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ĐKXĐ: \(x\ge-2;y\ge0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\) pt đầu trở thành:
\(a\left(a^2+1\right)=b\left(ab+1\right)\)
\(\Leftrightarrow a^3+a=ab^2+b\)
\(\Leftrightarrow a^3-ab^2+a-b=0\)
\(\Leftrightarrow a\left(a^2-b^2\right)+a-b=0\)
\(\Leftrightarrow a\left(a-b\right)\left(a+b\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+1\right)=0\)
\(\Leftrightarrow a-b=0\) (do \(a^2+ab+1>0;\forall a\ge0;b\ge0\))
\(\Leftrightarrow\sqrt{x+2}=\sqrt{y}\)
\(\Rightarrow y=x+2\)
Thế vào pt dưới:
\(x^2+\left(x+3\right)\left(x+3\right)=x+16\)
\(\Leftrightarrow2x^2+5x-7=0\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=3\\x=-\dfrac{7}{2}< -2\left(loại\right)\end{matrix}\right.\)
\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)
\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)
\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)
\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)
\(\left(\sqrt{x^2+5}+x\right)\left(\sqrt{y^2+5}+y\right)=5\)
⇔ \(\left(\sqrt{x^2+5}+x\right)\left(\sqrt{x^2+5}-x\right)\left(\sqrt{y^2+5}+y\right)=5\left(\sqrt{x^2+5}-x\right)\) ⇔ \(5\left(\sqrt{y^2+5}+y\right)=5\left(\sqrt{x^2+5}-x\right)\)
⇔ \(x+y=\sqrt{x^2+5}-\sqrt{y^2+5}\left(1\right)\)
Tương tự : \(x+y=\sqrt{y^2+5}-\sqrt{x^2+5}\left(2\right)\)
Cộng từng vế của ( 1 ; 2 ) , ta có : x + y = 0
\(\left\{{}\begin{matrix}\dfrac{x+2}{y-1}=\dfrac{x-4}{y+2}\\\dfrac{2x+3}{y-1}=\dfrac{4x+1}{2y+1}\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\left(x+2\right)\left(y+2\right)=\left(y-1\right)\left(x-\text{4}\right)\\\left(2x+3\right)\left(2y+1\right)=\left(y-1\right)\left(4x+1\right)\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}xy+2x+2y+4=xy-4y-x+4\\4xy+2x+6y+3=4xy-4x+y-1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}3x+6y=0\\6x+5y=-4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=-\dfrac{8}{7}\\y=\dfrac{4}{7}\end{matrix}\right.\)(TM)
\(\left\{{}\begin{matrix}5\left(x-y\right)-3\left(2x+3y\right)=12\\3\left(x+2y\right)-4\left(x+2y\right)=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}5x-5y-6x-9y=12\\3x+6y-4x-8y=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}-x-14y=12\\-x-2y=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=-\dfrac{26}{3}\\y=-\dfrac{7}{12}\end{matrix}\right.\)
Vậy HPT có nghiệm (x;y) = (\(-\dfrac{26}{3};-\dfrac{7}{12}\))
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
\(\left\{{}\begin{matrix}\dfrac{x+1}{x-1}+\dfrac{3y}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\dfrac{x-1+2}{x-1}+\dfrac{3y+6-6}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}1+\dfrac{2}{x-1}+3-\dfrac{6}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2}{x-1}-\dfrac{6}{y+2}=7-4=3\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{1}{y+2}=-1\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+2=1\\\dfrac{2}{x-1}-5=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-1\\\dfrac{2}{x-1}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x-1=\dfrac{2}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=\dfrac{11}{9}\end{matrix}\right.\left(nhận\right)\)
Đặt \(\left(x+\sqrt{x^2+5}\right)\left(y+\sqrt{y^2+5}\right)=5\)là A
Nhân 2 vế A cho \(\sqrt{x^2+5}-x\)ta được:
\(5.\left(y+\sqrt{y^2+5}\right)=5.\left(\sqrt{x^2+5}-x\right)\)
\(\Leftrightarrow y+\sqrt{y^2+5}=\sqrt{x^2+5}-x\)
\(\Leftrightarrow x+y=\sqrt{x^2+5}-\sqrt{y^2+5}\left(1\right)\)
Nhân 2 vế A cho \(\sqrt{y^2+5}-y\) ta được:
\(5.\left(x+\sqrt{x^2+5}\right)=5.\left(\sqrt{y^2+5}-y\right)\)
\(\Leftrightarrow x+\sqrt{x^2+5}=\sqrt{y^2+5}-y\)
\(\Leftrightarrow x+y=\sqrt{y^2+5}-\sqrt{x^2+5}\left(2\right)\)
từ (1) và (2) suy ra:
\(x+y-\left(x+y\right)=\sqrt{x^2+5}-\sqrt{y^2+5}-\left(\sqrt{y^2+5}-\sqrt{x^2+5}\right)\)
\(\Leftrightarrow2\left(\sqrt{x^2+5}-\sqrt{y^2+5}\right)=0\)
\(\Leftrightarrow\sqrt{x^2+5}-\sqrt{y^2+5}=0\)
\(\Rightarrow x+y=\sqrt{x^2+5}-\sqrt{y^2+5}=0\)
\(\left(\sqrt{x^2+5}+x\right)\left(\sqrt{y^2+5}+y\right)=5\)
Ta có: \(\hept{\begin{cases}\left(\sqrt{x^2+5}+x\right)\left(\sqrt{x^2+5}-x\right)=5\\\left(\sqrt{y^2+5}+y\right)\left(\sqrt{y^2+5}-y\right)=5\end{cases}}\)
Kết hợp với giải thiết ta được: \(\hept{\begin{cases}\sqrt{x^2+5}-x=\sqrt{y^2+5}+y\\\sqrt{y^2+5}-y=\sqrt{x^2+5}+x\end{cases}}\)
Cộng theo vế ta được: \(-\left(x+y\right)=x+y\)
\(\Rightarrow\)\(x+y=0\)