Đk: x \(\ge\)0; y \(\ge\)0

Ta có: \(A=x\sqrt{x}+y\sqrt{y}=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)\)

\(A=\left(\sqrt{x}+\sqrt{y}\right)^3-3\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\)

Với \(\sqrt{x}+\sqrt{y}=1\) => \(A=1^3-3\sqrt{xy}.1=1-3\sqrt{xy}\) (1)

Do \(\sqrt{xy}\le\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{4}\)(bđt cosi ) => \(1-3\sqrt{xy}\ge1-3\cdot\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{4}=1-\frac{3}{4}=\frac{1}{4}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\sqrt{x}=\sqrt{y}\\\sqrt{x}+\sqrt{y}=1\end{cases}}\) <=> \(x=y=\frac{1}{4}\)

Vậy MinA = 1/4 <=> x = y = 1/4

Lại có: \(\sqrt{x}+\sqrt{y}=1\) => \(\sqrt{y}=1-\sqrt{x}\le1\)  => \(\sqrt{y}-1\le0\)

          => \(\sqrt{x}=1-\sqrt{y}\le1\) ==> \(\sqrt{x}-1\le0\)

=> \(\left(\sqrt{x}-1\right)\left(\sqrt{y}-1\right)\ge0\) <=> \(xy-\left(\sqrt{x}+\sqrt{y}\right)+1\ge0\)

<=> \(xy-1+1\ge0\) <=> \(xy\ge0\) <=> \(\sqrt{xy}\ge\)0

Do đó: \(A=1-3\sqrt{xy}\le1-3.0=1\)

Dấu "=" xảy ra<=> \(\hept{\begin{cases}xy=0\\\sqrt{x}+\sqrt{y}=1\end{cases}}\) <=> \(\hept{\begin{cases}x=0\\y=1\end{cases}}\) hoặc \(\hept{\begin{cases}x=1\\y=0\end{cases}}\)

Vậy MaxA = 1 <=> \(\hept{\begin{cases}x=1\\y=0\end{cases}}\)hoặc \(\hept{\begin{cases}x=0\\y=1\end{cases}}\)

Đặt A = \(\sqrt{5-\sqrt{21}}-\sqrt{6-\sqrt{32}}\)

\(\sqrt{2}A=\sqrt{10-2\sqrt{7.3}}-\sqrt{2}\sqrt{6-2\sqrt{4.2}}\)

\(=\sqrt{7}-\sqrt{3}-\sqrt{2}\left(\sqrt{4}-\sqrt{2}\right)\)

\(=\sqrt{7}-\sqrt{3}-2\sqrt{2}+2\)

Vậy \(A=\frac{\sqrt{7}-\sqrt{3}-2\sqrt{2}+2}{\sqrt{2}}=\frac{\sqrt{14}-\sqrt{6}-8+2\sqrt{2}}{2}\)

\(A=\sqrt{5-\sqrt{21}}-\sqrt{6-4\sqrt{2}}\)

nên \(A=\sqrt{\frac{7}{2}-2\sqrt{\frac{7}{2}.\frac{3}{2}}+\frac{3}{2}}-\sqrt{4-4\sqrt{2}+2}=\sqrt{\left(\sqrt{\frac{7}{2}}-\sqrt{\frac{3}{2}}\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}\)

\(A=\frac{\sqrt{14}-\sqrt{6}}{2}-2+\sqrt{2}\)

\(\sqrt{3+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)

\(=\sqrt{3+2\sqrt{5-\sqrt{13+2\sqrt{12}}}}\)

\(=\sqrt{3+2\sqrt{5-\left(\sqrt{12}+1\right)}}\)

\(=\sqrt{3+2\sqrt{4-2\sqrt{3}}}=\sqrt{3+2\left(\sqrt{3}-1\right)}\)

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

\(\sqrt{3+2\sqrt{5-\sqrt{13+2\sqrt{12}}}}\)

=\(\sqrt{3+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\)

=\(\sqrt{3+2\sqrt{5-2\sqrt{3}-1}}\)

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

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

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

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

(d) cắt (d2) khi \(a\ne\frac{1}{2}\)

Thay x = 6 vào pt đt (d2) \(y=\frac{1}{2}.6+1=4\)

Vậy (d) cắt (d2) tại ( x ; y ) ; ( 6 ; 4 ) 

Thay x = 6 ; y = 4 vào ptđt (d) \(4=6a+b\)

đề có thiếu ko bạn, tìm a;b kiểu gì ? bạn kiểm tra lại đề nhé 

\(\sqrt{2-\sqrt{3}}-\sqrt{4-\sqrt{15}}=\frac{\sqrt{4-2\sqrt{3}}-\sqrt{8-2\sqrt{15}}}{\sqrt{2}}\)

\(=\frac{\sqrt{\sqrt{3}^2-2\sqrt{3}+1}-\sqrt{\sqrt{5}^2-2\sqrt{3}\sqrt{5}+\sqrt{3}^2}}{\sqrt{2}}\)

\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{2}}\)

\(=\frac{\sqrt{3}-1-\sqrt{5}+\sqrt{3}}{\sqrt{2}}\)

\(=\frac{2\sqrt{3}-1-\sqrt{5}}{\sqrt{2}}\)