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\(\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}+\sqrt{45-\sqrt{41}}}}:\left(\sqrt{3}-\sqrt{2}\right)\) ( 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.....................
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!!
\(A=\frac{8\sqrt{41}}{2\sqrt{45+4\sqrt{41}}}=\frac{8\sqrt{41}}{2\sqrt{41+4\sqrt{41}+4}}=\frac{8\sqrt{41}}{2\sqrt{\left(\sqrt{41}\right)^2+2\cdot\sqrt{41}\cdot2+2^2}}\)
\(=\frac{8\sqrt{41}}{2\sqrt{\left(\sqrt{41}+2\right)^2}}=\frac{8\sqrt{41}}{2\left(\sqrt{41}+2\right)}=\frac{8\sqrt{41}\left(\sqrt{41}-2\right)}{2\left(41-4\right)}=\frac{328-16\sqrt{41}}{74}=\frac{164-8\sqrt{41}}{37}\)
\(B=\left(\frac{2x+1}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{x+4}{x+\sqrt{x}+1}\right)\)
\(=\left(\frac{2x+1}{\sqrt{x}^3+1^3}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{x+\sqrt{x}+1-x-4}{x+\sqrt{x}+1}\right)\)
\(=\left(\frac{2x+1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}-3}{x+\sqrt{x}+1}\right)\)
\(=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{x+\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}}{\sqrt{x}-3}=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}=\frac{x+3\sqrt{x}}{x-9}\)
\(A=\frac{8\sqrt{41}}{\sqrt{\sqrt{41}^2+2.2.\sqrt{41}+2^2}+\sqrt{\sqrt{41}^2-2.2.\sqrt{41}+2^2}}.\frac{1}{\sqrt{3}-\sqrt{2}}\)
\(=\frac{8\sqrt{41}}{\sqrt{\left(\sqrt{41}+2\right)^2}+\sqrt{\left(\sqrt{41}-2\right)^2}}.\frac{\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\frac{8\sqrt{41}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{41}+2+\sqrt{41}-2}=\frac{8\sqrt{41}\left(\sqrt{3}+\sqrt{2}\right)}{2\sqrt{41}}=4\left(\sqrt{3}+\sqrt{2}\right)\)
\(M=\dfrac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}\)
\(M=\dfrac{8\sqrt{41}}{\sqrt{\left(\sqrt{41}+2\right)^2}+\sqrt{\left(\sqrt{41}-2\right)^2}}\)
\(M=\dfrac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}\)
\(M=\dfrac{8\sqrt{41}}{2\sqrt{41}}=\dfrac{8}{2}=4\)
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?!:)))
\(P=\dfrac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}\)
\(P=\dfrac{8\sqrt{41}}{\sqrt{\left(\sqrt{41}+2\right)^2}+\sqrt{\left(\sqrt{41}-2\right)^2}}\)
\(P=\dfrac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}=\dfrac{8\sqrt{41}}{2\sqrt{41}}=4\)
*P/S: đã nhỡ làm câu a, câu b bạn Phùng Khánh Linh làm rồi :)
\(P=\dfrac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}=\dfrac{8\sqrt{41}}{\sqrt{41+2.2\sqrt{41}+4}+\sqrt{41-2.2\sqrt{41}+4}}=\dfrac{8\sqrt{41}}{2\sqrt{41}}=4\) \(Q=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\dfrac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{3+2\sqrt{3}+1}}+\dfrac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{3-2\sqrt{3}+1}}=\dfrac{\left(2\sqrt{2}+\sqrt{6}\right)\left(3-\sqrt{3}\right)+\left(2\sqrt{2}-\sqrt{6}\right)\left(3+\sqrt{3}\right)}{9-3}=\dfrac{6\sqrt{2}-2\sqrt{6}+3\sqrt{6}-\sqrt{18}+6\sqrt{2}+2\sqrt{6}-3\sqrt{6}-\sqrt{18}}{6}=\dfrac{12\sqrt{6}-6\sqrt{2}}{6}=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)
à nó bị liền : \(M=\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}\)
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
a: \(=\left(\sqrt{3}-\sqrt{2}+\sqrt{2}\right)\cdot\sqrt{3}=\sqrt{3}\cdot\sqrt{3}=3\)
b: \(=\dfrac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}+\sqrt{41}+2}}\)
\(=\dfrac{8\sqrt{41}}{\sqrt{47+5\sqrt{41}}}\)
\(1.\sqrt{\left(5+\sqrt{7}\right)^2}-\sqrt{8-2\sqrt{7}}=5+\sqrt{7}-\sqrt{7-2\sqrt{7}+1}=5+\sqrt{7}-\sqrt{7}+1=6\)
\(2.\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-2\sqrt{3}}=\sqrt{3}+1-\sqrt{3-2\sqrt{3}+1}=\sqrt{3}+1-\sqrt{3}+1=2\)
\(3.VT=\sqrt{11}-\sqrt{20-6\sqrt{11}}=\sqrt{11}-\sqrt{11-2.3\sqrt{11}+9}=\sqrt{11}-\sqrt{11}+3=3=VP\)
Vậy , đẳng thức được chứng minh .
\(4.VT=\sqrt{41+12\sqrt{5}}-\sqrt{41-12\sqrt{5}}=\sqrt{36+2.6\sqrt{5}+5}-\sqrt{36-2.6\sqrt{5}+5}=6+\sqrt{5}-6+\sqrt{5}=2\sqrt{5}=VP\)
Vậy , đẳng thức được chứng minh .