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\(\left\{{}\begin{matrix}\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\\\sqrt{2x+y+1}+2\sqrt[3]{7x+12y+8}=2xy+y+5\end{matrix}\right.\)
Xét \(pt\left(1\right)\) dễ dàng suy ra \(x+y\ge0\)
\(VT=\sqrt{\left(x-y\right)^2+\left(2x+y\right)^2}+\sqrt{\left(x-y\right)^2+\left(2y+x\right)^2}\)
\(\ge\left|2x+y\right|+\left|2y+x\right|\ge3\left(x+y\right)\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x=y\\x,y\ge0\end{matrix}\right.\)
Thay vào \(pt\left(2\right)\) ta được:
\(\sqrt{3x+1}+2\sqrt[3]{19x+8}=2x^2+x+5\)
\(\Leftrightarrow\left[\sqrt{3x+1}-\left(x+1\right)\right]+2\left[\sqrt[3]{19x+8}-\left(x+2\right)\right]=2x^2-2x\)
\(\Leftrightarrow\left(x-x^2\right)\left[\dfrac{1}{\sqrt{3x+1}+x+1}+2\cdot\dfrac{x+7}{\sqrt[3]{\left(19x+8\right)^2}+\left(x+2\right)\sqrt[3]{19x+8}+\left(x+2\right)^2}+2\right]=0\)
Do \(x;y\ge0\) nên pt trong ngoặc luôn dương
\(\Rightarrow x-x^2=0\Rightarrow x\left(1-x\right)=0\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Mà \(x=y\)\(\Rightarrow\left[{}\begin{matrix}x=y=0\\x=y=1\end{matrix}\right.\) là nghiệm của hpt
b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)
\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)
\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)
\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)
\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)
1)Điều kiện: \(x + y > 0\)\((1) \Leftrightarrow (x + y)^2 - 2xy + \dfrac{2xy}{x + y} - 1 = 0 \\ \Leftrightarrow (x + y)^3 - 2xy(x + y) + 2xy -(x + y) = 0 \\ \Leftrightarrow (x+y)[(x+y)^2- 1]-2xy(x+y-1)=0 \\ \Leftrightarrow (x+y)(x+y+1)(x+y-1)-2xy(x+y-1)=0 \\ \Leftrightarrow (x + y - 1)[(x+y)(x + y + 1)-2xy] = 0 \\ \Leftrightarrow \left[ \begin{matrix}x + y = 1 \,\, (3) \\ x^2+y^2+x+y=0 \,\, (4) \end{matrix} \right.\)(4) vô nghiệm vì x + y > 0
Thế (3) vào (2) , giải được nghiệm của hệ :\((x =1 ; y = 0)\)và \((x = -2 ; y = 3)\)
\((1)\Leftrightarrow (x-2y)+(2x^3-4x^2y)+(xy^2-2y^3)=0\)\(\Leftrightarrow (x-2y)(1+2x^2+y^2)=0\)
\(\Leftrightarrow x=2y\)(vì \(1+2x^2+y^2>0, \forall x,y\))
Thay vào phương trình (2) giải dễ dàng.
