Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1,\(x^2-2y^2-xy=0\)
<=> \(\left(x-2y\right)\left(x+y\right)=0\)
<=> \(\orbr{\begin{cases}x=2y\\x=-y\end{cases}}\)
Sau đó bạn thế vào PT dưới rồi tính
3. ĐKXĐ \(x\le1\); \(x+2y+3\ge0\)
.\(2y^3-\left(x+4\right)y^2+8y+x^2-4x=0\)
<=> \(\left(2y^3-xy^2\right)+\left(x^2-4y^2\right)-\left(4x-8y\right)=0\)
<=> \(\left(x-2y\right)\left(-y^2+x+2y-4\right)=0\)
Mà \(-y^2+2y-4=-\left(y-1\right)^2-3\le-3\); \(x\le1\)nên \(-y^2+x+2y-4< 0\)
=> \(x=2y\)
Thế vào Pt còn lại ta được
\(\sqrt{\frac{1-x}{2}}+\sqrt{2x+3}=\sqrt{5}\)ĐK \(-\frac{3}{2}\le x\le1\)
<=> \(\frac{1-x}{2}+2x+3+2\sqrt{\frac{\left(1-x\right)\left(2x+3\right)}{2}}=5\)
<=> \(\sqrt{2\left(1-x\right)\left(2x+3\right)}=-\frac{3}{2}x+\frac{3}{2}\)
<=> \(\sqrt{2\left(1-x\right)\left(2x+3\right)}=-\frac{3}{2}\left(x-1\right)\)
<=> \(\orbr{\begin{cases}x=1\\\sqrt{2\left(2x+3\right)}=\frac{3}{2}\sqrt{1-x}\end{cases}}\)=> \(\orbr{\begin{cases}x=1\\x=-\frac{3}{5}\end{cases}}\)(TMĐK )
Vậy \(\left(x;y\right)=\left(1;\frac{1}{2}\right),\left(-\frac{3}{5};-\frac{3}{10}\right)\)
ôi trờiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
a,\(\hept{\begin{cases}x^2+y^2+\frac{2xy}{x+y}=1\\\sqrt{x+y}=x^2-y\end{cases}}\)
ĐK: \(x+y\ge0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+y\right)^2-2xy+\frac{2xy}{x+y}=1\left(1\right)\\\sqrt{x+y}=x^2-y\left(2\right)\end{cases}}\)
Đặt \(\hept{\begin{cases}x+y=a\\2xy=b\end{cases}\left(a\ge0\right)}\)
\(\left(1\right)\Leftrightarrow a^2-b+\frac{b}{a}=1\)
\(\Leftrightarrow a^3-ab-a+b=0\)
\(\Leftrightarrow\left(a-1\right)\left(a^2+a-b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=1\\a^2+a-b=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x+y=1\left(3\right)\\\left(x+y\right)^2+\left(x+y\right)-xy=0\left(4\right)\end{cases}}\)
Thay (3) vào (2) ta được
\(x^2-y=1\Leftrightarrow y=x^2-1\)
\(\Rightarrow1-x=x^2-1\Leftrightarrow x^2+x-2=0\Leftrightarrow\orbr{\begin{cases}x=1\Rightarrow y=0\\x=-2\Rightarrow y=3\end{cases}}\)
Giải (4)
Ta có \(\left(x+y\right)^2\ge4xy\Rightarrow\left(x+y\right)^2-xy>0\)
do đó (4) không xảy ra
Vậy..........
a: \(\Leftrightarrow\left\{{}\begin{matrix}35x-28y=21\\35x-45y=40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}17y=-19\\5x-4y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{19}{17}\\x=-\dfrac{5}{17}\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{8}{y}=18\\\dfrac{10}{x}+\dfrac{8}{y}=102\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{x}=120\\\dfrac{1}{x}-\dfrac{8}{y}=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{120}\\y=-\dfrac{44}{39}\end{matrix}\right.\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{30}{x-1}+\dfrac{3}{y+2}=3\\\dfrac{25}{x-1}+\dfrac{3}{y+2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}=1\\\dfrac{10}{y-1}+\dfrac{1}{y+2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=5\\\dfrac{1}{y+2}+2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=-3\end{matrix}\right.\)
d: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{135}{2x-y}+\dfrac{160}{x+3y}=35\\\dfrac{135}{2x-y}-\dfrac{144}{x+3y}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3y=8\\2x-y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+6y=16\\2x-y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=5\end{matrix}\right.\)
câu 1
\(\hept{\begin{cases}2x^2+3xy-2y^2-5\left(2x-y\right)=0\left(1\right)\\x^2-2xy-3y^2+15=0\left(2\right)\end{cases}}\)
pt (1) \(\Leftrightarrow\left(2x-y\right)\left(x+2y\right)-5\left(2x-y\right)=0\)
\(\Leftrightarrow\left(2x-y\right)\left(x+2y-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y=2x\\x=5-2y\end{cases}}\)
hpt \(\Leftrightarrow\hept{\begin{cases}y=2x\\x^2-2x.2x-3\left(2x\right)^2+15=0\end{cases}\left(3\right)}\) hoặc \(\hept{\begin{cases}x=5-2y\\\left(5-2y\right)^2-2\left(5-2y\right)y-3y^2+15=0\end{cases}\left(4\right)}\)
hpt (3) \(\Leftrightarrow\hept{\begin{cases}y=2x\\-15x^2+15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1;y=2\\x=-1;y=-2\end{cases}}}\)
hpt(4) \(\Leftrightarrow\hept{\begin{cases}x=5-2y\\5y^2-30y+40=0\end{cases}\Leftrightarrow\orbr{\begin{cases}y=2;x=1\\y=4;x=-3\end{cases}}}\)
Vậy hpt đã cho có nghiệm (1;2),(-1;-2),(-3;4)
\(\hept{\begin{cases}x+y+\frac{1}{x}+\frac{1}{y}=5\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=9\end{cases}}\left(ĐK:x\ne0;y\ne0\right)\).
