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

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

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

b/ \(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=2\left(4+\sqrt{15}\right)\left(4+\sqrt{15}\right)=2\left(16-15\right)\)

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

\(M=\sqrt{\frac{18-3\sqrt{3}-8\sqrt{3}+4}{11}}+\sqrt{\frac{5\sqrt{3}+6+20+8\sqrt{3}}{13}}\)

\(M=\sqrt{\frac{11\left(2-\sqrt{3}\right)}{11}}+\sqrt{\frac{13\left(2+\sqrt{3}\right)}{13}}=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)

\(M=\frac{1}{\sqrt{2}}\left(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\right)\)

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

\(M=\frac{1}{\sqrt{2}}\left(\sqrt{3}-1+\sqrt{3}+1\right)=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

1.\(=\sqrt{9+4\sqrt{5}-2+\sqrt{5}}\)

\(=\sqrt{7+5\sqrt{5}}\)

Câu 2,3 ttự.

4.\(=2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}\)

5.\(=\sqrt{3}-1+\sqrt{3}+1\)(Ptích trong căn thành bình phương)

\(=2\sqrt{3}\)

6.Đề sai bạn đánh thừa dấu căn.

Đề đúng:\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)

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

\(=2\)

a, Xét \(M^2=4-\sqrt{10-2\sqrt{5}}+4+\sqrt{10-2\sqrt{5}}-2\sqrt{\left(4-\sqrt{10-2\sqrt{5}}\right)\left(4+\sqrt{10-2\sqrt{5}}\right)}\)

\(=8-2\sqrt{4^2-10+2\sqrt{5}}\\ =8-2\sqrt{16-10+2\sqrt{5}}\\ =8-2\sqrt{6+2\sqrt{5}}\\ =8-2\sqrt{\left(\sqrt{5}+1\right)^2}\\ =8-2\left(\sqrt{5}+1\right)\\ =8-2\sqrt{5}-2=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\\ \Rightarrow M=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)

b,

\(P\sqrt{2}=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}\\ =\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{3}+1}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{3}+1}\\ =\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{3+\sqrt{3}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}\\ =\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{3}+1\right)}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{3}-1\right)}\\ =\frac{\sqrt{2}\left[\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)+\left(2-\sqrt{3}\right)\left(\sqrt{3}+1\right)\right]}{\sqrt{3}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\\ =\frac{\sqrt{2}\left(2\sqrt{3}+3-2-\sqrt{3}+2\sqrt{3}-3+2-\sqrt{3}\right)}{\sqrt{3}\left(3-1\right)}\\ =\frac{\sqrt{2}\left(2\sqrt{3}\right)}{\sqrt{3}\cdot2}=\sqrt{2}\\ \Rightarrow P=1\)

c,

\(Q\sqrt{2}=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\\ =\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\\ \Rightarrow Q=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{2}\cdot\sqrt{3}=6\)

Chúc bạn học tốt nha.

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

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

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

\(=\sqrt{4+2\sqrt{3}}\)

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

\(b,=\sqrt{6-2\sqrt{3+\sqrt{12+2\sqrt{12}+1}}}\)

\(=\sqrt{6-2\sqrt{3+\sqrt{12}+1}}\)

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

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

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

\(c,=\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{4+2.2\sqrt{3}+3}}}\)

\(=\sqrt{\sqrt{3}+\sqrt{48-10\left(2+\sqrt{3}\right)}}\)

\(=\sqrt{\sqrt{3}+\sqrt{28-10\sqrt{3}}}\)

\(=\sqrt{\sqrt{3}+\sqrt{25-2.5\sqrt{3}+3}}\)

\(=\sqrt{\sqrt{3}+5-\sqrt{3}}=\sqrt{5}\)

\(d,=\sqrt{23-6\sqrt{10+4\sqrt{2-2\sqrt{2}+1}}}\)

\(=\sqrt{23-6\sqrt{6+4\sqrt{2}}}\)

\(=\sqrt{23-6\sqrt{4+2.2\sqrt{2}+2}}\)

\(=\sqrt{23-6\sqrt{\left(2+\sqrt{2}\right)^2}}\)

\(=\sqrt{23-12-6\sqrt{2}}=\sqrt{11-6\sqrt{2}}\)

\(=\sqrt{9-2.3\sqrt{2}+2}=3-\sqrt{2}\)

a) Ta có: \(\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)

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

\(=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)

b) Ta có: \(\sqrt{6-2\sqrt{3+\sqrt{13+4\sqrt{3}}}}\)

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

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

\(=\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\)

c) Ta có: \(\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)