\(1,HPT\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)+\left(\dfrac{1}{y}-\dfrac{1}{x}\right)=0\\2y=x^3+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\dfrac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y\\2y=x^3+1\end{matrix}\right.\\ \Leftrightarrow2y=y^3+1\Leftrightarrow y^3-2y+1=0\\ \Leftrightarrow\left[{}\begin{matrix}y=0\\y=\dfrac{-1+\sqrt{5}}{2}\\y=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)
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 \(\left(x;y\right)=\left(0;0\right)\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+y}=a\ge0\\\sqrt{x-y}=b\ge0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}a-b=2\\\sqrt{\frac{a^4+b^4}{2}}+ab=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=2\\\sqrt{\frac{a^4+b^4}{2}}=4-ab\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a-b=2\\\frac{a^4+b^4}{2}=a^2b^2-8ab+16\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=2\\a^4+b^4-2a^2b^2=16\left(2-ab\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=2\\\left(a^2-b^2\right)^2=16\left(2-ab\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=2\\4\left(a+b\right)^2=16\left(2-ab\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=2\\\left(a+b\right)^2=8-4ab\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}b=a-2\\\left(2a-2\right)^2=8-4a\left(a-2\right)\end{matrix}\right.\)
\(\Rightarrow4a^2-8a+4=8-4a^2+8a\)
\(\Rightarrow2a^2-4a-1=0\Rightarrow\left[{}\begin{matrix}a=\frac{2+\sqrt{6}}{2}\\a=\frac{2-\sqrt{6}}{2}\left(l\right)\end{matrix}\right.\) \(\Rightarrow b=\frac{\sqrt{6}-2}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x+y}=\frac{2+\sqrt{6}}{2}\\\sqrt{x-y}=\frac{\sqrt{6}-2}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+y=\frac{5+2\sqrt{6}}{2}\\x-y=\frac{5-2\sqrt{6}}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\y=\sqrt{6}\end{matrix}\right.\)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ge-1\\y\ge0\end{matrix}\right.\)
Ta có : \(x+\sqrt{\left(x+1\right).y}=2y-1\)
\(\Leftrightarrow x+1+\sqrt{\left(x+1\right)y}-2y=0\)
\(\Leftrightarrow\left(\sqrt{x+1}-\sqrt{y}\right)\left(\sqrt{x+1}+2\sqrt{y}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=\sqrt{y}\left(1\right)\\\sqrt{x+1}+2\sqrt{y}=0\left(2\right)\end{matrix}\right.\)
Từ (2) ta có \(\left\{{}\begin{matrix}x+1=0\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=0\end{matrix}\right.\) (tm)
Thử lại ta có (x;y) = (-1;0) là 1 nghiệm của hệ phương trình
Từ (1) ta có : x + 1 = y
Khi đó \(\sqrt{2x+3}+\sqrt{y}=x^2-y\)
\(\Leftrightarrow\sqrt{2x+3}+\sqrt{x+1}=x^2-x-1\)
\(\Leftrightarrow\left(\sqrt{2x+3}-3\right)+\left(\sqrt{x+1}-2\right)=x^2-x-6\)
\(\Leftrightarrow\dfrac{2x-6}{\sqrt{2x+3}+3}+\dfrac{x-3}{\sqrt{x+1}+2}=\left(x-3\right)\left(x+2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\\dfrac{2}{\sqrt{2x+3}+3}+\dfrac{1}{\sqrt{x+1}+2}=x+2\end{matrix}\right.\)
Với x = 3 => y = 4 (tm)
Với \(\dfrac{2}{\sqrt{2x+3}+3}+\dfrac{1}{\sqrt{x+1}+2}=x+2\)
