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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,\(hpt\Leftrightarrow\left\{{}\begin{matrix}\left(x-2y\right)\left(x+y\right)=0\\\sqrt{2x}+\sqrt{y+1}=2\left(\circledast\right)\end{matrix}\right.\)
\(\left(x-2y\right)\left(x+y\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2y\\x=-y\end{matrix}\right.\)
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
a/ ĐKXĐ: \(x\ge4\)
Đặt \(\sqrt{x+4}+\sqrt{x-4}=a>0\)
\(\Rightarrow a^2=2x+2\sqrt{x^2-16}\)
Phương trình trở thành:
\(a=a^2-12\Leftrightarrow a^2-a-12=0\Rightarrow\left[{}\begin{matrix}a=4\\a=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+4}+\sqrt{x-4}=4\)
\(\Leftrightarrow2x+2\sqrt{x^2-16}=16\)
\(\Leftrightarrow\sqrt{x^2-16}=8-x\left(x\le8\right)\)
\(\Leftrightarrow x^2-16=x^2-16x+64\)
\(\Rightarrow x=5\)
b/ \(x\ge-\frac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{2x+1}=a\\\sqrt{4x^2-2x+1}=b\end{matrix}\right.\) ta được:
\(a+3b=3+ab\)
\(\Leftrightarrow ab-a-\left(3b-3\right)=0\)
\(\Leftrightarrow a\left(b-1\right)-3\left(b-1\right)=0\)
\(\Leftrightarrow\left(a-3\right)\left(b-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=3\\b=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+1}=3\\\sqrt{4x^2-2x+1}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x+1=9\\4x^2-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\\x=\frac{1}{2}\end{matrix}\right.\)
Bài 2:
a/ \(\left\{{}\begin{matrix}\left(x+2y\right)^2-4xy-5=0\\4xy\left(x+2y\right)+5\left(x+2y\right)-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+2y\right)^2-\left(4xy+5\right)=0\\\left(4xy+5\right)\left(x+2y\right)-1=0\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+2y=a\\4xy+5=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2-b=0\\ab=1\end{matrix}\right.\) \(\Rightarrow a^2-\frac{1}{a}=0\Rightarrow a^3-1=0\)
\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+2y=1\\4xy+5=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1-2y\\4y\left(1-2y\right)+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1-2y\\-8y^2+4y+4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=-1\\y=-\frac{1}{2}\Rightarrow x=2\end{matrix}\right.\)
b/Cộng vế với vế:
\(17x^2-2\left(4y^2+1\right)x+y^4+1=0\)
\(\Delta'=\left(4y^2+1\right)^2-17\left(y^4+1\right)=-y^4+8y^2-16\)
\(\Delta'=-\left(y^2-4\right)^2\ge0\Rightarrow y^2-4=0\Rightarrow\left[{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\)
- Với \(y=2\) \(\Rightarrow x^2-2x+1=0\Rightarrow x=1\)
\(\)- Với \(y=-2\Rightarrow x^2-2x-7=0\Rightarrow x=1\pm2\sqrt{2}\)
b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)
\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)
\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)
\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)
\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)
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a)
\(\left\{{}\begin{matrix}x^2+x+5< 0\\x^2-6x+1>0\end{matrix}\right.\)
\(\)Ta có
\(x^2+x+5=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{19}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}>0\)
=> Bất phương trình đàu tiên sai, hệ bất phương trình sai
b)
\(\left\{{}\begin{matrix}2x^2+x-6>0\\3x^2-10x+3\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3\right)\left(x+2\right)>0\\\left(x-3\right)\left(3x-1\right)\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>2\\x< -3\end{matrix}\right.\\\left[{}\begin{matrix}x\le-\dfrac{1}{3}\\x\ge3\end{matrix}\right.\end{matrix}\right.\)
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)
\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)
Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)
\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)
Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:
\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)
\(\Leftrightarrow10b+40=3\left(b+8\right)b\)
\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)
TH1: \(b=2\Leftrightarrow...\)
TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)
5.
ĐK: \(x\ge-4\)
Đặt \(\sqrt{\dfrac{x}{8}+\dfrac{1}{2}}=\dfrac{1}{2}y+1\Rightarrow x=2y^2+8y+4\)
Phương trình đã cho tương đương \(y=2x^2+8x+4\)
Ta có hệ \(\left\{{}\begin{matrix}x=2y^2+8y+4\left(1\right)\\y=2x^2+8x+4\end{matrix}\right.\)
Trừ vế theo vế ta được:
\(x-y=2y^2+8y-2x^2-8x\)
\(\Leftrightarrow\left(x-y\right)\left(2x+2y+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\2x+2y+9=0\end{matrix}\right.\)
TH1: \(x=y\)
\(\left(1\right)\Leftrightarrow2x^2+7x+4=0\)
\(\Leftrightarrow x=\dfrac{-7\pm\sqrt{17}}{4}\left(tm\right)\)
Trường hợp còn lại tương tự
4.
