\(\Rightarrow2y=x^3+3x\)

\(\Rightarrow2I=2x^4+x^3\left(x^3+3x\right)+6x^2+x\left(x^3+3x\right)-\left(x^3+3x\right)^2+2\)

\(=2x^4+x^6+3x^4+6x^2+x^4+3x^2-\left(x^6+6x^4+9x^2\right)+2\)

\(=2\)

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Đk: \(x\ge2\).

\(\sqrt{x^2-4}+2\sqrt{x-2}=0\Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}+2\right)=0\).

Thấy ngay: \(\sqrt{x+2}+2\ge0+2=2>0\)nên từ đây suy ra: \(\sqrt{x-2}=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)(thỏa mãn).

Vậy: \(x=2\).

Có nhận xét sau: 

\(\frac{1}{\left(a+1\right)\sqrt{a}+a\sqrt{a+1}}=\frac{1}{\sqrt{a}\sqrt{a+1}\left(\sqrt{a}+\sqrt{a+1}\right)}=\frac{\sqrt{a+1}-\sqrt{a}}{\sqrt{a^2+a}.\left(a+1-a\right)}\)

\(=\frac{\sqrt{a+1}-\sqrt{a}}{\sqrt{a}\sqrt{a+1}}=\frac{\sqrt{a+1}}{\sqrt{a}\sqrt{a+1}}-\frac{\sqrt{a}}{\sqrt{a}\sqrt{a+1}}=\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{a+1}}\). Từ đây suy ra:

\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-....-\frac{1}{\sqrt{2025}}=1-\frac{1}{45}=\frac{44}{45}\).

Vậy: \(A=\frac{44}{45}\).

suy ra 2x+1=125

suy ra 2x=124

suy ra x=62

Trả lời :

Chọn B. 1 nghiệm

Nghiệm đó là : \(\left(x ; y\right)=\left(\frac{7}{5} ;\frac{ 11}{5}\right)\)

~HT~

\(\hept{\begin{cases}x+3y=8\\3x-y=2\end{cases}\Leftrightarrow\hept{\begin{cases}3x+9y=24\\3x-y=2\end{cases}\Leftrightarrow\hept{\begin{cases}10y=22\\x+3y=8\end{cases}\Leftrightarrow}}\hept{\begin{cases}y=\frac{22}{10}=\frac{11}{5}\\x+3y=8\end{cases}}}\)

Thay (1) vào (2) ta được : \(x+\frac{33}{5}=8\Leftrightarrow x=8-\frac{33}{5}=\frac{7}{5}\)

Vậy hệ phương trình có một nghiệm ( x ; y ) = ( 7/5 ; 11/5 )

=> Chọn B 

\(4\sqrt{2}-6+2\sqrt{2x}-3\sqrt{x}+6=0\)

\(\sqrt{32}+\sqrt{8x}-\sqrt{9x}=0\)

\(\sqrt{32}+\sqrt{x}\left(\sqrt{8}-3\right)=0\)

\(\sqrt{x}\left(\sqrt{8}-3\right)=-\sqrt{32}\)

\(\sqrt{x}=16+12\sqrt{2}\)

\(x=\left(16+12\sqrt{2}\right)^2\)

sửa đề : Đặt \(A=\sqrt{x+2\sqrt{2x-4}}-\sqrt{x-2\sqrt{2x-4}}\)

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

\(=2x-2\sqrt{x^2-8x+16}=2x-2\sqrt{\left(x-4\right)^2}\)

\(=2x-2\left(x-4\right)=2x-2x+8=8\Rightarrow A=\sqrt{8}=2\sqrt{2}\)

Bạn ơi lỡ x<4 thì  \(\sqrt{\left(x-4\right)^2}=4-x\)chứ

Đặt \(A=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)

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

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

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

Vậy \(\Rightarrow A=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)

Tổng quát: \(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}=1+\frac{\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}=1+\frac{1}{n^2\left(n+1\right)^2}+\frac{2}{n\left(n+1\right)}\)

\(=\left(1+\frac{1}{n\left(n+1\right)}\right)^2=\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^2\)

\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\left|1+\frac{1}{n}-\frac{1}{n+1}\right|=1+\frac{1}{n}-\frac{1}{n+1}\)

Áp dụng ta được: 

\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2009^2}+\frac{1}{2010^2}}\)

\(=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2009}-\frac{1}{2010}\)

\(=2008+\frac{1}{2}-\frac{1}{2010}\)

\(=2008\frac{502}{1005}\)

Mình hết k cho bạn đc rồi để mai mik k nha