\(ĐK:x+y\ge0;x-y\ge0;x,y\ge0\)

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

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

\(\Leftrightarrow y=x-1\)

Thế vào \(PT\left(2\right)\Leftrightarrow\sqrt{x}+\sqrt{x-1}=1\left(x\ge1\right)\Leftrightarrow\sqrt{x}-1+\sqrt{x-1}=0\)

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

Vậy ...

Từ pt thứ nhất: \(\Leftrightarrow x+1+\sqrt{\left(x+1\right)^2+1}=\left(-y\right)+\sqrt{\left(-y\right)^2+1}\)

Xét hàm \(f\left(t\right)=t+\sqrt{t^2+1}\Rightarrow f'\left(t\right)=1+\dfrac{t}{\sqrt{t^2+1}}=\dfrac{t+\sqrt{t^2+1}}{\sqrt{t^2+1}}\)

\(f'\left(t\right)>\dfrac{t+\sqrt{t^2}}{\sqrt{t^2+1}}=\dfrac{t+\left|t\right|}{\sqrt{t^2+1}}\ge0\Rightarrow f'\left(t\right)>0\) ; \(\forall t\)

\(\Rightarrow f\left(t\right)\) đồng biến trên R

\(\Rightarrow x+1=-y\Rightarrow y=-x-1\)

Thế xuống pt dưới:

\(x^3-\left(3x^2-2x-8\right)\sqrt{2x^2+x-1}=0\)

Bạn coi lại đề, pt vô tỉ này ko giải được

1) Ta có: \(\left\{{}\begin{matrix}3\sqrt{x}-\sqrt{y}=5\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}9\sqrt{x}-3\sqrt{y}=15\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}11\sqrt{x}=33\\3\sqrt{x}-\sqrt{y}=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{y}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)

2) Ta có: \(\left\{{}\begin{matrix}\sqrt{x+3}-2\sqrt{y+1}=2\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2\sqrt{x+3}+4\sqrt{y+1}=-4\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{y+1}=0\\\sqrt{x+3}-2\sqrt{y+1}=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y+1}=0\\\sqrt{x+3}=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y+1=0\\x+3=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

4. Đk: \(x,y\ge0\)

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

Ta có: \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}\ge0+1=1\\\sqrt{y}+\sqrt{x+1}\ge0+1=1\end{matrix}\right.\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}\sqrt{x}=0,\sqrt{x+1}=1\\\sqrt{y}=0,\sqrt{y+1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)<tmđk>

Vậy hệ pt có nghiệm \(\left(x,y\right)=\left(0;0\right)\)

Điều kiện: \(\left\{ \begin{array}{l} x > - 2\\ y > 1\\ x + y > 0 \end{array} \right.\)

Hệ phương trình tương đương: \(\left\{ \begin{array}{l} \sqrt {\dfrac{{x + y}}{{x + 2}}} + \sqrt {\dfrac{{x + y}}{{y - 1}}} = 2\\ {\left( {\dfrac{{x + 2}}{{x + y}}} \right)^2} + \left( {\dfrac{{y - 1}}{{x + y}}} \right)^2 = 2 \end{array} \right.\). Đặt \(\left\{ \begin{array}{l} a = \sqrt {\dfrac{{x + y}}{{x + 2}}} \\ b = \sqrt {\dfrac{{x + y}}{{y - 1}}} \end{array} \right.\) (với \(a,b > 0\))

Ta có hệ phương trình: \(\left\{ \begin{array}{l} a + b = 2\\ \dfrac{1}{{{a^4}}} + \dfrac{1}{{{b^4}}} = 2 \end{array} \right.\left( * \right)\)

Áp dụng BĐT AM - GM, ta có:

\(\begin{array}{l} 2 = a + b \geqslant 2\sqrt {ab} \Rightarrow ab \leqslant 1\\ 2 = \dfrac{1}{{{a^4}}} + \dfrac{1}{{{b^4}}} \geqslant 2\sqrt {\dfrac{1}{{{a^4}}}.\dfrac{1}{{{b^4}}}} \Rightarrow ab \geqslant 1 \end{array}\)

Thế nên \(\left( * \right) \Leftrightarrow a = b = 1\)

Ta lại có hệ phương trình: \(\left\{ \begin{array}{l} \dfrac{{x + y}}{{x + 2}} = 1\\ \dfrac{{x + y}}{{y - 1}} = 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = - 1\\ y = 2 \end{array} \right.\)