Vì \(x\ge-1\) nên \(\dfrac{2}{\sqrt{2x+3}+3}\le\dfrac{1}{2};\dfrac{1}{\sqrt{x+1}+2}\le\dfrac{1}{2}\)
nên \(VT\le\dfrac{1}{2}+\dfrac{1}{2}=1\)
lại có \(VP\ge1\) khi x \(\ge-1\)
Dấu "=" xảy ra khi x = -1 => y = 0 (tm)
Vậy (x;y) = (-1;0) ; (3;4)
đk: \(\left\{{}\begin{matrix}x\ge-1\\y\ge0\\x^2>y\end{matrix}\right.\)
pt đầu \(\Leftrightarrow\sqrt{\left(x+1\right)y}=2y-x-1\)
\(\Rightarrow\left(x+1\right)y=4y^2+x^2+1+2x-4xy-4y\)
\(\Leftrightarrow x^2+4y^2-5xy+2x-5y+1=0\)
\(\Leftrightarrow\left(x-y\right)\left(x-4y\right)+\left(x-y\right)+\left(x-4y\right)+1=0\)
\(\Leftrightarrow\left(x-y+1\right)\left(x-4y+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=x+1\\x=4y-1\end{matrix}\right.\)
TH1: \(y=x+1\) thay vào pt thứ hai, ta được
\(\sqrt{2x+3}+\sqrt{x+1}=x^2-x-1\)
\(\Leftrightarrow\left(\sqrt{2x+3}-3\right)+\left(\sqrt{x+1}-2\right)=x^2-x-6\)
\(\Leftrightarrow\dfrac{2x-6}{\sqrt{2x+3}+3}+\dfrac{x-3}{\sqrt{x+1}+2}-\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\dfrac{2}{\sqrt{2x+3}+3}+\dfrac{1}{\sqrt{x+1}+2}-x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\\dfrac{2}{\sqrt{2x+3}+3}+\dfrac{1}{\sqrt{x+1}+2}-x+2=0\end{matrix}\right.\)
TH1.1: \(x=3\Rightarrow y=x+1=4\) (nhận)
TH1.2:\(\dfrac{2}{\sqrt{2x+3}+3}+\dfrac{1}{\sqrt{x+1}+2}-x+2=0\) (chỗ này mai mình nghĩ tiếp)
TH2: \(x=4y-1\). Thay vào pt thứ hai, ta được
\(\sqrt{8y+1}+\sqrt{y}=16y^2-9y+1\)
\(\Leftrightarrow\left(\sqrt{8y+1}-1\right)+\sqrt{y}=16y^2-9y\)
\(\Leftrightarrow\dfrac{8y}{\sqrt{8y+1}+1}+\dfrac{y}{\sqrt{y}}-16y^2+9y=0\)
\(\Leftrightarrow y\left(\dfrac{8}{\sqrt{8y+1}+1}+\dfrac{1}{\sqrt{y}}-16y+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\\\dfrac{8}{\sqrt{8y+1}+1}+\dfrac{1}{\sqrt{y}}-16y+9=0\end{matrix}\right.\)
TH2.1: \(y=0\) \(\Rightarrow x=4y-1=-1\) (nhận)
TH2.2: \(\dfrac{8}{\sqrt{8y+1}+1}+\dfrac{1}{\sqrt{y}}-16y+9=0\)
(đoạn này để mai mình nghĩ tiếp nhé, ta tìm được các nghiệm \(\left(x;y\right)=\left(-1;0\right);\left(3;4\right)\))
1/ \(\left\{{}\begin{matrix}x^3+y^3=1\left(1\right)\\x^2y+2xy^2+y^3=2\left(2\right)\end{matrix}\right.\)
Lấy (1). 2 - (2) ta được:
\(2x^3+y^3-x^2y-2xy^2=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)\left(2x-y\right)=0\)
Đến đây dễ rồi nhé ^^
2/ Ta viết lại pt thứ 2 của hệ:
\(y^2-4\left(x+2\right)y+16+16x-5x^2=0\)
\(\Leftrightarrow y^2-4\left(x+2\right)y+4\left(x+2\right)^2-9x^2=0\)
\(\Leftrightarrow\left[y-2\left(x+2\right)\right]^2-\left(3x\right)^2=0\)
\(\Leftrightarrow\left(x+y-4\right)\left(y-5x-4\right)=0\)
Bạn làm tiếp nhé!
3/ Ta viết lại pt thứ nhất của hệ
\(x^2-x\left(2y-3\right)+y^2-3y-4=0\)
\(\Leftrightarrow x^2-x\left(2y-3\right)+\dfrac{4y^2-12y+9}{4}-\dfrac{25}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{2y+3}{2}\right)^2-\left(\dfrac{5}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-y-4\right)\left(x-y+1\right)=0\)
Bạn làm tiếp được chứ?
4/ Viết lại pt thứ 2 của hệ
\(\left(y+\sqrt{x}\right)^2-\left(y\sqrt{x}\right)^2=0\)
\(\Leftrightarrow\left(y-\sqrt{x}-y\sqrt{x}\right)\left(y-\sqrt{x}+y\sqrt{x}\right)=0\)