ĐKXĐ; ...

Đặt \(\left\{{}\begin{matrix}\sqrt{x+5}=a>0\\\sqrt{y-2}=b\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=7\\\sqrt{a^2-7}+\sqrt{b^2+7}=7\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{a^2-7}+\sqrt{\left(7-a\right)^2+7}=7\)

\(\Leftrightarrow\sqrt{a^2-14a+56}=7-\sqrt{a^2-7}\) (\(a\le\sqrt{56}\))

\(\Leftrightarrow a^2-14a+56=42+a^2-14\sqrt{a^2-7}\)

\(\Leftrightarrow\sqrt{a^2-7}=a-1\) 

\(\Leftrightarrow a^2-7=a^2-2a+1\Leftrightarrow a=4\Rightarrow b=3\)

\(\Rightarrow x;y\)

\(ĐK:x\le6;y\ge3\\ \left\{{}\begin{matrix}x^2+2y=xy+4\left(1\right)\\x^2-x-3-x\sqrt{6-x}=\left(y-3\right)\sqrt{y-3}\left(2\right)\end{matrix}\right.\)

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

Từ đó thế vào PT(2)

Với \(x=y-2\Leftrightarrow x+2=y\)

\(\left(2\right)\Leftrightarrow x^2-x+3-x\sqrt{6-x}=\left(x-1\right)\sqrt{x-1}\left(1\le x\le6\right)\\ \Leftrightarrow2x^2-2x+6-2x\sqrt{6-x}=2\left(x-1\right)\sqrt{x-1}\\ \Leftrightarrow\left(x-\sqrt{6-x}\right)^2+x\left(x-1\right)=2\left(x-1\right)\sqrt{x-1}\\ \Leftrightarrow\left(x-\sqrt{6-x}\right)^2+\left(x-1\right)\left(x-2\sqrt{x-1}\right)=0\\ \Leftrightarrow\left(\dfrac{x^2-6+x}{x+\sqrt{6-x}}\right)^2+\dfrac{\left(x-1\right)\left(x^2-4x+4\right)}{x^2+2\sqrt{x-1}}=0\\ \Leftrightarrow\left[\dfrac{\left(x-2\right)\left(x+3\right)}{x+\sqrt{6-x}}\right]^2+\dfrac{\left(x-1\right)\left(x-2\right)^2}{x^2+2\sqrt{x-1}}=0\\ \Leftrightarrow\left(x-2\right)^2\left[\left(\dfrac{x+3}{x+\sqrt{6-x}}\right)^2+\dfrac{x-1}{x^2+2\sqrt{x-1}}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\\left(\dfrac{x+3}{x+\sqrt{6-x}}\right)^2+\dfrac{x-1}{x^2+2\sqrt{x-1}}=0\left(1\right)\end{matrix}\right.\)

Dễ thấy \(\left(1\right)>0\) với \(x\ge1\)

Do đó \(x=2\Leftrightarrow y=4\)

Vậy HPT có nghiệm \(\left(x;y\right)=\left(2;4\right)\)

Giải hệ phương trình :(4.x^2 + 1).x + (y − 3) √5 − 2y = 04.x^2 + y^2 + 2.√3 − 4x = 7(x, y ∈ R)  - Hoc24

ĐKXĐ: ...

\(\Leftrightarrow\left\{{}\begin{matrix}7\sqrt{16-y^2}=x^2+5x-6\\2\left(y-4\right)^2=-x^2-4x+5\end{matrix}\right.\)

\(\Rightarrow7\sqrt{16-y^2}+2\left(y-4\right)^2=x-1\)

Do \(7\sqrt{16-y^2}+2\left(y-4\right)^2\ge0\Rightarrow x-1\ge0\Rightarrow x\ge1\)

\(\Rightarrow\left(x+2\right)^2+2\left(y-4\right)^2\ge\left(x+2\right)^2\ge9\)

Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}x=1\\y=4\end{matrix}\right.\)

Vậy hệ có cặp nghiệm duy nhất nói trên

Đặt vế trái là P

\(P=\dfrac{x^4}{\dfrac{x^2}{y}+\dfrac{1}{y}}+\dfrac{y^4}{\dfrac{y^2}{z}+\dfrac{1}{z}}+\dfrac{z^4}{\dfrac{z^2}{x}+\dfrac{1}{x}}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}\)

\(P\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{x^3z+y^3x+z^3y+xy+yz+zx}\)

Ta có:

\(x^2y^2+y^2z^2+z^2x^2\ge\dfrac{1}{3}\left(xy+yz+zx\right)^2\ge\dfrac{1}{3}.3\sqrt[3]{xy.yz.zx}\left(xy+yz+zx\right)\)

\(\Rightarrow3\left(x^2y^2+y^2z^2+z^2x^2\right)\ge3\left(xy+yz+zx\right)\) (1)

