\(2,ĐK:x\ge4;y\ge1\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x-4}=a\\\sqrt{y-1}=b\end{matrix}\right.\left(a,b\ge0\right)\)

\(HPT\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\a^2+b^2=58\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2ab+58=16\\a^2+b^2=58\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}ab=-21\\a+b=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=4-b\\b^2-4b-21=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}b=7\Rightarrow a=-3\\b=-3\Rightarrow a=7\end{matrix}\right.\left(loại\right)\)

Vậy hệ vô nghiệm

\(1,\\ \forall x=0\\ HPT\Leftrightarrow1=19\left(\text{vô lí}\right)\\ \forall x\ne0\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x^3}+y^3=19\\\dfrac{y}{x^2}+\dfrac{y^2}{x}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(\dfrac{1}{x}+y\right)^3-3\cdot\dfrac{y}{x}\left(\dfrac{1}{x}+y\right)=19\\\dfrac{y}{x}\left(\dfrac{1}{x}+y\right)=-6\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}+y=a\\\dfrac{y}{x}=b\end{matrix}\right.\)

\(HPT\Leftrightarrow\left\{{}\begin{matrix}a^3-3ab=19\\ab=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+y=1\\\dfrac{y}{x}=-6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}1+xy=x\\y=-6x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3};y=-2\\x=-\dfrac{1}{2};y=3\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(\dfrac{1}{3};-2\right);\left(-\dfrac{1}{2};3\right)\)

1: \(\left\{{}\begin{matrix}x\sqrt{2}-3y=1\\2x+y\sqrt{2}=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-3\sqrt{2}\cdot y=\sqrt{2}\\2x+y\sqrt{2}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4\sqrt{2}\cdot y=\sqrt{2}+2\\2x+y\sqrt{2}=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=\dfrac{2+\sqrt{2}}{-4\sqrt{2}}=\dfrac{-\sqrt{2}-1}{4}\\2x=-2-y\sqrt{2}=-2+\sqrt{2}\cdot\dfrac{\sqrt{2}+1}{4}=\dfrac{-6+\sqrt{2}}{4}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=\dfrac{-\sqrt{2}-1}{4}\\x=\dfrac{-6+\sqrt{2}}{8}\end{matrix}\right.\)

2: \(\left\{{}\begin{matrix}5x\sqrt{3}+y=2\sqrt{2}\\x\sqrt{6}-y\sqrt{2}=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x\sqrt{6}+y\sqrt{2}=4\\x\sqrt{6}-y\sqrt{2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x\cdot\sqrt{6}=6\\x\sqrt{6}-y\sqrt{2}=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{1}{\sqrt{6}}=\dfrac{\sqrt{6}}{6}\\y\sqrt{2}=x\sqrt{6}-2=1-2=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{\sqrt{6}}{6}\\y=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

1/PT (1) cho ta nhân tử x - y - 1:)

\(\left\{{}\begin{matrix}\left(17-3x\right)\sqrt{5-x}+\left(3y-14\right)\sqrt{4-y}=0\left(1\right)\\2\sqrt{2x+y+5}+3\sqrt{3x+2y+11}=x^2+6x+13\left(2\right)\end{matrix}\right.\)

ĐK: \(x\le5;y\le4\); \(2x+y+5\ge0;3x+2y+11\ge0\)

PT (1) \(\Leftrightarrow\left(17-3x\right)\left(\sqrt{5-x}-\sqrt{4-y}\right)-3\left(x-y-1\right)\sqrt{4-y}=0\)

\(\Leftrightarrow\left(3x-17\right)\left(\frac{x-y-1}{\sqrt{5-x}+\sqrt{4-y}}\right)-3\left(x-y-1\right)\sqrt{4-y}=0\)

\(\Leftrightarrow\left(x-y-1\right)\left(\frac{3x-17}{\sqrt{5-x}+\sqrt{4-y}}-3\sqrt{4-y}\right)=0\)

Dễ thấy cái ngoặc to < 0

Do đó x= y + 1

Thay xuống PT (2):\(y^2+8y+20=2\sqrt{3y+7}+3\sqrt{5y+14}\)\(\left(y+1\right)\left(y+2\right)=y^2+3y+2\)

ĐK: \(y\ge-\frac{7}{3}\) (để các căn thức được thỏa mãn)

PT (2) \(\Leftrightarrow y^2+3y+2+2\left(y+3-\sqrt{3y+7}\right)+3\left(y+4-\sqrt{5y+14}\right)=0\)

\(\Leftrightarrow\left(y^2+3y+2\right)\left(1+\frac{2}{y+3+\sqrt{3y+7}}+\frac{3}{y+4+\sqrt{5y+14}}\right)=0\)

Cái ngoặc to > 0 =>...

P/s: Is that true? Ko đúng thì chịu thua-_- Mất nửa tiếng đồng hồ để gõ bài này đấy:(

2/ĐK: \(x\ge-y;y\ge0\)

PT (1) \(\Leftrightarrow x\left(x+y\right)+\sqrt{x+y}=2y^2+\sqrt{2y}\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)+y\left(x-y\right)+\sqrt{x+y}-\sqrt{2y}=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+2y+\frac{1}{\sqrt{x+y}+\sqrt{2y}}\right)=0\)

Cái ngoặc to \(\ge y+\frac{1}{\sqrt{x+y}+\sqrt{2y}}>0\).

Do đó x = y \(\ge0\)

Thay xuống pt dưới: \(x^3-5x^2+14x-4=6\sqrt[3]{x^2-x+1}\)

Lập phương hai vế lên ra pt bậc 6, tuy nhiên cứ yên tâm, nghiệm rất đẹp: x = 1:)

Em đưa kết quả luôn: \(\left(x-1\right)\left(x^2-4x+7\right)\left(x^6-10x^5+56x^4-160x^3+272x^2-64x+40\right)=0\)

P/s: khúc cuối em ko còn cách nào khác nên đành lập phương:((

a.

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)^2-3\left(2x-y\right)=0\\x+2y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(2x-y-3\right)=0\\x+2y=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-y=0\\x+2y=0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-y-3=0\\x+2y=0\end{matrix}\right.\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{6}{5}\\y=-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)

b.

ĐKXĐ: \(\dfrac{2x-y}{x+y}>0\)

Đặt \(\sqrt{\dfrac{2x-y}{x+y}}=t>0\) pt đầu trở thành:

\(t+\dfrac{1}{t}=2\Leftrightarrow t^2-2t+1=0\)

\(\Leftrightarrow t=1\Leftrightarrow\sqrt{\dfrac{2x-y}{x+y}}=1\)

\(\Leftrightarrow2x-y=x+y\Leftrightarrow x=2y\)

Thay xuống pt dưới:

\(6y+y=14\Rightarrow y=2\)

\(\Rightarrow x=4\)