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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
b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)
\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)
\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)
\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)
\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)
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 \(y=-2\Rightarrow x^2-2x-7=0\Rightarrow x=1\pm2\sqrt{2}\)
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a/ \(\left\{{}\begin{matrix}\left(2x+y\right)^2-5\left(4x^2-y^2\right)+6\left(2x-y\right)^2=0\\2x+y+\dfrac{1}{2x-y}=3\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}2x+y=a\\2x-y=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2-5ab+6b^2=0\left(1\right)\\a+\dfrac{1}{b}=3\left(2\right)\end{matrix}\right.\)
\(\Rightarrow\left(1\right)\Leftrightarrow\left(2b-a\right)\left(3b-a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\a=3b\end{matrix}\right.\)
Thế vô (2) làm tiếp sẽ ra
b/ \(\left\{{}\begin{matrix}2x^3+y\left(x+1\right)=4x^2\left(1\right)\\5x^4-4x^6=y^2\left(2\right)\end{matrix}\right.\)
\(\Rightarrow\left(1\right)\Leftrightarrow2x^3+y=4x^2-xy\)
\(\Leftrightarrow4x^6+4x^3y+y^2=16x^4-8x^3y+x^2y^2\)
\(\Leftrightarrow4x^6+4x^3y+5x^4-4x^6=16x^4-8x^3y+x^2y^2\)
\(\Leftrightarrow11x^4-12x^3y+x^2y^2=0\)
\(\Leftrightarrow x^2\left(11x^2-12xy+y^2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\11x^2-12xy+y^2=0\end{matrix}\right.\)
Tới đây thì đơn giản rồi làm nốt nhé.
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 \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)