giải hệ phương trình x√y+y√x=30 x√x+y√y=35

$(I)\begin{cases} x\sqrt{y} + y\sqrt{x } = 30\\ x\sqrt{x} + y\sqrt{y} = 35\\ \end{cases}$ (ĐK: `{(x \ge 0),(y \ge 0):}` $\\$ Đặt $\begin{cases} \sqrt{y} = a\\ \sqrt{x} = b\\ \end{cases} (*)$ $\\$khi đó, pt `(I)` có dạng: $\\$$\begin{cases} b^2a + a^2b = 30\\ b^3 + a^3 = 35\\ \end{cases}$ $<=>\begin{cases} 7b^2a + 7a^2b = 210^{(1)}\\ 6b^3 + 6a^3 = 210^{(2)}\\ \end{cases}$ $\\$ Lấy $^{(1)} - ^{(2)},$ ta có: `7b^2a + 7a^2b - (6b^3 + 6a^3) = 210 - 210 ` $\\$ `<=> 7ab(b + a) - 6(b + a)(b^2 - ba + a^2) = 0`$\\$ `<=> (b + a) (-6a^2 + 13ab - 6b^2) = 0` $\\$ `<=> 13ab - 6a^2 - 6b^2 = 0` $\\$ Đặt `a = cb,` ta có: `13cb^2 - 6(cb)^2 - 6b^2 = 0` $\\$ `<=> b^2(13c - 6c^2 - 6) = 0` $\\$ `<=> 6c^2 - 13x + 6 = 0` $\\$ `<=> 6c^2 - 4c - 9c + 6 = 0` $\\$ `<=> 2c(3c - 2) - 3(3c - 2) = 0` $\\$ `<=> (3c - 2)(2c - 3) = 0` $\\$ <=>\(\left[ \begin{array}{l}3c - 2 = 0\\2c - 3 = 0\end{array} \right.\) <=>\(\left[ \begin{array}{l}3c = 2\\ 2c = 3\end{array} \right.\) <=>\(\left[ \begin{array}{l}c = \dfrac{2}{3}\\c = \dfrac{3}{2}\end{array} \right.\) $\\$ Với `c = 3/2 <=> a = {3b}/{2}` $\\$ `<=> 6.({3b}/{2})^3 + 6b^3 = 210` $\\ $ `<=>{81b^3}/{4} + 6b^3 = 210 ` $\\$ `<=> {81b^3 + 24b^3}/{4} = 840/4` $\\$ `<=> 105b^3 = 840 <=> b^3 = 8` $\\$ `<=>b = 2` $\\$ `=> a = 3`$\\$ Thay `a = 3; b= 2` vào $(*),$ ta có: `{(\sqrt{y} = 3),(\sqrt{x} = 2):}` `<=>{(y = 9 (t.m)),(x = 4 (t.m)):}` $\\$ Với `c = 2/3, <=> a = {2b}/{3}` $\\$ `<=> 6.({2b}/{3})^3 + 6b^3 = 210` $\\ $ `<=>{16b^3}/{9} + 6b^3 = 210 ` $\\$ `<=> {16b^3 + 54b^3}/{9} = 1890/9` $\\$ `<=> 70b^3 = 1890 <=> b^3 = 27` $\\$ `<=>b = 3` $\\$ `=> a = 2`$\\$ Thay `a = 2; b= 3` vào $(*),$ ta có: `{(\sqrt{y} = 2),(\sqrt{x} = 3):}` `<=>{(y = 4 (t.m)),(x = 9 (t.m)):}` $\\$ `KL: (x;y) = {(4; 9) , (9; 4)}`