cho hệ phương trình:
\(\left\{{}\begin{matrix}x^2+y^2=5\\x^4-x^2y^2+y^4=13\end{matrix}\right.\)
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Giải các hệ phương trình saua,left{{}begin{matrix}sqrt{3}x-ysqrt{2}x-sqrt{2}ysqrt{3}end{matrix}right. b, left{{}begin{matrix}5left(x-yright)-3left(2x+3yright)123left(x+2yright)-4left(x+2yright)5end{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.
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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)
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\(\left\{{}\begin{matrix}5\left(x-y\right)-3\left(2x+3y\right)=12\\3\left(x+2y\right)-4\left(x+2y\right)=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}5x-5y-6x-9y=12\\3x+6y-4x-8y=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}-x-14y=12\\-x-2y=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=-\dfrac{26}{3}\\y=-\dfrac{7}{12}\end{matrix}\right.\)
Vậy HPT có nghiệm (x;y) = (\(-\dfrac{26}{3};-\dfrac{7}{12}\))
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Giải hệ phương trình sau bằng phương pháp thế
a)
\(\left\{{}\begin{matrix}\sqrt{5}+2)x+y=3-\sqrt{5}\\-x+2y=6-2\sqrt{5}\end{matrix}\right.\)
b)
\(\left\{{}\begin{matrix}5\left(x+2y\right)=3x-1\\2x+4=3\left(x-5y\right)-12\end{matrix}\right.\)
giải hệ pt bằng phương pháp thế:1) left{{}begin{matrix}x+y3x+2y5end{matrix}right.2) left{{}begin{matrix}x-y3y2x+1end{matrix}right.3) left{{}begin{matrix}2x+3y4y-x-2end{matrix}right.4) left{{}begin{matrix}xy+2x3y+8end{matrix}right.5) left{{}begin{matrix}2x-y13x-4y2end{matrix}right.giúp mk vs ạ mai mk hc rồi
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giải hệ pt bằng phương pháp thế:
1) \(\left\{{}\begin{matrix}x+y=3\\x+2y=5\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}x-y=3\\y=2x+1\end{matrix}\right.\)
3) \(\left\{{}\begin{matrix}2x+3y=4\\y-x=-2\end{matrix}\right.\)
4) \(\left\{{}\begin{matrix}x=y+2\\x=3y+8\end{matrix}\right.\)
5) \(\left\{{}\begin{matrix}2x-y=1\\3x-4y=2\end{matrix}\right.\)
giúp mk vs ạ mai mk hc rồi
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\3-y+2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}x-2x-1=3\\y=2x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=2\left(-2\right)+1=-3\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}2x+3x-6=4\\y=x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\\ 4,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y+2=3y+8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\\ 5,\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1+y}{2}\\\dfrac{3+3y}{2}-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1+y}{2}\\3+3y-8y=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{y+1}{2}\\y=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-\dfrac{1}{5}\end{matrix}\right.\)
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Giải hệ phương trình sau bằng phương pháp thế
1) \(\left\{{}\begin{matrix}x-2y=4\\-2x+5y=-3\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}2x+y=10\\5x-3y=3\end{matrix}\right.\)
3) \(\left\{{}\begin{matrix}x+2y=4\\-3x+y=7\end{matrix}\right.\)
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)
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Giải hệ phương trình sau bằng cách cộng hệ số
1) \(\left\{{}\begin{matrix}x-y=5\\2x+y=11\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}3x+2y=1\\3x+y=2\end{matrix}\right.\)
3) \(\left\{{}\begin{matrix}x-y=2\\3x+2y=11\end{matrix}\right.\)
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
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Giải các hệ phương trình sau : a, left{{}begin{matrix}x^2+xyy^2+13x+yy^2+3end{matrix}right.b,left{{}begin{matrix}x^2-y^24x-2y-3x^2+y^25end{matrix}right.c, left{{}begin{matrix}x^2+x-xy-2y^2-2y0x^2+y^21end{matrix}right.d,left{{}begin{matrix}2left(y+zright)yzxy+yz+zx108xyz180end{matrix}right.
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Giải các hệ phương trình sau :
a, \(\left\{{}\begin{matrix}x^2+xy=y^2+1\\3x+y=y^2+3\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}x^2-y^2=4x-2y-3\\x^2+y^2=5\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}x^2+x-xy-2y^2-2y=0\\x^2+y^2=1\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}2\left(y+z\right)=yz\\xy+yz+zx=108\\xyz=180\end{matrix}\right.\)
Giải hệ phương trình:
\(\left\{{}\begin{matrix}x^5=x^4-2x^2y+2\\y^5=y^4-2y^2z+2\\z^5=z^4-2z^2x+2\end{matrix}\right.\)
Giải phương trình:
1. \(\left\{{}\begin{matrix}5x-2y=-9\\4x+3y=2\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}2x+y-4=0\\x+2y-5=0\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}2x+3y-7=0\\x+2y-4=0\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}5x+6y=17\\9x-y=7\end{matrix}\right.\)
1)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}15x-6y=-27\\8x+6y=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2y=5x+9\\23x=-23\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(-1;2\right)\)
2)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x+y=4\\2x+4y=10\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-3y=-6\\x=5-2y\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(1;2\right)\)
3)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=14\\3x+6y=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=4-x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(2;1\right)\)
4)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}5x+6y=17\\54x-6y=42\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}59x=59\\y=9x-7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(1;2\right)\)
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3 tháng 1 lúc 22:03
giải hệ phương trìnha) left{{}begin{matrix}x+2y2-2x+y1end{matrix}right.b) left{{}begin{matrix}3x-2y42x+y5end{matrix}right.c) left{{}begin{matrix}2y-x22x-y-1end{matrix}right.giúp tui giải bài này với tui c.ơn trước
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giải hệ phương trình
a) \(\left\{{}\begin{matrix}x+2y=2\\-2x+y=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}3x-2y=4\\2x+y=5\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}2y-x=2\\2x-y=-1\end{matrix}\right.\)
giúp tui giải bài này với
tui c.ơn trước
b)\(\left\{{}\begin{matrix}3x-2y=4\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2\left(5-2x\right)=4\\y=5-2x\end{matrix}\right.\)\(\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3x-10+4x=4\\y=5-2x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7x=14\\y=5-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy nghiệm duy nhất của hpt là: (2;1)
c) \(\left\{{}\begin{matrix}2y-x=2\\2x-y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-2\\2\left(2y-2\right)-y=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-2\\4y-4-y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-2\\3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)
Vậy nghiệm duy nhất của hpt là: (0;1)
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a) \(\left\{{}\begin{matrix}x+2y=2\left(1\right)\\-2x+y=1\left(2\right)\end{matrix}\right.\)
Từ (1): \(x=2-2y\) (3)
Thế (3) vào (2), ta được: \(-2\left(2-2y\right)+y=1< =>-4+4y+y=1\)
\(\Leftrightarrow y=1\)\(\Rightarrow\)\(x=2-2.1=0\)
Vậy nghiệm duy nhất của hpt là: (0;1)
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