\(=\sqrt{\sqrt{3}+\sqrt{48-10\left(2+\sqrt{3}\right)}}\)

\(=\sqrt{\sqrt{3}+\sqrt{28-10\sqrt{3}}}\)

\(=\sqrt{\sqrt{3}+5-\sqrt{3}}\)

\(=\sqrt{5}\)

d) Ta có: \(\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}\)

\(=\sqrt{23-6\sqrt{10+4\left(\sqrt{2}-1\right)}}\)

\(=\sqrt{23-6\sqrt{6-4\sqrt{2}}}\)

\(=\sqrt{23-6\left(2-\sqrt{2}\right)}\)

\(=\sqrt{11+6\sqrt{2}}\)

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

b)\(\sqrt{m+2\sqrt{m-1}}+\sqrt{m-2\sqrt{m-1}}\)

\(\Leftrightarrow\sqrt{m-1+2\sqrt{m-1}+1}+\sqrt{m-1-2\sqrt{m-1}+1}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{m-1}+1\right)^2}+\sqrt{\left(\sqrt{m-1}-1\right)^2}\)

\(\Leftrightarrow\sqrt{m-1}+1+\sqrt{m-1}-1\Leftrightarrow2\sqrt{m-1}\)

Câu 1 phá từng lớp ra :VD\(9+4\sqrt{2}\) =\((\sqrt{2}+2)^2\)

Câu 2:m+2\(\sqrt{m-1}\) =m-1+1+2\(\sqrt{m-1}\) =\((\sqrt{m-1} -1)^2 \)

a)\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

\(\Leftrightarrow\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)

\(\Leftrightarrow\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)

\(\Leftrightarrow\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)

\(\Leftrightarrow\sqrt{13+30\sqrt{2}+30}\)

\(\Leftrightarrow\sqrt{43+30\sqrt{2}}\)

\(\Leftrightarrow\sqrt{\left(3\sqrt{2}+5\right)^2}\)

\(\Leftrightarrow3\sqrt{2}+5\)

tth làm sau có để thì để tên khác đi

mình đang cần gấp làm nhanh nha mọi người

2: \(=\sqrt{2}-1-\sqrt{2}=-1\)

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

\(=\dfrac{7+4\sqrt{3}-7+4\sqrt{3}}{1}=8\sqrt{3}\)

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

Câu 4: a) ĐK: \(x^2\ge9\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)

b) ĐK: \(x^2-3x+2\ge0\Leftrightarrow\left[{}\begin{matrix}x\le1\\x\ge2\end{matrix}\right.\)

c) Đk: \(-3\le x< 5\)

d) x + 3 và 5 - x đồng dấu. Xét hai trường hợp:

\(\left\{{}\begin{matrix}x+3\ge0\\5-x>0\left(\text{do mẫu phải khác 0}\right)\end{matrix}\right.\Leftrightarrow-3\le x< 5\)

\(\left\{{}\begin{matrix}x+3< 0\\5-x< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -3\\x>5\end{matrix}\right.\) do x ko thể đồng thời thỏa mãn cả hai nên loại.

Câu 1:

a) Đặt \(A=x+\sqrt{\left(x+2\right)^2}\cdot\left(x-2\right)\)

\(A=x+\left|x+2\right|\cdot\left(x-2\right)\)

+) Với \(x\ge-2\):

\(A=x+\left(x+2\right)\left(x-2\right)=x+x^2-4\)

+) Với \(x< -2\):

\(A=x-\left(x+2\right)\left(x-2\right)=x-x^2+4\)

b) \(B=\sqrt{m^2-6m+9-2m}\)

\(B=\sqrt{m^2-8m+9}\)

Bạn xem lại đề nhé :)

c) \(C=1+\sqrt{\frac{\left(x-1\right)^2}{x-1}}\)

\(C=1+\sqrt{x-1}\)

d) \(D=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)

\(D=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)

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

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

+) Xét \(x\ge8\):

\(D=\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)

+) Xét \(4< x< 8\):

\(D=\sqrt{x-4}+2+2-\sqrt{x-4}=4\)

Vậy....

Câu 2:

a) Ta có: \(\sqrt{5}< \sqrt{9}=3\)

b) \(2\sqrt{2}< \left(2+1\right)\sqrt{2}=3\sqrt{2}\)

c) \(-4\sqrt{5}>-4\sqrt{6}>-6\sqrt{6}\)

d) Xét hiệu: \(2\sqrt{3}-5-\sqrt{3}+4=\sqrt{3}-1>\sqrt{1}-1=0\)

Nên \(2\sqrt{3}-5>\sqrt{3}-4\)

e) Tương tự