Nhớ giải rõ ra nhé

\(\sqrt{53-20\sqrt{4-\sqrt{9+4\sqrt{2}}}}=\sqrt{53-20\sqrt{4-\sqrt{\left(2\sqrt{2}\right)^2+2.2\sqrt{2}.1+1}}}=\sqrt{53-20\sqrt{4-\sqrt{\left(2\sqrt{2}+1\right)^2}}}=\sqrt{53-20\sqrt{4-2\sqrt{2}-1}}=\sqrt{53-20\sqrt{2-2\sqrt{2}+1}}=\sqrt{53-20\sqrt{\left(\sqrt{2}-1\right)^2}}=\sqrt{53-20\left(\sqrt{2}-1\right)}=\sqrt{53-20\sqrt{2}-20}=\sqrt{25-2.5.2\sqrt{2}+8}=\sqrt{\left(5-2\sqrt{2}\right)^2}=5-2\sqrt{2}\)

\(\sqrt{53-20\sqrt{4}+\sqrt{9-4\sqrt{2}}}\)

=\(\sqrt{53-40+\sqrt{\left(2\sqrt{2}-1\right)^2}}\)

=\(\sqrt{13+2\sqrt{2}-1}\)=\(\sqrt{12+2\sqrt{2}}\)

Câu 1: Sửa lạ đề chút nhé : 4x + 1  -> 4x -1 

 Đặt A = \(\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}\)

=>  \(\sqrt{2}.A\)= ​\(\sqrt{4x-1+2\sqrt{4x-1}+1}+\sqrt{4x-1-2\sqrt{4x-1}+1}\)​

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

= \(\left|\sqrt{4x-1}+1\right|+\left|\sqrt{4x-1}-1\right|\)

Vì \(\frac{1}{4}< x< \frac{1}{2}\Rightarrow0< 4x-1< 1\Rightarrow0< \sqrt{4x-1}< 1\)

nên \(\sqrt{2}A=\)\(\sqrt{4x-1}+1+1-\sqrt{4x-1}\)=2

=> \(A=2:\sqrt{2}=\sqrt{2}\)

Câu 2. Có: \(9-4\sqrt{2}=8-2.2\sqrt{2}+1=\left(2\sqrt{2}-1\right)^2\)

=> \(\sqrt{9-4\sqrt{2}}=2\sqrt{2}-1\)

=> ​\(4+\sqrt{9-4\sqrt{2}}=4+2\sqrt{2}-1=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)​

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

=> \(53-20\sqrt{4+\sqrt{9-4\sqrt{2}}}=53-20\left(\sqrt{2}+1\right)=33-2.10\sqrt{2}=5^2-2.5.2\sqrt{2}+8=\left(5-2\sqrt{2}\right)^2\)

=> \(\sqrt{53-20\sqrt{4+\sqrt{9-4\sqrt{2}}}}=5-2\sqrt{2}\)

\(\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}\)

\(1,=2\sqrt{3}-3\sqrt{3}+4\sqrt{3}=3\sqrt{3}\\ 2,=\left(2\sqrt{6}+2\sqrt{5}-4\sqrt{5}\right):5=\dfrac{2\sqrt{6}}{5}-\dfrac{2\sqrt{5}}{5}\\ 3,=6\sqrt{3}-\dfrac{4\sqrt{3}}{3}-4\sqrt{3}-\dfrac{5\sqrt{3}}{3}=2\sqrt{3}-\dfrac{9\sqrt{3}}{3}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\\ 4,Sửa:\dfrac{1}{\sqrt{5}-\sqrt{3}}-\dfrac{1}{\sqrt{5}+\sqrt{3}}\\ =\dfrac{\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)

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

2) \(=\left(2\sqrt{6}+2\sqrt{5}-4\sqrt{5}\right)=\dfrac{2\sqrt{6}}{5}+\dfrac{2\sqrt{5}}{5}-\dfrac{4\sqrt{5}}{5}\)

3) \(=6\sqrt{3}-\dfrac{4\sqrt{3}}{3}-4\sqrt{3}-\dfrac{5\sqrt{3}}{3}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\)

4) \(=\dfrac{\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}}{5-3}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)

giải giúp mình bài này ới ạ mình đng cần gấp 

Cho biểu thức 

c=(căng x-2/căng x+2+căng x+2/căng x-2)nhân căng x+2/2 - 4 căng x/căng x-2

a)

 \(P=\frac{\sqrt{a}}{\sqrt{a}+3}+\frac{2\sqrt{a}}{\sqrt{a}-3}-\frac{3a+9}{a-9}\)

\(P=\frac{\sqrt{a}}{\sqrt{a}+3}+\frac{2\sqrt{a}}{\sqrt{a}-3}-\frac{3a+9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)

\(P=\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}+\frac{\sqrt{a}\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}-\frac{3a+9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)

\(P=\frac{a-3\sqrt{a}+3+3\sqrt{a}-3a-9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)

\(P=\frac{-2a-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)

\(P=\frac{-2a-3}{a-9}\)

b) Để \(P=\frac{1}{3}\Rightarrow\frac{-2a-3}{a-9}=\frac{1}{3}\)

\(\Rightarrow3\left(-2a-3\right)=a-9\)

\(\Rightarrow-6a-9=a-9\)

\(\Rightarrow-6a-a=-9+9\)

\(\Rightarrow-7a=0\left(L\right)\)

Vậy ko có gt của a để P=1/3 ( mk ko chắc.....)

a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)

\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)

\(=4-3\cdot A\)

\(\Leftrightarrow A^3+3A-4=0\)

\(\Leftrightarrow A^3-A+4A-4=0\)

\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)

\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)

\(\Leftrightarrow A=1\)