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

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

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

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

\(\Leftrightarrow x=y\)

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

Tìm m để pt sau có nghiệm: \(\left(5+x\right)\left(2-x\right)+m=\sqrt{x^2+3x}\)

ĐKXĐ: \(x\le3;y\ge2\)

Từ pt dưới ta có \(y=x+7\) thay lên pt trên được:

\(\sqrt{3-x}+\sqrt{x+5}=4\)

Áp dụng BĐT Bunhia: \(\sqrt{3-x}+\sqrt{x+5}\le\sqrt{\left(1+1\right)\left(3-x+x+5\right)}=4\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{3-x}=\sqrt{x+5}\Leftrightarrow3-x=x+5\Rightarrow x=-1\)

\(\Rightarrow y=6\)

Vậy nghiệm của hệ là \(\left\{{}\begin{matrix}x=-1\\y=6\end{matrix}\right.\)

\(ĐK:x+y\ge0;x-y\ge0;x,y\ge0\)

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

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

\(\Leftrightarrow y=x-1\)

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 ...

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

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

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

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

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

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

a:

ĐKXĐ: y+1>=0

=>y>=-1

 \(\left\{{}\begin{matrix}2\left(x^2-2x\right)+\sqrt{y+1}=0\\3\left(x^2-2x\right)-2\sqrt{y+1}+7=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2\left(x^2-2x\right)+\sqrt{y+1}=0\\3\left(x^2-2x\right)-2\sqrt{y+1}=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4\left(x^2-2x\right)+2\sqrt{y+1}=0\\3\left(x^2-2x\right)-2\sqrt{y+1}=-7\end{matrix}\right.\)

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

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

=>\(\left\{{}\begin{matrix}x^2-2x+1=0\\2\sqrt{y+1}=-3+7=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\sqrt{y+1}=2\end{matrix}\right.\)

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

b: \(\left\{{}\begin{matrix}5\left|x-1\right|-3\left|y+2\right|=7\\2\sqrt{4x^2-8x+4}+5\sqrt{y^2+4y+4}=13\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5\left|x-1\right|-3\left|y+2\right|=7\\2\cdot\sqrt{\left(2x-2\right)^2}+5\cdot\sqrt{\left(y+2\right)^2}=13\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5\left|x-1\right|-3\left|y+2\right|=7\\4\left|x-1\right|+5\left|y+2\right|=13\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}20\left|x-1\right|-12\left|y+2\right|=28\\20\left|x-1\right|+25\left|y+2\right|=65\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-37\left|y+2\right|=-37\\4\left|x-1\right|+5\left|y+2\right|=13\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left|y+2\right|=1\\4\left|x-1\right|=13-5=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|y+2\right|=1\\\left|x-1\right|=2\end{matrix}\right.\)

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

c: ĐKXĐ: \(\left\{{}\begin{matrix}x< >-1\\y< >-4\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3x+3-3}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x+2-2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3-\dfrac{3}{x+1}-\dfrac{2}{y+4}=4\\2-\dfrac{2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3}{x+1}+\dfrac{2}{y+4}=3-4=-1\\\dfrac{2}{x+1}+\dfrac{5}{y+4}=2-9=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{6}{x+1}+\dfrac{4}{y+4}=-2\\\dfrac{6}{x+1}+\dfrac{15}{y+4}=-21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-11}{y+4}=19\\\dfrac{3}{x+1}+\dfrac{2}{y+4}=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y+4=-\dfrac{11}{19}\\\dfrac{3}{x+1}+2:\dfrac{-11}{19}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{11}{19}-4=-\dfrac{87}{19}\\\dfrac{3}{x+1}=-1-2:\dfrac{-11}{19}=-1+2\cdot\dfrac{19}{11}=\dfrac{27}{11}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{87}{19}\\x+1=\dfrac{11}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{87}{19}\\x=\dfrac{2}{9}\end{matrix}\right.\)(nhận)

d:

ĐKXĐ: x<>1 và y<>-2

\(\left\{{}\begin{matrix}\dfrac{x+1}{x-1}+\dfrac{3y}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\dfrac{x-1+2}{x-1}+\dfrac{3y+6-6}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}1+\dfrac{2}{x-1}+3-\dfrac{6}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{2}{x-1}-\dfrac{6}{y+2}=7-4=3\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-\dfrac{1}{y+2}=-1\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+2=1\\\dfrac{2}{x-1}-5=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-1\\\dfrac{2}{x-1}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x-1=\dfrac{2}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=\dfrac{11}{9}\end{matrix}\right.\left(nhận\right)\)