\(\left\{{}\begin{matrix}\sqrt{3y-1}+x^2=3y+1+\sqrt{x}\\2\sqrt{x^2+3x-y}-\sqrt{4x+y^2}=x+1\end{matrix}\right.\)
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. \(\left\{{}\begin{matrix}x^3-12z^2+48z-64=0\\y^3-12x^2+48x-64=0\\z^3-12y^2+48y-64=0\end{matrix}\right.\)
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,\(\left\{{}\begin{matrix}1+4x^2y^2+5xy=10x^2\\1-2xy+3y=2x\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}\sqrt{x+1}\sqrt{y+1}=\sqrt{4-x+5y}\\x^2+y+2=\sqrt{5\left(2x-y+1\right)}+\sqrt{3x+2}\end{matrix}\right.\)
giải HPT
a) \(\left\{{}\begin{matrix}\left(x+3\right)\left(y-5\right)=xy\\\left(2x-y\right)\left(y+15\right)=2xy\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\sqrt{4x}-3y+4z^2=-2\\\sqrt{3x}+2y-3z^2=1\\-3\sqrt{x}+y+2z^2=4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^3y\left(1+y\right)+x^2y^2\left(2+y\right)+xy^3=30\\x^2y+x\left(1+y+y^2\right)+y=11\end{matrix}\right.\)
Ta có hpt \(\left\{{}\begin{matrix}xy+3y-5x-15=xy\\2xy+30x-y^2-15y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}5x=3y-15\\6\left(3y-15\right)-y^2-15y=0\end{matrix}\right.\)
Ta có pt (2) \(\Leftrightarrow3y-y^2-80=0\Leftrightarrow y^2-3y+80=0\left(VN\right)\)
=> hpy vô nghiệm
c) Ta có hpt \(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left(xy+x+y\right)=30\\xy\left(x+y\right)+xy+x+y=11\end{matrix}\right.\)
Đặt j\(xy\left(x+y\right)=a;xy+x+y=b\), ta có hpt
\(\left\{{}\begin{matrix}ab=30\\a+b=11\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=5;b=6\\a=6;b=5\end{matrix}\right.\)
với a=5;b=6, ta có \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}xy=1;x+y=5\\xy=5;x+y=1\end{matrix}\right.\)
đến đây thì thế y hoặc x ra pt bậc 2, còn TH còn lại bn tự giải nhé !
b) Ta có hpt <=> \(\left\{{}\begin{matrix}2\sqrt{x}-3y+2=-4z^2\\2\sqrt{3x}+4y-2=6z^2\\-3\sqrt{x}+y-4=-2z^2\end{matrix}\right.\)
cộng 3 vế của 3 pt, ta có \(\left(2\sqrt{3}-1\right)\sqrt{x}=4\Leftrightarrow\sqrt{x}=\dfrac{4}{2\sqrt{3}-1}\Leftrightarrow x=\dfrac{16}{\left(2\sqrt{3}-1\right)^2}\)
đến đây thay căn(x)=...vào và đặt z^2=m, ta sẽ ra 1 hệ mới chỉ có 2 ẩn y và m bậc 1 , lát thế vào sẽ ra bậc 2 thì dễ rồi !
