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Chương 3: PHƯƠNG TRÌNH, HỆ PHƯƠNG TRÌNH
Các câu hỏi tương tự
\(\left(4,5-x\right)^4+\left(5,5-x\right)^4=626\)
7 tháng 11 2019 lúc 15:26
giải pt
a) \(\left|x^2-5x-4\right|=\left|x^2-4\right|\)
b) \(\left|x-1\right|+3\left|x-3\right|=6\)
c) \(\left|\frac{x^2-6x-4}{x^2-4}\right|=1\)
d) \(\left|x-1\right|-2\left|x-2\right|=x^2-x-3\)
giải hệ phương trình
1 , left{{}begin{matrix}left(x+yright)left(x-1right)left(x-yright)left(x+1right)+2xyleft(y-xright)left(y-1right)left(y+xright)left(y-2right)-2xyend{matrix}right.
2, left{{}begin{matrix}2left(frac{1}{x}+frac{1}{2y}right)+3left(frac{1}{x}-frac{1}{2y}right)^29left(frac{1}{x}+frac{1}{2y}right)-6left(frac{1}{x}-frac{1}{2y}right)^2-3end{matrix}right.
3 , left{{}begin{matrix}frac{xy}{x+y}frac{2}{3}frac{yz}{y+z}frac{6}{5}frac{zx}{z+x}frac{3}{4}end{matrix}right.
4 , left{{}begin...
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giải hệ phương trình
1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\)
3 , \(\left\{{}\begin{matrix}\frac{xy}{x+y}=\frac{2}{3}\\\frac{yz}{y+z}=\frac{6}{5}\\\frac{zx}{z+x}=\frac{3}{4}\end{matrix}\right.\)
4 , \(\left\{{}\begin{matrix}2xy-3\frac{x}{y}=15\\xy+\frac{x}{y}=15\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}x+y+3xy=5\\x^2+y^2=1\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}x+y+xy=11\\x^2+y^2+3\left(x+y\right)=28\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)
8, \(\left\{{}\begin{matrix}x+y+xy=11\\xy\left(x+y\right)=30\end{matrix}\right.\)
9 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
1.Giải hpt : a,\(\left\{{}\begin{matrix}\left(x+y+3\right)\sqrt{x-2y}+2y+4=0\\\left(x-y\right)\left(x^2+4\right)=y^2+1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+y^2=\dfrac{1}{2}\\x^3+3xy^2=\dfrac{1}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(1+x^2\right)^2\left(1+\dfrac{1}{y^4}\right)=8\\\left(1+y^2\right)^2\left(1+\dfrac{1}{x^4}\right)=8\end{matrix}\right.\)
Em cảm ơn ạ !!!
a)left{{}begin{matrix}2left|x-6right|+3left|y-1right|55left|x-6right|-4left|y+1right|1end{matrix}right.
b) left{{}begin{matrix}2left|x+yright|-left|x-yright|93left|x+yright|+2left|x-yright|+17end{matrix}right.
c)left{{}begin{matrix}4left|x+yright|+3left|x-yright|83left|x+yright|-5left|x-yright|6end{matrix}right.
d) left{{}begin{matrix}x^2-xy242x-3y1end{matrix}right.
e) left{{}begin{matrix}3x-4y+10xy3left(x+yright)-9end{matrix}right.
f) left{{}begin{matrix}2x+3y53x^2-y^2+2y4end{matrix}right.
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a)\(\left\{{}\begin{matrix}2\left|x-6\right|+3\left|y-1\right|=5\\5\left|x-6\right|-4\left|y+1\right|=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2\left|x+y\right|-\left|x-y\right|=9\\3\left|x+y\right|+2\left|x-y\right|+17\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}4\left|x+y\right|+3\left|x-y\right|=8\\3\left|x+y\right|-5\left|x-y\right|=6\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}x^2-xy=24\\2x-3y=1\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}3x-4y+1=0\\xy=3\left(x+y\right)-9\end{matrix}\right.\)
f) \(\left\{{}\begin{matrix}2x+3y=5\\3x^2-y^2+2y=4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x-1\right).6+\left(y-2\right).\left(-2\right)=0\\\left(x+1\right).4+\left(y-1\right).\left(-3\right)=0\end{matrix}\right.\)
Giải hpt
26 tháng 10 2019 lúc 16:35
giải pt
a) x^2+4x-3left|x+2right|+40
b) left(x+2right)^2-3left|x+2right|-40
c) left(x^2-3right)^2-6left|x^2-3right|+50
d) frac{x^2-4x+4}{x^2-2x+1}+frac{left|2x-4right|}{x-1}3
e) left|frac{2x-1}{x+2}right|-2left|frac{x+2}{2x-1}right|1
f) x^2+frac{1}{x^2}-102left|x-frac{1}{x}right|
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giải pt
a) \(x^2+4x-3\left|x+2\right|+4=0\)
b) \(\left(x+2\right)^2-3\left|x+2\right|-4=0\)
c) \(\left(x^2-3\right)^2-6\left|x^2-3\right|+5=0\)
d) \(\frac{x^2-4x+4}{x^2-2x+1}+\frac{\left|2x-4\right|}{x-1}=3\)
e) \(\left|\frac{2x-1}{x+2}\right|-2\left|\frac{x+2}{2x-1}\right|=1\)
f) \(x^2+\frac{1}{x^2}-10=2\left|x-\frac{1}{x}\right|\)
Giải hệ phương trình
left{{}begin{matrix}x^2+y^2+xy13x^4+y^4+x^2y^291end{matrix}right.
Leftrightarrowleft{{}begin{matrix}left(x+yright)^2-xy13left(x^2+y^2right)^2-left(xyright)^291end{matrix}right.
Leftrightarrowleft{{}begin{matrix}left(x+yright)^213+xyleft[left(x+yright)^2-2xyright]^2-left(xyright)^291end{matrix}right.
Leftrightarrowleft{{}begin{matrix}left(x+yright)^2-xy13left(13-xyright)^2-left(xyright)^291end{matrix}right.
Leftrightarrowleft{{}begin{matrix}xy3left(x+yright)^216end{matri...
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Giải hệ phương trình
\(\left\{{}\begin{matrix}x^2+y^2+xy=13\\x^4+y^4+x^2y^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=13\\\left(x^2+y^2\right)^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=13+xy\\\left[\left(x+y\right)^2-2xy\right]^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=13\\\left(13-xy\right)^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy=3\\\left(x+y\right)^2=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\) hoặc x+y = -4
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=-4\\xy=3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)
Mọi người có thể giải thích từ dấu tương đương thứ 3 xuống 4. tại sao lại như vậy k?