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 

Bài này đi thi vio mk cũng gặp ..

    bằng 1 ak
 

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!!

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

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

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

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

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

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

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

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

ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt[4]{56-x}=a\ge0\\\sqrt[4]{x^2+41}=b\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=5\\a^4+b^4=97\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\\left(a^2+b^2\right)^2-2a^2b^2=97\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\\left[\left(a+b\right)^2-2ab\right]^2-2a^2b^2=97\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\\left(25-2ab\right)^2-2a^2b^2=97\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\2a^2b^2-100ab+528=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\\left[{}\begin{matrix}ab=44\\ab=6\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}a+b=5\\ab=44\end{matrix}\right.\) (vô nghiệm)

TH2: \(\left\{{}\begin{matrix}a+b=5\\ab=6\end{matrix}\right.\) theo Viet đảo a;b là nghiệm:

\(t^2-5t+6=0\Rightarrow...\)

help

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

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

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

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

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

Vậy M = 4

Học tốt nhé :)

Dấu căn dưới mẫu có vt lộn ko z bạn?

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

Đề sau khi đã sửa: \(M=\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}\)

Giải:

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

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

Vậy M = 4

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

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