\(S=\frac{\sqrt{3}-1}{3-1}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2019^2-\left(2019^2-2\right)}\)

\(S=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2}\)

\(S=\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+...+\sqrt{2019^2}-\sqrt{2019^2-2}\right)\)

\(S=\frac{1}{2}\left(-1+\sqrt{2019^2}\right)\)

\(S=\frac{\left(2019-1\right)}{2}=1009\)

\(S=\frac{1-\sqrt{3}}{1-3}+\frac{\sqrt{3}-\sqrt{5}}{3-5}+\frac{\sqrt{5}-\sqrt{7}}{5-7}+...+\frac{2019-\sqrt{2019^2-2}}{2019^2-2019^2-2}.\)

\(S=\frac{1-\sqrt{3}}{-2}+\frac{\sqrt{3}-\sqrt{5}}{-2}+\frac{\sqrt{5}-\sqrt{7}}{-2}+...+\frac{2019-\sqrt{2019^2-2}}{-2}.\)

\(-2S=1-\sqrt{3}+\sqrt{3}-\sqrt{5}+\sqrt{5}...+2019-\sqrt{2019^2-2}\)

\(-2S=1-\sqrt{2019^2-2}\Rightarrow S=\frac{\sqrt{2019^2-2}-1}{2}\)

• a,de bieu thuc 2x^2+3x-1 /2x+1 nguyen thi2x+1 phai khac 0suy ra 2x khac -1suy ra x khac -1/2b,de bieu thuc 2x^2+4x-1/2x+1 nguyen thi 2x+1 khac 0suy ra 2x khac -1suy ra x khac -1/2c,de bieu thuc 2x^2+3/x-2 nguyen thi x-2 khac 0suy ra x khac 2 

\(a,=\frac{\left(2x+1\right)\left(x+1\right)-2}{2x+1}\in Z\Leftrightarrow2⋮2x+1mà:2x+1le\Rightarrow2x+1\in\left\{1;-1\right\}\Leftrightarrow x\in\left\{-1;0\right\}\)

ĐK: \(x^2-1\ge0\) (1)

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

\(\Leftrightarrow6\sqrt{x^2-1}+9\left(x^2-1\right)=-x^2+1\)

\(\Leftrightarrow6\sqrt{x^2-1}+10\left(x^2-1\right)=0\)

\(\Leftrightarrow\sqrt{x^2-1}\left(6+10\sqrt{x^2-1}\right)=0\)

\(\Leftrightarrow\sqrt{x^2-1}=0\)

\(\Leftrightarrow x^2-1=0\Leftrightarrow x=\pm1\)Thỏa mãn đk (1)

Vậy ...

\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)

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

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

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

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

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

\(=\frac{5+2\sqrt{6}+5-2\sqrt{6}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)

\(=10\)

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

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

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

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

\(A=x^6+2x\left(x^2+y\right)+x^2+y^2+26\) 

   \(=x^6+2x^2+2xy+x^2+y^2+26\) 

    \(=x^6+2x^2+\left(x+y\right)^2+26\ge26\forall x;y\) 

Dấu "=" xảy ra<=> \(x=0\) và \(\left(x+y\right)^2=0\Rightarrow y=0\) 

Vậy A_min =26 tại x=y=0

B=\(y^2-2xy+3x^2+2y-14x+1949\)

 \(=\left(y^2-2xy+x^2+2y-2x+1\right)+\left(2x^2-12x+18\right)+1930\)

 \(=\left(x-y-1\right)^2+2\left(x-3\right)^2+1930\)

  \(\ge1930\)

MinB=1930 khi \(\hept{\begin{cases}x=y+1\\x=3\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=2\end{cases}}}\)

1/y = -1/2 - x 

thay 1 phần y vào vế 2 xong tìm x rồi thay x vào vế 1 tìm y ~  ~ 

chi tiết đc kh bạn ơi

\(A=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4+\sqrt{5}}.\)\(\sqrt{4+\sqrt{5}}.\sqrt{4-\sqrt{5}}\)

\(=\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4+\sqrt{5}}.\)\(\sqrt{\left(4-\sqrt{5}\right)\left(4+\sqrt{5}\right)}\)

\(=\left(\sqrt{10}-\sqrt{6}\right)\)\(\sqrt{4+\sqrt{5}}.\sqrt{16-15}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}.\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(=5-3=2\)

\(\Rightarrow A\)là số hữu tỉ