Vậy hệ phương trình có nghiệm là \((-1;2)\)

Đk: \(\left\{{}\begin{matrix}x>-2\\y>1\\x+y>0\end{matrix}\right.\)

hpt\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{x+y}{x+2}}+\sqrt{\dfrac{x+y}{y-1}}=2\\2\left(x+y\right)^2=\left(x+2\right)^2+\left(y-1\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{x+y}{x+2}}+\sqrt{\dfrac{x+y}{y-1}}=2\\\left(\dfrac{x+2}{x+y}\right)^2+\left(\dfrac{y-1}{x+y}\right)^2=2\end{matrix}\right.\)

Đặt \(a=\sqrt{\dfrac{x+y}{x+2}},b=\sqrt{\dfrac{x+y}{y-1}}\left(a,b>0\right)\)

Ta có hệ: \(\left\{{}\begin{matrix}a+b=2\\\dfrac{1}{a^4}+\dfrac{1}{b^4}=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\a^4+b^4=2a^4b^4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\\left[\left(a+b\right)^2-2ab\right]^2-2a^2b^2=2a^4b^4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\\left(4-2ab\right)^2-2a^2b^2=2a^4b^4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\a^4b^4=a^2b^2-8ab+8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\a^2b^2\left(a^2b^2-1\right)+8\left(ab-1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\\left(ab-1\right)\left[a^2b^2\left(ab+1\right)+8\right]=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\ab-1\end{matrix}\right.\left(a,b>0\right)\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{x+y}{x+2}}=1\\\sqrt{\dfrac{x+y}{y-1}}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=x+2\\x+y=y-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

ĐKXĐ: \(x;y\ge0\)

Với \(x=0\) hoặc \(y=0\) đều ko là nghiệm

Với \(x;y>0\) hệ tương đương:

\(\left\{{}\begin{matrix}1+\dfrac{1}{x+y}=\dfrac{2}{\sqrt{3x}}\\1-\dfrac{1}{x+y}=\dfrac{4\sqrt{2}}{\sqrt{7y}}\end{matrix}\right.\)

Lần lượt cộng vế với vế và trừ vế cho vế ta được:

\(\left\{{}\begin{matrix}1=\dfrac{1}{\sqrt{3x}}+\dfrac{2\sqrt{2}}{\sqrt{7y}}\\\dfrac{1}{x+y}=\dfrac{1}{\sqrt{3x}}-\dfrac{2\sqrt{2}}{\sqrt{7y}}\end{matrix}\right.\)

Nhân vế với vế:

\(\dfrac{1}{x+y}=\dfrac{1}{3x}-\dfrac{8}{7y}\)

\(\Leftrightarrow\dfrac{y}{3}-\dfrac{8x}{7}=1\)

\(\Rightarrow y=\dfrac{24x+21}{7}\)

Rồi thế vào 1 trong các pt đầu 

Nhưng em có nhầm đề ko mà con số xấu kinh khủng vậy nhỉ? Số \(\sqrt{7}\) kia cho xấu 1 cách ko cần thiết, nó ko ảnh hưởng đến cách giải mà chỉ khiến cho việc tính toán khó khăn 1 cách cơ học khá vớ vẩn

\(PT\left(1\right)\Leftrightarrow\left(x^2+y^2\right)\left(x+y\right)+2xy=x+y\\ \Leftrightarrow\left[\left(x+y\right)^2-2xy\right]\left(x+y\right)+2xy-\left(x+y\right)=0\\ \Leftrightarrow\left(x+y\right)^3-2xy\left(x+y\right)+2xy-\left(x+y\right)=0\\ \Leftrightarrow\left(x+y\right)\left[\left(x+y\right)^2-1\right]-2xy\left(x+y-1\right)=0\\ \Leftrightarrow\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)-2xy\left(x+y-1\right)=0\\ \Leftrightarrow\left(x+y-1\right)\left[\left(x+y\right)\left(x+y+1\right)-2xy\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}x+y-1=0\\x^2+2xy+x+y^2+y+1=0\left(3\right)\end{matrix}\right.\\ \left(3\right)\Leftrightarrow\left(x+y\right)^2+\left(x+y\right)+\dfrac{1}{4}+\dfrac{3}{4}=0\\ \Leftrightarrow\left(x+y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vô.n_o\right)\)

