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b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)
\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)
\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)
\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)
\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)
5,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x\left(x+y\right)\left(x+2\right)=0\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14}=x-2\end{matrix}\right.\)
Thay từng TH rồi làm nha bạn
3,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x-y=\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\frac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\)
thay nhá
Bài 1:ĐKXĐ: \(2x\ge y;4\ge5x;2x-y+9\ge0\)\(\Rightarrow2x\ge y;x\le\frac{4}{5}\Rightarrow y\le\frac{8}{5}\)
PT(1) \(\Leftrightarrow\left(x-y-1\right)\left(2x-y+3\right)=0\)
+) Với y = x - 1 thay vào pt (2):
\(\frac{2}{3+\sqrt{x+1}}+\frac{2}{3+\sqrt{4-5x}}=\frac{9}{x+10}\) (ĐK: \(-1\le x\le\frac{4}{5}\))
Anh quy đồng lên đê, chắc cần vài con trâu đó:))
+) Với y = 2x + 3...
\(\left\{{}\begin{matrix}\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\\\sqrt{2x+y+1}+2\sqrt[3]{7x+12y+8}=2xy+y+5\end{matrix}\right.\)
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
Xét VT của (1):
\(3VT\)
\(=\sqrt{5x^2+2xy+2y^2}.\sqrt{2^2+2^2+1^2}+\sqrt{2x^2+2xy+5y^2}.\sqrt{2^2+2^2+1^2}\)
\(=\sqrt{\left(x+y\right)^2+4x^2+y^2}.\sqrt{2^2+2^2+1^2}+\sqrt{\left(x+y\right)^2+x^2+4y^2}.\sqrt{2^2+2^2+1^2}\)
\(\ge\left[2\left(x+y\right)+4x+y\right]+\left[2\left(x+y\right)+x+4y\right]=9x+9y\)
\(\Rightarrow VT\ge3x+3y=VT\)
Đẳng thức xảy ra \(\Leftrightarrow...\Leftrightarrow x=y\)
Sau đó thay \(y=x\) vào pt (2) ta được:
\(\sqrt{3x+1}+2\sqrt[3]{19x+8}=2x^2+x+5\)
\(\Leftrightarrow\left(2x^2-\sqrt{3x+1}\right)+\left(x-5-2\sqrt[3]{19x+8}\right)=0\)
\(\Leftrightarrow\dfrac{4x^2-3x-1}{2x^2+\sqrt{3x+1}}+\dfrac{\left(x+5\right)^3-8\left(19x+8\right)}{\left(x-5\right)^2+2\left(x-5\right)\sqrt[3]{19x+8}+4\sqrt[3]{\left(19x+8\right)^2}}=0\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(4x+1\right)}{2x^2+\sqrt{3x+1}}+\dfrac{ \left(x-1\right)\left(x^2+16x-61\right)}{\left(x-5\right)^2+2\left(x-5\right)\sqrt[3]{19x+8}+4\sqrt[3]{\left(19x+8\right)^2}}=0\)
\(\Leftrightarrow\left(x-1\right)\left[\dfrac{4x+1}{2x^2+\sqrt{3x+1}}+\dfrac{x^2+16x-61}{\left(x-5\right)^2+2\left(x-5\right)\sqrt[3]{19x+8}+4\sqrt[3]{\left(19x+8\right)^2}}\right]=0\)
\(\Leftrightarrow x=1\Rightarrow y=1\)
Từ pt thứ nhất: \(\Leftrightarrow x+1+\sqrt{\left(x+1\right)^2+1}=\left(-y\right)+\sqrt{\left(-y\right)^2+1}\)
Xét hàm \(f\left(t\right)=t+\sqrt{t^2+1}\Rightarrow f'\left(t\right)=1+\dfrac{t}{\sqrt{t^2+1}}=\dfrac{t+\sqrt{t^2+1}}{\sqrt{t^2+1}}\)
\(f'\left(t\right)>\dfrac{t+\sqrt{t^2}}{\sqrt{t^2+1}}=\dfrac{t+\left|t\right|}{\sqrt{t^2+1}}\ge0\Rightarrow f'\left(t\right)>0\) ; \(\forall t\)
\(\Rightarrow f\left(t\right)\) đồng biến trên R
\(\Rightarrow x+1=-y\Rightarrow y=-x-1\)
Thế xuống pt dưới:
\(x^3-\left(3x^2-2x-8\right)\sqrt{2x^2+x-1}=0\)
Bạn coi lại đề, pt vô tỉ này ko giải được
a. ĐKXĐ: ..