Giải giúp mình vài hệ pt này nha
thanks nhiều
1.\(\left\{{}\begin{matrix}\sqrt{x^2+x+y+1}+x+\sqrt{y^2+x+y+1}+y=15\\\sqrt{x^2+x+y+1}-x+\sqrt{y^2+x+y+1}-y=2\end{matrix}\right.\)
2.\(\left\{{}\begin{matrix}\left(1-\dfrac{12}{y+3x}\right)\sqrt{x}=2\\\left(1+\dfrac{12}{y+3x}\right)\sqrt{y}=6\end{matrix}\right.\)
3.\(\left\{{}\begin{matrix}x^3+y^3=8\\x+y+2xy=2\end{matrix}\right.\)
4.\(\left\{{}\begin{matrix}x^3+1=2y\\y^3+1=2x\end{matrix}\right.\)
5.\(\left\{{}\begin{matrix}x^3-3x=y^3-3y\\x^6+y^6=1\end{matrix}\right.\)
6.\(\left\{{}\begin{matrix}x^2-2xy+3y^2=9\\2x^2-13xy+15y^2=0\end{matrix}\right.\)
1. Đề này là 18 chứ không phải 15 nhé
\(\left\{{}\begin{matrix}\sqrt{x^2+x+y+1}+x+\sqrt{y^2+x+y+1}+y=18\left(1\right)\\\sqrt{x^2+x+y+1}-x+\sqrt{y^2+x+y+1}-y=2\left(2\right)\end{matrix}\right.\)
Lấy (1) + (2) và (1) - (2) ta được hệ mới
\(\left\{{}\begin{matrix}\sqrt{x^2+x+y+1}+\sqrt{y^2+x+y+1}=10\\x+y=8\end{matrix}\right.\)
\(\Rightarrow x=8-y\)
\(\Rightarrow\sqrt{x^2+9}+\sqrt{y^2+9}=10\)\(\Leftrightarrow\sqrt{x^2+9}=10-\sqrt{y^2+9}\)
\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\x^2+9=100-20\sqrt{y^2+9}+y^2+9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\x^2=100-20\sqrt{y^2+9}+y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\\left(8-y\right)^2=100-20\sqrt{y^2+9}+y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\9y^2-72y+144=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\)
2. Dễ thấy x = y = 0 không phải là nghiệm của phương trình
HPT\(\Leftrightarrow\left\{{}\begin{matrix}1-\dfrac{12}{y+3x}=\dfrac{2}{\sqrt{x}}\left(1\right)\\1+\dfrac{12}{y+3x}=\dfrac{6}{\sqrt{y}}\left(2\right)\end{matrix}\right.\)
Lấy (1) + (2) ; (1) - (2) ta được
\(\left\{{}\begin{matrix}1=\dfrac{1}{\sqrt{x}}+\dfrac{3}{\sqrt{y}}\left(3\right)\\\dfrac{12}{y+3x}=\dfrac{3}{\sqrt{y}}-\dfrac{1}{\sqrt{x}}\left(4\right)\end{matrix}\right.\)
Lấy ( 3) nhân (4)
\(\dfrac{12}{y+3x}=\dfrac{9}{y}-\dfrac{1}{x}=\dfrac{9x-y}{xy}\)
\(\Leftrightarrow27x^2-6xy-y^2=0\Leftrightarrow\left(9x+y\right)\left(3x-y\right)=0\)
\(\Rightarrow y=3x\)
đến đây thì dễ rồi
3. Đây là hệ đối xứng loại I
\(\left\{{}\begin{matrix}\left(x+y\right)^3-3xy\left(x+y\right)=8\\\left(x+y\right)+2xy=2\end{matrix}\right.\)
Đặt S = a + b ; P = ab (\(S^2\ge4P\) )
xong giải ra thôi mà
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 ạ
1. \(\left\{{}\begin{matrix}x\sqrt{2}-3y=1\\2x+y\sqrt{2}=-2\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}5x\sqrt{3}+y=2\sqrt{2}\\x\sqrt{6}-y\sqrt{2}=2\end{matrix}\right.\)
1: \(\left\{{}\begin{matrix}x\sqrt{2}-3y=1\\2x+y\sqrt{2}=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-3\sqrt{2}\cdot y=\sqrt{2}\\2x+y\sqrt{2}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4\sqrt{2}\cdot y=\sqrt{2}+2\\2x+y\sqrt{2}=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{2+\sqrt{2}}{-4\sqrt{2}}=\dfrac{-\sqrt{2}-1}{4}\\2x=-2-y\sqrt{2}=-2+\sqrt{2}\cdot\dfrac{\sqrt{2}+1}{4}=\dfrac{-6+\sqrt{2}}{4}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{-\sqrt{2}-1}{4}\\x=\dfrac{-6+\sqrt{2}}{8}\end{matrix}\right.\)
2: \(\left\{{}\begin{matrix}5x\sqrt{3}+y=2\sqrt{2}\\x\sqrt{6}-y\sqrt{2}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5x\sqrt{6}+y\sqrt{2}=4\\x\sqrt{6}-y\sqrt{2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x\cdot\sqrt{6}=6\\x\sqrt{6}-y\sqrt{2}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{1}{\sqrt{6}}=\dfrac{\sqrt{6}}{6}\\y\sqrt{2}=x\sqrt{6}-2=1-2=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{\sqrt{6}}{6}\\y=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)