Giải hệ phương trình
\(\left\{{}\begin{matrix}2y^2-4xy+3y-4x-1=3\sqrt{\left(y^2-1\right)\left(y-2x\right)}\\\sqrt{y+1}+\sqrt{y-2x}=\sqrt{2\left(y-x+1\right)}\end{matrix}\right.\)
Giải các hệ phương trình sau:
a) \(\left\{{}\begin{matrix}4x^2-4xy-14x-3y^2+y+10=0\\5\sqrt{xy}+2x+2y=6\sqrt{y}-8\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2x^4+3x^2y+4x^2-2y^2+3y+2=0\\\sqrt{x\left(y-1\right)}+2y+2\sqrt{y-1}=3x+2\sqrt{x}+2\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}x^6+3x^2-y^3-6y^2-15y-14=0\\\sqrt{xy+2x-y-2}+6x-2y=10\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}xy+x+y=x^2-2y^2\\x\sqrt{2y}-y\sqrt{x-1}=2x-2y\end{matrix}\right.\)
Giải hệ
a) \(\left\{{}\begin{matrix}x^2\left(y^2+1\right)+2y\left(x^2+x+1\right)=3\\\left(x^2+x\right)\left(y^2+y\right)=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\left(6x+5\right)\sqrt{2x+1}-2y-3y^3=0\\y+\sqrt{x}=\sqrt{2x^2+4x-23}\end{matrix}\right.\)
Giải bất pt
\(\dfrac{9}{\left|x-5\right|-3}\ge\left|x-2\right|\)
Giải hệ phương trình:
a) \(\left\{{}\begin{matrix}\sqrt{3y^2+13}-\sqrt{15-2x}=\sqrt{x+1}\\y^4-2x^2y+7y^2=\left(x+1\right)\left(8-x\right)\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\sqrt{x+y}-\sqrt{x-y}=2\\\sqrt{x^2+y^2+1}-\sqrt{x^2-y^2}=3\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\sqrt{2x+y+1}-\sqrt{x+y}=3\\\sqrt{3\left(x+y\right)^2+1}+\sqrt{x-5}=5\end{matrix}\right.\)
Giải hệ phương trình
\(\left\{{}\begin{matrix}2x^3-4x^2+3x-1=2x^3\left(2-y\right)\sqrt{3-2y}\\\left(\sqrt{x\sqrt{3-2y}-\sqrt{x}}\right)^2\left(\sqrt{x\sqrt{3-2y}+2}+\sqrt{x+1}\right)=4\end{matrix}\right.\)
bài này mình chưa giải dc triệt để ở cái cuối
\(2x^3-4x^2+3x-1=2x^3\left(2-y\right)\sqrt{3-2y}\) \(\left(y\le\dfrac{3}{2}\right)\)
\(\Leftrightarrow4x^3-8x^2+6x-2=2x^3\left(4-2y\right)\sqrt{3-2y}\left(1\right)\)
\(đặt:\sqrt{3-2y}=a\ge0\Rightarrow a^2+1=4-2y\)
\(\left(1\right)\Leftrightarrow4x^3-8x^2+6x-2=2x^3.\left(a^2+1\right)a\)
\(\Leftrightarrow4x^3-8x^2+6x-2-2x^3\left(a^2+1\right)a\)
\(\Leftrightarrow-2\left(xa-x+1\right)\left[\left(xa\right)^2+x^2a+2x^2-xa-2x+1\right]=0\)
\(\Leftrightarrow x.a-x+1=0\Leftrightarrow x\left(a-1\right)=-1\Leftrightarrow x=\dfrac{-1}{a-1}\)
\(\left(\sqrt{x\sqrt{3-2y}-\sqrt{x}}\right) ^2=x\sqrt{3-2y}-\sqrt{x}\)
\(=\dfrac{-a}{a-1}-\sqrt{\dfrac{-1}{a-1}}\)
\(\left(\sqrt{x\sqrt{3-2y}+2}+\sqrt{x+1}\right)=\sqrt{\dfrac{-a}{a-1}+2}+\sqrt{\dfrac{a-2}{a-1}}\)
\(\Rightarrow\left(\dfrac{-a}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right)\left(\sqrt{\dfrac{-a}{a-1}+2}+\sqrt{\dfrac{a-2}{a-1}}\right)-4=0\)
\(\Leftrightarrow\left(-\dfrac{a}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right).2\sqrt{\dfrac{a-2}{a-1}}=4\)
\(\Leftrightarrow\left(-\dfrac{a}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right)\sqrt{\dfrac{a-2}{a-1}}=2\)
\(\Leftrightarrow\left(-1+\dfrac{-1}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right)\sqrt{1-\dfrac{1}{a-1}}=2\)(3)
\(đặt:1-\dfrac{1}{a-1}=u\Rightarrow\sqrt{-\dfrac{1}{a-1}}=\sqrt{u-1}\)
\(\left(3\right)\Leftrightarrow\left(u-2-\sqrt{u-1}\right)\sqrt{u}=2\)
bình phương lên tính được u
\(\Rightarrow u=.....\Rightarrow a\Rightarrow y=...\Rightarrow x=....\)
Với \(x=0\) không phải nghiệm
Với \(x>0\) chia 2 vế cho pt đầu cho \(x^3\)
\(\Rightarrow2-\dfrac{4}{x}+\dfrac{3}{x^2}-\dfrac{1}{x^3}=2\left(2-y\right)\sqrt{3-2y}\)
\(\Leftrightarrow1-\dfrac{1}{x}+\left(1-\dfrac{1}{x}\right)^3=\sqrt{3-2y}+\sqrt{\left(3-2y\right)^3}\)
Xét hàm \(f\left(t\right)=t+t^3\Rightarrow f'\left(t\right)=1+3t^2>0\Rightarrow f\left(t\right)\) đồng biến
\(\Rightarrow1-\dfrac{1}{x}=\sqrt{3-2y}\)