Từ đó em thế vô PT(2) thôi

Thế \(x+y-1=0\Leftrightarrow y=x-1\) vào PT(2)

\(\Leftrightarrow\sqrt{x+x-1}=x^2-x+1\\ \Leftrightarrow\sqrt{2x-1}=x^2-x+1\left(x\ge\dfrac{1}{2}\right)\\ \Leftrightarrow\sqrt{2x-1}-1=x^2-x\\ \Leftrightarrow\dfrac{2x-2}{\sqrt{2x-1}+1}-x\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(\dfrac{2}{\sqrt{2x-1}+1}-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x\sqrt{2x-1}+x=2\left(4\right)\end{matrix}\right.\)

Giải (4) ta được \(x=1\Leftrightarrow y=0\)

Vậy ...

part full :v

*Th 1: \(x+y=2\)

\(Pt\left(1\right)\Leftrightarrow3y\sqrt{2-y^2}=x+\dfrac{4}{x+1}\)

xét \(VT=3y\sqrt{2-y^2}=3\sqrt{y^2\left(2-y^2\right)}\le3.\dfrac{y^2+2-y^2}{2}=3\)(theo AM-GM)

\(VT=x+\dfrac{4}{x+1}=\left(x+1\right)+\dfrac{4}{x+1}-1\ge2\sqrt{\dfrac{4\left(x+1\right)}{x+1}}-1=4-1=3\)(theo AM-GM)

do đó \(VT\le3;VF\ge3\)

\(VT=VF\Leftrightarrow\left\{{}\begin{matrix}y^2=2-y^2\\x+1=\dfrac{4}{x+1}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\pm1\\\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\end{matrix}\right.\)(tmđkxđ)(4 cặp)

*TH 2 \(\left(x+1\right)\sqrt{2-y^2}=1\Leftrightarrow x+1=\dfrac{1}{\sqrt{2-y^2}}\)(\(-\sqrt{2}< y< \sqrt{2}\))

thế vào Pt(1) , bình phương giải (nhác làm quá)

\(Pt\left(2\right)\Leftrightarrow\left(x+y-2\right)\left[\left(x+1\right)\sqrt{2-y^2}-1\right]=0\)

\(1,HPT\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)+\left(\dfrac{1}{y}-\dfrac{1}{x}\right)=0\\2y=x^3+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\dfrac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y\\2y=x^3+1\end{matrix}\right.\\ \Leftrightarrow2y=y^3+1\Leftrightarrow y^3-2y+1=0\\ \Leftrightarrow\left[{}\begin{matrix}y=0\\y=\dfrac{-1+\sqrt{5}}{2}\\y=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(0;0\right);\left(\dfrac{-1+\sqrt{5}}{2};\dfrac{-1+\sqrt{5}}{2}\right);\left(\dfrac{-1-\sqrt{5}}{2};\dfrac{-1-\sqrt{5}}{2}\right)\)

\(2,HPT\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2\left(x^2+y^2\right)}+2\sqrt{xy}=16\\x+y+2\sqrt{xy}=16\end{matrix}\right.\\ \Leftrightarrow\sqrt{2\left(x^2+y^2\right)}=x+y\\ \Leftrightarrow\left(x-y\right)^2=0\Leftrightarrow x=y\\ \Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4\)

Vậy \(\left(x;y\right)=\left(4;4\right)\)

\(3,\text{Sửa: }\left\{{}\begin{matrix}\sqrt{x^2+3}+\left|y\right|=\sqrt{3}\left(1\right)\\\sqrt{y^2+5}+\left|x\right|=\sqrt{x^2+5}\left(2\right)\end{matrix}\right.\)

Ta thấy \(\sqrt{x^2+3}\ge\sqrt{3};\left|y\right|\ge0\Leftrightarrow VT\left(1\right)\ge\sqrt{3}=VP\left(1\right)\)

Dấu \("="\Leftrightarrow x=y=0\)

Thay vào \(\left(2\right)\Leftrightarrow\sqrt{5}+0=\sqrt{5}\left(tm\right)\)

Vậy \(\left(x;y\right)=\left(0;0\right)\)

*Công thức: Biến đổi x theo y và ngc lại và dùng các quy tắc.

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

Cộng 2 pt ta đc: x=1

Thay vào (1):\(\Leftrightarrow y=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{6}}{3}\)

Vậy (x;y)\(=\left(1;\frac{\sqrt{6}}{3}\right)\)

Những câu sau làm ttự.

#Walker