\(\Leftrightarrow\hept{\begin{cases}\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)=5\\\left(x^2+\frac{1}{x^2}\right)+\left(y^2+\frac{1}{y^2}\right)=9\end{cases}}\).
Đặt \(x+\frac{1}{x}=a,y+\frac{1}{y}=b\)thì \(x^2+\frac{1}{x^2}=a^2-2;y^2+\frac{1}{y^2}=b^2-2\). Hệ phương trình trở thành:
\(\hept{\begin{cases}a+b=5\\a^2-2+b^2-2=9\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b=5\\a^2+b^2=13\end{cases}}\).
Đặt \(a+b=C,ab=D\)thì \(\left(a+b\right)^2=C^2\Leftrightarrow a^2+b^2+2D=C^2\Leftrightarrow a^2+b^2=C^2-2D\). Hệ phương trình trở thành:
\(\hept{\begin{cases}C=5\\C^2-2D=13\end{cases}}\Leftrightarrow\hept{\begin{cases}C=5\\25-2D=13\end{cases}}\Leftrightarrow\hept{\begin{cases}C=5\\D=6\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a+b=5\\ab=6\end{cases}}\Leftrightarrow\hept{\begin{cases}a=5-b\\ab=6\end{cases}}\Leftrightarrow\hept{\begin{cases}a=5-b\\\left(5-b\right)b=6\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=5-b\\-b^2+5b-6=0\end{cases}}\).
Xét phương trình \(\left(b-3\right)\left(b-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}b=3\\b=2\end{cases}}\).
Với \(b=3\)thì \(a=2\)
Với \(b=2\)thì \(a=3\)
- Với \(b=3\)và \(a=2\):
+ Với \(b=3\)thì \(y+\frac{1}{y}=3\Leftrightarrow\frac{y^2+1}{y}=\frac{3y}{y}\)
\(\Rightarrow y^2+1=3y\)
\(\Leftrightarrow y^2-3y+1=0\)
\(\Leftrightarrow\left(y-\frac{3}{2}\right)^2=\frac{5}{4}\).
\(\Leftrightarrow\orbr{\begin{cases}y-\frac{3}{2}=\frac{\sqrt{5}}{2}\\y-\frac{3}{2}=\frac{-\sqrt{5}}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=\frac{3+\sqrt{5}}{2}\\y=\frac{3-\sqrt{5}}{2}\end{cases}}\)(thỏa mãn ĐKXĐ).
+ Với \(a=2\)thì \(x+\frac{1}{x}=2\Leftrightarrow\frac{x^2+1}{x}=\frac{2x}{x}\).
\(\Rightarrow x^2+1=2x\Leftrightarrow x^2-2x+1=0\).
\(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)(thỏa mãn ĐKXĐ).
- Với \(b=2\)và \(a=3\):
+ Với \(b=2\)thì \(y+\frac{1}{y}=2\Leftrightarrow\frac{y^2+1}{y}=\frac{2y}{y}\).
\(\Rightarrow y^2+1=2y\)
\(\Leftrightarrow y^2-2y+1=0\Leftrightarrow\left(y-1\right)^2=0\)
\(\Leftrightarrow y-1=0\Leftrightarrow y=1\)(thỏa mãn ĐKXĐ).
+ Với \(a=3\)thì \(x+\frac{1}{x}=3\Leftrightarrow\frac{x^2+1}{x}=\frac{3x}{x}\).
\(\Rightarrow x^2+1=3x\).
\(\Leftrightarrow x^2-3x+1=0\)..
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3+\sqrt{5}}{2}\\x=\frac{3-\sqrt{5}}{2}\end{cases}}\)(thỏa mãn ĐKXĐ).
Vậy hệ phương trình có tậpnghiệm \(\left(x;y\right)=\left\{1;\frac{3\pm\sqrt{5}}{2}\right\};\left\{\frac{3\pm\sqrt{5}}{2};1\right\}\).