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2\left(2x+5y\right)}-\sqrt{2\left(x+y\right)}=4\\x+2y+\dfrac{2\sqrt{\left(x+y\right)\left(2x+5y\right)}}{3}=24\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\sqrt{2\left(2x+5y\right)}=a\ge0\\\sqrt{2\left(x+y\right)}=b\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=4\\\dfrac{a^2+b^2}{6}+\dfrac{ab}{3}=24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=4\\\left(a+b\right)^2=144\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=4\\\left[{}\begin{matrix}a+b=12\\a+b=-12\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(a;b\right)=\left(8;4\right)\\\left(a;b\right)=\left(-4;-8\right)\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2\left(2x+5y\right)=64\\2\left(x+y\right)=16\end{matrix}\right.\) \(\Leftrightarrow...\)
b.
Thế pt trên xuống dưới:
\(x^4+6y^4=\left(x+2y\right)\left(x^3+3y^3-2xy^2\right)\)
\(\Leftrightarrow2x^3y-2x^2y^2-xy^3=0\)
\(\Leftrightarrow xy\left(2x^2-2xy-y^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\y=0\\y=-\left(1+\sqrt{3}\right)x\\y=\left(-1+\sqrt{3}\right)x\end{matrix}\right.\)
Thế vào pt đầu ...
Đề cho hơi xấu, nếu pt đầu là \(x^3+3y^3-2x^2y=1\) thì đẹp hơn nhiều
\(1,HPT\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)+\left(\dfrac{1}{y}-\dfrac{1}{x}\right)=0\\2y=x^3+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\dfrac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y\\2y=x^3+1\end{matrix}\right.\\ \Leftrightarrow2y=y^3+1\Leftrightarrow y^3-2y+1=0\\ \Leftrightarrow\left[{}\begin{matrix}y=0\\y=\dfrac{-1+\sqrt{5}}{2}\\y=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)
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 \(\left(x;y\right)=\left(0;0\right)\)
\(ĐK:x\ge\dfrac{1}{5};y\ge\dfrac{3}{8}\)
\(PT\left(1\right)\Leftrightarrow\dfrac{3x^2-3y^2}{\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}}=3\left(x+y\right)\\ \Leftrightarrow3\left(x+y\right)\left(\dfrac{x-y}{\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+y=0\\\dfrac{x-y}{\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x-y=\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}\\ \Leftrightarrow\left(x-y\right)=\dfrac{3\left(x^2-y^2\right)}{\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}}\\ \Leftrightarrow\left(x-y\right)\left[\dfrac{3\left(x+y\right)}{\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}}-1\right]=0\)
\(\Leftrightarrow x=y\)
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 \(\left(x;y\right)\in\left\{\left(4+\sqrt{13};4+\sqrt{13}\right);\left(4-\sqrt{13};4-\sqrt{13}\right)\right\}\)
bạn làm nhầm rồi hay sao đấy
mình tìm ra cách rồi là
Từ pt(1) \(\sqrt{\left(2x+y\right)^2+\left(x-y\right)^2}+\sqrt{\left(2y+x\right)^2+\left(x-y\right)^2}=3\left(x+y\right)\)
Đặt a=2x+y;b=2y+x\(\Rightarrow\) 3(x+y)=a+b;x-y=a-b
rồi bình phương ra