{81}}\sqrt{x-2}=-4\)

\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)

\(\Leftrightarrow-\sqrt{x-2}=-4\) \(\Leftrightarrow x-2=16\)

\(\Leftrightarrow x=18\) (thỏa)

Vậy...

b)\(\sqrt{9x^2+12x+4}=4x\)(Đk:\(9x^2+12x+4\ge0\))

\(\Leftrightarrow\left\{{}\begin{matrix}4x\ge0\\9x^2+12x+4=16x^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+12x+4=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+14x-2x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x-2\right)\left(-7x-2\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{7}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow x=2\) (tm đk)

Vậy...

c) \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}\) (đk: \(x\ge1\))

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

\(\Leftrightarrow\sqrt{x-1}=\dfrac{1}{2}\) \(\Leftrightarrow x=\dfrac{5}{4}\) (tm)

Vậy...

1. \(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}=2\sqrt{2}\)

2. \(\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}=\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-3\right)^2}\\ =3-\sqrt{3}+2\sqrt{3}-3=\sqrt{3}\)

3. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(3+\sqrt{6}\right)^2}\\ =2\sqrt{6}-3+3+\sqrt{6}=3\sqrt{6}\)

1.\(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\)

\(=\sqrt{5}+\sqrt{2}-\left(\sqrt{5}-\sqrt{2}\right)=2\sqrt{2}\)

2. \(\sqrt{12-6\sqrt{3}+\sqrt{21-12\sqrt{3}}}=\sqrt{12-6\sqrt{3}+\sqrt{\left(3-2\sqrt{3}\right)^2}}\)

\(=\sqrt{12-6\sqrt{3}+2\sqrt{3}-3}=\sqrt{9-4\sqrt{3}}\)

3. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(\sqrt{6}-3\right)^2}\)

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

\(A=4-\sqrt{21-8\sqrt{5}}=4-\sqrt{4^2-8\sqrt{5}+\left(\sqrt{5}\right)^2}.\)

\(A=4-\sqrt{\left(4-\sqrt{5}\right)^2}=4-\left(4-\sqrt{5}\right)\)

=> \(A=\sqrt{5}\)

\(\sqrt{\left(2\sqrt{2-1}\right)^2}-\sqrt{17+12\sqrt{2}}\\ =\left|2\sqrt{2}-1\right|-\sqrt{9+2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}\\ =2\sqrt{2}-1-\sqrt{\left(3+2\sqrt{2}\right)^2}\\=2\sqrt{2}-1-\left(3+2\sqrt{2}\right)\\ =2\sqrt{2}-1-3-2\sqrt{2}\\ =-4\)

__

\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\\ =\left|2-\sqrt{5}\right|+\sqrt{9-2\cdot3\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\\ =2-\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}\\ =2-\sqrt{5}+3-\sqrt{5}\\ =5-2\sqrt{5}\)

__

\(\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{19+6\sqrt{2}}\\ =\left|4-3\sqrt{2}\right|-\sqrt{18+2\cdot3\cdot\sqrt{2}+1}\\ =4-3\sqrt{2}-\sqrt{\left(3\sqrt{2}+1\right)^2}\\ =4-3\sqrt{2}-3\sqrt{2}-1\\ =3-6\sqrt{2}\)