Thế vào pt dưới:
\(\left(\sqrt{x\left(1-\dfrac{1}{x}\right)-\sqrt{x}}\right)^2\left(\sqrt{x\left(1-\dfrac{1}{x}\right)+2}+\sqrt{x+1}\right)=4\)
\(\Leftrightarrow\left(x-\sqrt{x}-1\right)\left(\sqrt{x+1}+\sqrt{x+1}\right)=4\)
\(\Leftrightarrow\left(x-\sqrt{x}-1\right)\sqrt{x+1}=2\)
Phương trình này ko có nghiệm đẹp, chắc bạn ghi nhầm đề bài của pt dưới
... giải ra \(1-\dfrac{1}{x}=\sqrt{3-2y}\)
Thế xuống pt dưới:
\(\left(\sqrt{x\left(1-\dfrac{1}{x}\right)+2}\right)^2\left(\sqrt{x\left(1-\dfrac{1}{x}\right)+2}+\sqrt{x-1}\right)^4=4\)
\(\Leftrightarrow\left(x+1\right)\left(\sqrt{x+1}+\sqrt{x-1}\right)^4=4\)
Có vẻ đề bài vẫn sai
Do \(x\ge1\) theo ĐKXĐ nên \(x+1\ge2\) ; \(\left(\sqrt{x+1}+\sqrt{x-1}\right)^4\ge\left(\sqrt{2}+0\right)^4=4\)
\(\Rightarrow\left(x+1\right)\left(\sqrt{x+1}+\sqrt{x-1}\right)^4\ge8>4\) nên pt vô nghiệm
giải hệ phương trình:
1, \(\left\{{}\begin{matrix}\sqrt{3+2x^2y-x^4y^2}+x^2\left(1-2x^2\right)=y^4\\1+\sqrt{1+\left(x-y\right)^2}=-x^2\left(x^4+1-2x^2-2xy^2\right)\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\sqrt{x-1}+\sqrt{x}\left(3\sqrt{x}-y\right)+x\sqrt{x}=3y+\sqrt{y-1}\\3xy^2+4=4x^2+2y+x\end{matrix}\right.\)
Giải các hệ phương trình sau
a,\(\left\{{}\begin{matrix}\sqrt{3}x-y=\sqrt{2}\\x-\sqrt{2}y=\sqrt{3}\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}5\left(x-y\right)-3\left(2x+3y\right)=12\\3\left(x+2y\right)-4\left(x+2y\right)=5\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}\dfrac{x+2}{y-1}=\dfrac{x-4}{y+2}\\\dfrac{2x+3}{y-1}=\dfrac{4x+1}{2y+1}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{x+2}{y-1}=\dfrac{x-4}{y+2}\\\dfrac{2x+3}{y-1}=\dfrac{4x+1}{2y+1}\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\left(x+2\right)\left(y+2\right)=\left(y-1\right)\left(x-\text{4}\right)\\\left(2x+3\right)\left(2y+1\right)=\left(y-1\right)\left(4x+1\right)\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}xy+2x+2y+4=xy-4y-x+4\\4xy+2x+6y+3=4xy-4x+y-1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}3x+6y=0\\6x+5y=-4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=-\dfrac{8}{7}\\y=\dfrac{4}{7}\end{matrix}\right.\)(TM)
\(\left\{{}\begin{matrix}5\left(x-y\right)-3\left(2x+3y\right)=12\\3\left(x+2y\right)-4\left(x+2y\right)=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}5x-5y-6x-9y=12\\3x+6y-4x-8y=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}-x-14y=12\\-x-2y=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=-\dfrac{26}{3}\\y=-\dfrac{7}{12}\end{matrix}\right.\)
Vậy HPT có nghiệm (x;y) = (\(-\dfrac{26}{3};-\dfrac{7}{12}\))
Giải hệ phương trình:
1. \(\left\{{}\begin{matrix}x+3=2\sqrt{\left(3y-x\right)\left(y+1\right)}\\\sqrt{3y-2}-\sqrt{\dfrac{x+5}{2}}=xy-2y-2\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\sqrt{2y^2-7y+10-x\left(y+3\right)}+\sqrt{y+1}=x+1\\\sqrt{y+1}+\dfrac{3}{x+1}=x+2y\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}\sqrt{4x-y}-\sqrt{3y-4x}=1\\2\sqrt{3y-4x}+y\left(5x-y\right)=x\left(4x+y\right)-1\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}9\sqrt{\dfrac{41}{2}\left(x^2+\dfrac{1}{2x+y}\right)}=3+40x\\x^2+5xy+6y=4y^2+9x+9\end{matrix}\right.\)
5. \(\left\{{}\begin{matrix}\sqrt{xy+\left(x-y\right)\left(\sqrt{xy}-2\right)}+\sqrt{x}=y+\sqrt{y}\\\left(x+1\right)\left[y+\sqrt{xy}+x\left(1-x\right)\right]=4\end{matrix}\right.\)
6. \(\left\{{}\begin{matrix}x^4-x^3+3x^2-4y-1=0\\\sqrt{\dfrac{x^2+4y^2}{2}}+\sqrt{\dfrac{x^2+2xy+4y^2}{3}}=x+2y\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}x^3-12z^2+48z-64=0\\y^3-12x^2+48x-64=0\\z^3-12y^2+48y-64=0\end{matrix}\right.\)