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c. ĐKXĐ: ...
\(x^2+y^2+2xy-2xy+\dfrac{2xy}{x+y}-1=0\)
\(\Leftrightarrow\left(x+y\right)^2-1-2xy\left(1-\dfrac{1}{x+y}\right)=0\)
\(\Leftrightarrow\left(x+y-1\right)\left(x+y+1\right)-\dfrac{2xy\left(x+y-1\right)}{x+y}=0\)
\(\Leftrightarrow\left(x+y-1\right)\left(x+y+1-\dfrac{2xy}{x+y}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=1\\x^2+y^2+x+y=0\left(vô-nghiệm\right)\end{matrix}\right.\)
Thế \(y=1-x\) xuống pt dưới:
\(\sqrt{x+1-x}=x^2-\left(1-x\right)\)
\(\Leftrightarrow x^2+x-2=0\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=0\\x=-2\Rightarrow y=3\end{matrix}\right.\)
d.
ĐKXĐ: \(x>-2;y>1;x+y>0\)
\(\left\{{}\begin{matrix}\sqrt{\dfrac{x+y}{x+2}}+\sqrt{\dfrac{x+y}{y-1}}=2\\2\left(x+y\right)^2=\left(x+2\right)^2+\left(y-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{x+y}{x+2}}+\sqrt{\dfrac{x+y}{y-1}}=2\\\left(\dfrac{x+2}{x+y}\right)^2+\left(\dfrac{y-1}{x+y}\right)^2=2\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\sqrt{\dfrac{x+y}{x+2}}=a>0\\\sqrt{\dfrac{x+y}{y-1}}=b>0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2\\\dfrac{1}{a^4}+\dfrac{1}{b^4}=2\end{matrix}\right.\)
Ta có: \(\dfrac{1}{a^4}+\dfrac{1}{b^4}\ge\dfrac{1}{8}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^4\ge\dfrac{1}{8}\left(\dfrac{4}{a+b}\right)^4=\dfrac{1}{8}.\left(\dfrac{4}{2}\right)^4=2\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x+y}{x+2}=1\\\dfrac{x+y}{y-1}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
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...
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.
\(1,ĐK:x,y\ne0\\ HPT\Leftrightarrow\left\{{}\begin{matrix}2x^2y^2=y^3+1\\2x^2y^2=x^3+1\end{matrix}\right.\\ \Leftrightarrow x^3+1=y^3+1\\ \Leftrightarrow x^3=y^3\Leftrightarrow x=y\)
Thay vào PT 1
\(\Leftrightarrow2x^4=x^3+1\\ \Leftrightarrow2x^4-x^3-1=0\\ \Leftrightarrow2x^4-2x^3+x-1=0\\ \Leftrightarrow\left(2x^3+1\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^3=-\dfrac{1}{2}\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=\sqrt[3]{-\dfrac{1}{2}}\\x=y=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(\sqrt[3]{-\dfrac{1}{2}};\sqrt[3]{-\dfrac{1}{2}}\right);\left(1;1\right)\)
\(2,ĐK:x,y\ge1\\ HPT\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)+\sqrt{y-1}=\dfrac{1}{2}\\2\left(y-1\right)+\sqrt{x-1}=\dfrac{1}{2}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-1}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\)
\(HPT\Leftrightarrow\left\{{}\begin{matrix}2a^2+b=\dfrac{1}{2}\\2b^2+a=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow2\left(a-b\right)\left(a+b\right)-\left(a-b\right)=0\\ \Leftrightarrow\left(a-b\right)\left(2a+2b-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=b\\2a+2b=1\end{matrix}\right.\)
Với \(a=b\Leftrightarrow x-1=y-1\Leftrightarrow x=y\)
Thay vào \(PT\left(1\right)\Leftrightarrow2x+\sqrt{x-1}=\dfrac{5}{2}\Leftrightarrow2\sqrt{x-1}=5-4x\)