ĐK: \(-\dfrac{\sqrt{6}}{2}\le x\le\dfrac{\sqrt{6}}{2}\)
Dễ thấy x=0 không phải nghiệm của phương trình, phương trình trở thành:
\(\sqrt{-4x^4+4x^2+3}=2x^2+\dfrac{1}{2x^2}\)
Ta có \(VT=\sqrt{-4x^4+4x^2+3}=\sqrt{-\left(2x^2-1\right)^2+4}\le2\)
\(VP=2x^2+\dfrac{1}{2x^2}\ge2\sqrt{2x^2.\dfrac{1}{2x^2}}=2\)
\(\Rightarrow\sqrt{-4x^4+4x^2+3}\le2x^2+\dfrac{1}{2x^2}\)
Đẳng thức xảy ra khi \(x=\dfrac{\pm\sqrt{2}}{2}\left(tm\right)\)
Vậy \(x=\dfrac{\pm\sqrt{2}}{2}\)
7.
Nếu \(x=0\Rightarrow y=5\Rightarrow\left(x;y\right)=\left(0;5\right)\) là nghiệm của hệ
Nếu \(x\ne0\)
\(\left\{{}\begin{matrix}\sqrt{2x^2+2}+\left(x^2+1\right)^2+2y-10=0\left(1\right)\\\left(x^2+1\right)^2+x^2y\left(y-4\right)=0\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\dfrac{\left(x^2+1\right)^2}{x^2}+y\left(y-4\right)=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2+y\left(y-4\right)=0\)
Áp dụng BĐT Cosi:
\(\left(x+\dfrac{1}{x}\right)^2+y\left(y-4\right)\ge2^2+y^2-4y=\left(y-2\right)^2\ge0\)
Đẳng thức xảy ra khi \(\left(x;y\right)=\left(\pm1;2\right)\)
Thử lại vào phương trình \(\left(1\right)\) ta được \(\left(x;y\right)=\left(\pm1;2\right)\)
Vậy \(\left(x;y\right)\in\left\{\left(\pm1;2\right);\left(0;5\right)\right\}\)
6.
\(\left\{{}\begin{matrix}4x^3+xy^2=3x-y\\4xy+y^2=2\left(1\right)\end{matrix}\right.\)
Từ hệ phương trình suy ra: \(2\left(4x^3+xy^2\right)=\left(4xy+y^2\right)\left(3x-y\right)\)
\(\Leftrightarrow8x^3+2xy^2=12x^2y-xy^2-y^3\)
\(\Leftrightarrow8x^3-12x^2y+3xy^2+y^3=0\)
\(\Leftrightarrow\left(x-y\right)\left(8x^2-4xy-y^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\8x^2-4xy-y^2=0\end{matrix}\right.\)
TH1: \(x=y\)
\(\left(1\right)\Leftrightarrow5y^2=2\)
\(\Leftrightarrow y=\dfrac{\pm\sqrt{10}}{5}\)
\(\Rightarrow\left(x;y\right)\in\left\{\left(\dfrac{\sqrt{10}}{5};\dfrac{\sqrt{10}}{5}\right);\left(-\dfrac{\sqrt{10}}{5};-\dfrac{\sqrt{10}}{5}\right)\right\}\)
TH2: \(8x^2-4xy-y^2=0\left(2\right)\)
Cộng vế theo vế \(\left(1\right);\left(2\right)\) ta được:
\(8x^2=2\)
\(\Leftrightarrow x=\pm\dfrac{1}{2}\)
Nếu \(x=\dfrac{1}{2}\Rightarrow y=-1\pm\sqrt{3}\)
Nếu \(x=-\dfrac{1}{2}\Rightarrow y=1\pm\sqrt{3}\)
Thử lại rồi kết luận nghiệm
7.
ĐK: \(x^2-3y\ge0\)
\(\left\{{}\begin{matrix}\sqrt{x^2-3y}\left(2x+y+1\right)+2x+y-5=0\\5x^2+y^2+4xy-3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+y+1\right)\left(\sqrt{x^2-3y}+1\right)=6\\\left(2x+y\right)^2+\left(x^2-3y\right)=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+1\right)\left(b+1\right)=6\\a^2+b^2=5\end{matrix}\right.\) \(\left(a=2x+y;b=\sqrt{x^2-3y};b\ge0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}2ab+2\left(a+b\right)=10\\\left(a+b\right)^2-2ab=5\end{matrix}\right.\)
Cộng vế theo vế hai phương trình ta được:
\(\left(a+b\right)^2+2\left(a+b\right)-15=0\)
\(\Leftrightarrow\left(a+b-3\right)\left(a+b+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=3\\a+b=-5\end{matrix}\right.\)
TH1: \(a+b=3\Rightarrow ab=2\)
\(\Leftrightarrow...\)
TH2: \(a+b=-5\Rightarrow ab=10\)
\(\Leftrightarrow...\)
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