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

   \(=x^6+2x^2+2xy+x^2+y^2+26\) 

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

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 

\(a,8-2\sqrt{7}=\sqrt{7}^2-2\sqrt{7}+1^2=\left(\sqrt{7}-1\right)^2\)

\(b,8-2\sqrt{15}=\sqrt{5}^2-2.\sqrt{3}.\sqrt{5}+\sqrt{3}^2=\left(\sqrt{5}-\sqrt{3}\right)^2\)

\(c,8+4\sqrt{3}=2^2+2.2.\sqrt{3}+\sqrt{3}^2=\left(2+\sqrt{3}\right)^2\)

E=(4x^2-4x+1)+(9y^2+6y+1)+(16z^2+8z+1)+1

E=(2x-1)^2+(3y-1)^2+(4z+1)^2+1

Vì (2x-1)^2>=0

      ........>=0

       .........>=0

nên E>= 1.dấu = xảy ra khi x=1/2

  y=1/3

z=1/4

ĐKXĐ \(x\ge0\)

Đặt \(\sqrt{x}+\sqrt{x+2}=a\left(a\ge0\right)\)

=> \(a^2=2x+2+2\sqrt{x^2+2x}\)

Khi đó PT

<=> \(a^2-3a-4=0\)

<=> \(\orbr{\begin{cases}a=4\left(tmĐK\right)\\a=-1\left(kotmĐK\right)\end{cases}}\)

=> \(\sqrt{x}+\sqrt{x+2}=4\)

<=> \(2x+2+2\sqrt{x^2+2x}=16\)

<=> \(\sqrt{x^2+2x}=7-x\)

<=> \(\hept{\begin{cases}x\le7\\x^2+2x=49-14x+x^2\end{cases}}\)

=> \(x=\frac{49}{16}\left(tmĐKXĐ\right)\)

Vậy \(x=\frac{49}{16}\)

@To Kill A Mockingbird @ Làm các bước mong là em hiểu^^

Đk: \(x\ge0\)(1)

pt <=> \(2\sqrt{x^2+2x}-3\left(\sqrt{x}+\sqrt{x+2}\right)=2-2x\)

Đặt: \(\sqrt{x}+\sqrt{x+2}=t\left(đk:t\ge0\right)\)

ta có: \(t^2=x+2\sqrt{x\left(x+2\right)}+x+2\)

<=> \(t^2=2x+2+2\sqrt{x^2+2x}\)

\(\Leftrightarrow2\sqrt{x^2+2x}=t^2-2x-2\)

Thay vào ta có:

\(t^2-2x-2-3t=2-2x\)

\(\Leftrightarrow t^2-3t-4=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=4\\t=-1\left(loai\right)\end{cases}}\)

Với t=4 ta có phương trình:

 \(\sqrt{x}+\sqrt{x+2}=4\)

\(\Leftrightarrow2x+2+2\sqrt{x^2+2x}=4^2\)

\(\Leftrightarrow\sqrt{x^2+2x}=7-x\)

\(\Leftrightarrow\hept{\begin{cases}7-x\ge0\\x^2+2x=49-14x+x^2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\le7\\x=\frac{49}{16}\end{cases}\Leftrightarrow}x=\frac{49}{16}\)( thỏa mãn đk (1))

Vậy ...