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b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)
\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)
\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)
\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)
\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)
Bài 1:
Đặt $\sqrt[4]{y^3-1}=a; \sqrt{x}=b$ $(a,b\geq 0$)
Khi đó hệ PT trở thành:
\(\left\{\begin{matrix} a+b=3\\ b^4+a^4+1=82\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a+b=3\\ a^4+b^4=81\end{matrix}\right.\)
Có: \(a^4+b^4=81\)
\(\Leftrightarrow (a^2+b^2)^2-2a^2b^2=81\)
\(\Leftrightarrow [(a+b)^2-2ab]^2-2a^2b^2=81\)
\(\Leftrightarrow (9-2ab)^2-2a^2b^2=81\)
\(\Leftrightarrow 2a^2b^2-36ab=0\)
\(\Leftrightarrow ab(ab-18)=0\Rightarrow \left[\begin{matrix} ab=0\\ ab=18\end{matrix}\right.\)
Nếu $ab=0$. Kết hợp với $a+b=3$ suy ra $(a,b)=(3,0); (0,3)$
$\Rightarrow (x,y)=(0, \sqrt[4]{82}); (9, 1)$
Nếu $ab=18$. Kết hợp với $a+b=3$ và định lý Vi-et đảo suy ra $a,b$ là nghiệm của pt: $X^2-3X+18=0$
Dễ thấy pt này vô nghiệm nên loại
Vậy......
Bài 2:
ĐK: ..........
Đặt $\sqrt{x+\frac{1}{y}}=a; \sqrt{x+y-3}=b$ $(a,b\geq 0$)
HPT \(\Leftrightarrow \left\{\begin{matrix} a+b=3\\ a^2+b^2+3=8\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a+b=3\\ a^2+b^2=5\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} a+b=3\\ (a+b)^2-2ab=5\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a+b=3\\ ab=2\end{matrix}\right.\)
Áp dụng định lý Vi-et đảo thì $a,b$ là nghiệm của pt $X^2-3X+2=0$
$\Rightarrow (a,b)=(2,1); (1,2)$
Nếu $(a,b)=(2,1)$
\(\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=4\\ x+y-3=1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=4\\ x+y=4\end{matrix}\right.\Rightarrow y=\frac{1}{y}\Rightarrow y=\pm 1\)
$y=1\rightarrow x=3$
$y=-1\rightarrow y=5$
Nếu $(a,b)=(1,2)$
\(\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=1\\ x+y-3=4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=1\\ x+y=7\end{matrix}\right.\Rightarrow y-\frac{1}{y}=6\)
\(\Rightarrow y^2-6y-1=0\Rightarrow y=3\pm \sqrt{10}\)
Nếu $y=3+\sqrt{10}\rightarrow x=4-\sqrt{10}$
Nếu $y=3-\sqrt{10}\rightarrow x=4+\sqrt{10}$
Vậy...........
a. \(\left\{{}\begin{matrix}3x-5y=6\\4x+7y=-8\end{matrix}\right.\)
\(x=\dfrac{2}{41}\) ; \(y=\dfrac{-48}{41}\)
b. \(\left\{{}\begin{matrix}\text{−2x+3y=5}\\5x+2y=4\end{matrix}\right.\)
\(x=\dfrac{2}{19};y=\dfrac{33}{19}\)
c.\(\left\{{}\begin{matrix}\text{2x−3y+4z=−5}\\-4x+5y-z=6\\3x+4y-3z=7\end{matrix}\right.\)
\(x=\dfrac{22}{101};y=\dfrac{131}{101};z=\dfrac{-39}{101}\)
d. \(\left\{{}\begin{matrix}\text{− x + 2 y − 3 z = 2}\\2x+y+2z=-3\\-2x-3y+z=5\end{matrix}\right.\)
\(x=-4;y=\dfrac{11}{7};z=\dfrac{12}{7}\)
a)x=0,05 ; y=-1,17
b.x=0,11 ; y=1,74
c.x=0,22 ;y=1,29 z=-0.39
d.x=-4 y=1,57 z=1,71
\(\left\{{}\begin{matrix}\sqrt{x-y}-\sqrt{x-3y}=-1\\7x-5y=19+2\sqrt{x-y}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-y}=a\left(a\ge0\right)\\\sqrt{x-3y}=b\left(b\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=-1\\\frac{7\left(3a^2-b^2\right)}{2}-\frac{5\left(a^2-b^2\right)}{2}=19+2a\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=a+1\\7\left(2a^2-2a-1\right)+5\left(2a+1\right)=4a+38\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=a+1\\14a^2-8a-40=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=3\\a=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-y}=2\\\sqrt{x-3y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=4\\x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3}{2}\\y=-\frac{5}{2}\end{matrix}\right.\)
Vậy...
DKXĐ: \(x\ge y;x\ge3y\)
PT (1) \(\Leftrightarrow\sqrt{x-y}+1=\sqrt{x-3y}\)
\(\Leftrightarrow x-y+1+2\sqrt{x-y}=x-3y\)
\(\Leftrightarrow2\sqrt{x-y}=-\left(2y+1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y+1\le0\\4\left(x-y\right)=\left(2y+1\right)^2\end{matrix}\right.\Rightarrow x=\frac{\left(4y^2+8y+1\right)}{4}\text{và }y\le-\frac{1}{2}\)
Thay vào PT (2) \(\Leftrightarrow\frac{7}{4}\left(4y^2+8y+1\right)-5y=19+2\sqrt{\frac{\left(4y^2+8y+1\right)}{4}-y}\)
Or: \(7\,{y}^{2}+9\,y-{\frac{69}{4}}=-(2y+1)\)
Do đó y = -5/2 hoặc 13/14 (loại)
Từ đó x = 3/2 và y =-5/2
P/s: Em không chắc lắm..
\(x+y=\sqrt{x+y}\Leftrightarrow\left[{}\begin{matrix}x+y=0\\x+y=1\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+y=0\\2x-5y=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=1\\2x-5y=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{12}{7}\\y=-\frac{5}{7}\end{matrix}\right.\)