\(\Leftrightarrow4x-4=25-40x+16x^2\\ \Leftrightarrow16x^2-44x+29=0\\ \Leftrightarrow\left[{}\begin{matrix}x=y=\dfrac{11+\sqrt{5}}{8}\left(tm\right)\\x=y=\dfrac{11-\sqrt{5}}{8}\left(tm\right)\end{matrix}\right.\)
Với \(2a+2b=1\Leftrightarrow b=\dfrac{1}{2}-a\Leftrightarrow\sqrt{y-1}=\dfrac{1}{2}-\sqrt{x-1}\)
Thay vào \(PT\left(1\right)\Leftrightarrow2x+\dfrac{1}{2}-\sqrt{x-1}=\dfrac{5}{2}\Leftrightarrow2x-2=\sqrt{x-1}\)
\(\Leftrightarrow4x^2-8x+4=x-1\\ \Leftrightarrow4x^2-9x+5=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\Rightarrow y=1\left(tm\right)\\x=1\Rightarrow y=\dfrac{5}{4}\left(tm\right)\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(\dfrac{11+\sqrt{5}}{8};\dfrac{11+\sqrt{5}}{8}\right);\left(\dfrac{11-\sqrt{5}}{8};\dfrac{11-\sqrt{5}}{8}\right);\left(\dfrac{5}{4};1\right);\left(1;\dfrac{5}{4}\right)\)
1,\(hpt\Leftrightarrow\left\{{}\begin{matrix}\left(x-2y\right)\left(x+y\right)=0\\\sqrt{2x}+\sqrt{y+1}=2\left(\circledast\right)\end{matrix}\right.\)
\(\left(x-2y\right)\left(x+y\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2y\\x=-y\end{matrix}\right.\)
Th1:\(x=2y\) Thay vào \(\left(\circledast\right)\) , ta có :
\(\sqrt{4y}+\sqrt{y+1}=2\)
\(\Leftrightarrow2-2\sqrt{y}=\sqrt{y+1}\)\(\Leftrightarrow3y-8\sqrt{y}+3=0\)
Giải pt thu được (x;y)
Th2:x=-y thay vào \(\left(\circledast\right)\), ta có
\(\sqrt{-2x}+\sqrt{y+1}=2\)
Xét đk ta thấy:\(y\le0;y\ge-1\)(vô nghiệm)
Vậy ....
2,\(hpt\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-1\right)\left(x+y^2\right)=0\\\sqrt{x}+\sqrt{y+1}=2\end{matrix}\right.\)
\(\left(x-y-1\right)\left(x+y^2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=y+1\\x=-y^2\end{matrix}\right.\)
Th1:\(x=y+1\)
Thay vào ta có:\(\sqrt{x}+\sqrt{x}=2\Leftrightarrow x=1\)\(\Leftrightarrow y=0\)
Th2:\(x=-y^2\)thay vào ta có:
\(\sqrt{-y^2}+\sqrt{y+1}=2\)
vì \(-y^2\le0\) mà nhận thấy y=0 ko là nghiệm của pt
\(\Rightarrow\)Pt vô nghiệm
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: \(\left\{{}\begin{matrix}a+b=2\\\dfrac{1}{a^4}+\dfrac{1}{b^4}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\a^4+b^4=2a^4b^4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\\left[\left(a+b\right)^2-2ab\right]^2-2a^2b^2=2a^4b^4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\\left(4-2ab\right)^2-2a^2b^2=2a^4b^4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\a^4b^4=a^2b^2-8ab+8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\a^2b^2\left(a^2b^2-1\right)+8\left(ab-1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\\left(ab-1\right)\left[a^2b^2\left(ab+1\right)+8\right]=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\ab-1\end{matrix}\right.\left(a,b>0\right)\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{x+y}{x+2}}=1\\\sqrt{\dfrac{x+y}{y-1}}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=x+2\\x+y=y-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)