\(A=\frac{8\sqrt{41}}{\sqrt{\sqrt{41}^2+2.2.\sqrt{41}+2^2}+\sqrt{\sqrt{41}^2-2.2.\sqrt{41}+2^2}}.\frac{1}{\sqrt{3}-\sqrt{2}}\)

\(=\frac{8\sqrt{41}}{\sqrt{\left(\sqrt{41}+2\right)^2}+\sqrt{\left(\sqrt{41}-2\right)^2}}.\frac{\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)

\(=\frac{8\sqrt{41}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{41}+2+\sqrt{41}-2}=\frac{8\sqrt{41}\left(\sqrt{3}+\sqrt{2}\right)}{2\sqrt{41}}=4\left(\sqrt{3}+\sqrt{2}\right)\)

bạn ơi giải thích giúp mình chỗ dấu = số 3 với

sao mẫu nhân với\(\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)\) mà lại ra có

\(\sqrt{41}+2+2-\sqrt{41}\)ạ

\(\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}+\sqrt{45-\sqrt{41}}}}:\left(\sqrt{3}-\sqrt{2}\right)\) ( đề)

\(=\frac{8\sqrt{41}}{\sqrt{41}+2-\sqrt{41}-2}:\left(\sqrt{3}-\sqrt{2}\right)\)

\(=2\sqrt{41}:\left(\sqrt{3}-\sqrt{2}\right)\)

\(=2\sqrt{123}+2\sqrt{82}\)

vậy.....................

Dấu căn dưới mẫu có vt lộn ko z bạn?

à nó bị liền  : \(M=\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}\)

A = \(\frac{8\sqrt{41}}{2\sqrt{2^2+2.2.\sqrt{41}+\sqrt{41}^2}}\)

A = \(\frac{8\sqrt{41}}{2\sqrt{\left(2+\sqrt{41}\right)^2}}\)

A = \(\frac{8\sqrt{41}}{2\left|2+\sqrt{41}\right|}\)

A = \(\frac{8\sqrt{41}}{4+2\sqrt{41}}\)

B = \(\left(\frac{2x+1}{\sqrt{x}^3+1^3}-\frac{1}{\sqrt{x}-1}\right):\frac{x+\sqrt{x}+1+x+4}{x+\sqrt{x}+1}\)

B = \(\left(\frac{2x+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}\right).\frac{x+\sqrt{x}+1}{2x+\sqrt{x}+5}\)

Bạn tự làm tiếp nhé, mỏi tay quá!!

\(A=\frac{8\sqrt{41}}{2\sqrt{45+4\sqrt{41}}}=\frac{8\sqrt{41}}{2\sqrt{41+4\sqrt{41}+4}}=\frac{8\sqrt{41}}{2\sqrt{\left(\sqrt{41}\right)^2+2\cdot\sqrt{41}\cdot2+2^2}}\)

\(=\frac{8\sqrt{41}}{2\sqrt{\left(\sqrt{41}+2\right)^2}}=\frac{8\sqrt{41}}{2\left(\sqrt{41}+2\right)}=\frac{8\sqrt{41}\left(\sqrt{41}-2\right)}{2\left(41-4\right)}=\frac{328-16\sqrt{41}}{74}=\frac{164-8\sqrt{41}}{37}\)

\(B=\left(\frac{2x+1}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{x+4}{x+\sqrt{x}+1}\right)\)

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

\(=\left(\frac{2x+1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}-3}{x+\sqrt{x}+1}\right)\)

\(=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{x+\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}}{\sqrt{x}-3}=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}=\frac{x+3\sqrt{x}}{x-9}\)

Xét \(\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}=\frac{8\sqrt{41}}{\sqrt{\left(\sqrt{41}+2\right)^2}+\sqrt{\left(\sqrt{41}-2\right)^2}}=\frac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}=\frac{8\sqrt{41}}{2\sqrt{41}}=4\)
Phương trình trên tương đương:
x³+4x+5=0
<=>x(x²-1)+5(x+1)=0
<=>x(x-1)(x+1)+5(x+1)=0
<=>(x+1)(x²-x+5)=0
<=>x+1=0 hoặc x²-x+5=0(vô nghiệm)
<=>x=-1
Vậy pt trên có nghiệm là x=-1

x=-1 nha 

\(M=\dfrac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}\)

\(M=\dfrac{8\sqrt{41}}{\sqrt{\left(\sqrt{41}+2\right)^2}+\sqrt{\left(\sqrt{41}-2\right)^2}}\)

\(M=\dfrac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}\)

\(M=\dfrac{8\sqrt{41}}{2\sqrt{41}}=\dfrac{8}{2}=4\)

Vậy M = 4

Học tốt nhé :)

Bạn ơi,đâu có câu thức:a\(^2\)+b\(^2\) đâu?Chỉ có công thức a\(^{2^{ }}\)-b\(^2\) thôi mà?!:)))

\(A=\dfrac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}\cdot\dfrac{1}{\sqrt{3}-\sqrt{2}}\)

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

\(P=\dfrac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}\)

\(P=\dfrac{8\sqrt{41}}{\sqrt{\left(\sqrt{41}+2\right)^2}+\sqrt{\left(\sqrt{41}-2\right)^2}}\)

\(P=\dfrac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}=\dfrac{8\sqrt{41}}{2\sqrt{41}}=4\)

*P/S: đã nhỡ làm câu a, câu b bạn Phùng Khánh Linh làm rồi :)

\(P=\dfrac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}=\dfrac{8\sqrt{41}}{\sqrt{41+2.2\sqrt{41}+4}+\sqrt{41-2.2\sqrt{41}+4}}=\dfrac{8\sqrt{41}}{2\sqrt{41}}=4\) \(Q=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\dfrac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{3+2\sqrt{3}+1}}+\dfrac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{3-2\sqrt{3}+1}}=\dfrac{\left(2\sqrt{2}+\sqrt{6}\right)\left(3-\sqrt{3}\right)+\left(2\sqrt{2}-\sqrt{6}\right)\left(3+\sqrt{3}\right)}{9-3}=\dfrac{6\sqrt{2}-2\sqrt{6}+3\sqrt{6}-\sqrt{18}+6\sqrt{2}+2\sqrt{6}-3\sqrt{6}-\sqrt{18}}{6}=\dfrac{12\sqrt{6}-6\sqrt{2}}{6}=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)