\(x^4+x^2z^2\ge2\sqrt{x^6z^3}=2x^3z\)

\(y^4+x^2y^2\ge2y^3x\) ; \(z^4+y^2z^2\ge2z^3y\)

\(\Rightarrow x^4+y^4+z^4+x^2y^2+y^2z^2+z^2x^2\ge2\left(x^3z+y^3x+z^3y\right)\) (2)

Lại có: \(x^4+x^4+x^4+z^4\ge4x^3z\) ; \(3y^4+x^4\ge4y^3x\) ; \(3z^4+y^4\ge4z^3y\)

\(\Rightarrow x^4+y^4+z^4\ge x^3z+y^3x+z^3y\) (3)

Cộng vế (1);  (2) và (3):

\(2\left(x^2+y^2+z^2\right)^2\ge3\left(x^3z+y^3x+z^3y+xy+yz+zx\right)\)

\(\Rightarrow P\ge\dfrac{3}{2}\)

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

\(ĐK:x\ge\dfrac{1}{5};y\ge\dfrac{3}{8}\)

\(PT\left(1\right)\Leftrightarrow\dfrac{3x^2-3y^2}{\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}}=3\left(x+y\right)\\ \Leftrightarrow3\left(x+y\right)\left(\dfrac{x-y}{\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+y=0\\\dfrac{x-y}{\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}}=1\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x-y=\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}\\ \Leftrightarrow\left(x-y\right)=\dfrac{3\left(x^2-y^2\right)}{\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}}\\ \Leftrightarrow\left(x-y\right)\left[\dfrac{3\left(x+y\right)}{\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}}-1\right]=0\)

\(\Leftrightarrow x=y\)

Với \(x+y=0\Leftrightarrow x=-y\), thay vào PT 2

\(\Leftrightarrow3\left(-y\right)\left(y-7\right)+10=\sqrt{10\left(-y\right)-2}+2\sqrt{8y-3}\\ \Leftrightarrow3y\left(7-y\right)+10=\sqrt{-10y-2}+2\sqrt{8y-3}\)

ĐK: \(\left\{{}\begin{matrix}-10y-2\ge0\\8y-3\ge0\end{matrix}\right.\Leftrightarrow y\in\varnothing\)

Với \(x-y=0\Leftrightarrow x=y\), thay vào PT 2

\(\Leftrightarrow3x^2-21x+10=\sqrt{10x-2}+2\sqrt{8x-3}\left(x\ge\dfrac{3}{8}\right)\\ \Leftrightarrow3x^2-24x+9=\sqrt{10x-2}-\left(x+1\right)+2\sqrt{8x-3}-2x\)

\(\Leftrightarrow3\left(x^2-8x+3\right)=\dfrac{-x^2+8x-3}{\sqrt{10x-2}+\left(x+1\right)}+\dfrac{2\left(-x^2+8x-3\right)}{\sqrt{8x-3}+x}\\ \Leftrightarrow\left(x^2-8x+3\right)\left(3+\dfrac{1}{\sqrt{10x-2}+x+1}+\dfrac{2}{\sqrt{8x-3}+x}\right)=0\)

Dễ thấy ngoặc lớn vô nghiệm với \(x\ge\dfrac{3}{8}>0\)

\(\Leftrightarrow x^2-8x+3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4+\sqrt{13}\left(n\right)\\x=4-\sqrt{13}\left(n\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=4+\sqrt{13}\\y=4-\sqrt{13}\end{matrix}\right.\)

Vậy HPT có nghiệm \(\left(x;y\right)\in\left\{\left(4+\sqrt{13};4+\sqrt{13}\right);\left(4-\sqrt{13};4-\sqrt{13}\right)\right\}\)

ĐKXĐ: ...

Trừ vế cho vế:

\(2\left(x-y\right)+\sqrt{4-y}-\sqrt{4-x}=0\)

\(\Leftrightarrow2\left(x-y\right)+\dfrac{x-y}{\sqrt{4-y}+\sqrt{4-x}}=0\)

\(\Leftrightarrow\left(x-y\right)\left(2+\dfrac{1}{\sqrt{4-y}+\sqrt{4-x}}\right)=0\)

\(\Leftrightarrow x=y\)

Thế vào pt đầu:

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

\(\Leftrightarrow\sqrt{4-x}=1-2x\) (\(x\le\dfrac{1}{2}\))

\(\Leftrightarrow4-x=1-4x+4x^2\)

\(\Leftrightarrow4x^2-3x-3=0\)

\(\Leftrightarrow..\)

Với \(y=3\) ko phải nghiệm

Với \(y\ne3\)

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

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

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

Trừ vế:

\(\Rightarrow\sqrt{x}+\sqrt{x+3}=3\)

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

\(\Leftrightarrow\dfrac{x-1}{\sqrt{x}+1}+\dfrac{x-1}{\sqrt{x+3}+2}=0\)

\(\Rightarrow x;y\)