Cm: \(\frac{3y\left(x+1\right)-6x-6}{3y-6}\)=\(\frac{2\left(y+3\right)+2xy+6x}{2y+6}\)
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Chứng minh đẳng thức
\(\frac{3y\left(x+1\right)-6x-6}{3y-6}=\frac{2\left(y+3\right)+2xy+6}{2y+6}\)
\(y\ne2;y\ne-3\)
bn chưa bít làm nhé mk chưa hok tới bài đó mà
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24 tháng 4 2020 lúc 22:15
Nhờ các ae CTV và AD giúp :
Chứng minh các đẳng thức sau :
\(\frac{3xy+3y-6x-6}{3x-6}=\frac{2y+2xy+6x+6}{2y+6}\left(x\ne2;y\ne-3\right)\)
Bài này dễ nên tạm thời mình làm trong nick phụ nha
\(ĐT\Leftrightarrow\frac{\left(3y-6\right)\left(x+1\right)}{3x-6}=\frac{\left(2y+6\right)\left(x+1\right)}{2y+6}\)(quá đúng luôn á)
Thử cho x= 4; y = 5 xem, => not true
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Giải hpt:
1, \(\left\{{}\begin{matrix}x^2+y+x^3y+x^2y+xy=\frac{-5}{4}\\x^4+y^2+xy\left(1+2x\right)=\frac{-5}{4}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^4+2x^2y+x^2y^2=-2x+9\\x^2+2xy=6x+6\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}x-\frac{1}{x}=y-\frac{1}{y}\\2y=x^3+1\end{matrix}\right.\)
3) ta xét phương trình thứ nhất
\(x-\frac{1}{x}=y-\frac{1}{y}\)
<=>\(x-y-\frac{1}{x}+\frac{1}{y}=0\)
<=>\(x-y-\left(\frac{1}{x}-\frac{1}{y}\right)=0\)
<=>\(x-y-\left(\frac{y-x}{xy}\right)=0\)
<=>\(\left(x-y\right)\left(1+\frac{1}{xy}\right)=0\)
<=>\(x=y\) hoặc xy=-1
Với x=y thay vào phương trình thứ hai ta có
\(2x=x^3+1
\)
<=> \(x^3-2x+1=0\)
<=>\(x^3-x^2+x^2-x-x+1=0\)
<=>\(\left(x-1\right)\left(x^2+x-1\right)=0\)
<=> \(x=1\) hoặc \(x^2+x-1=0\)
\(x^2+x-1=0\) <=> \(x=\frac{-1+\sqrt{5}}{2}\)
hoặc \(x=\frac{-1-\sqrt{5}}{2}\)
Đối với xy=-1 thì y=-1/x thay vào phương trình 2 giải bình thường
\(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2xy+5x-6y-15=2xy-2x+7y-7\\12xy-24x+3y-6=12xy+18x-2y-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x-6y-15=-2x+7y-7\\-24x+3y-6=18x-2y-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x-6y+2x-7y=-7+15\\-24x+3y-18x+2y=-3+6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7x-13y=8\\-42x+5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}42x-78y=48\\-42x+5y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-73y=51\\7x-13y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-51}{73}\\7x=8+13y=8+13\cdot\dfrac{-51}{73}=-\dfrac{79}{73}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-79}{511}\\y=-\dfrac{51}{73}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{-79}{511}\\y=-\dfrac{51}{73}\end{matrix}\right.\)
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\(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2xy+5x-6y-15=2xy-2x+7y-7\\12xy-24x+3y-6=12xy+18x-2y-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7x-13y=8\\-42x+5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}42x-78y=48\\-42x+5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-73y=51\\7x-13y=8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{51}{73}\\x=-\dfrac{79}{511}\end{matrix}\right.\)
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\(\left\{{}\begin{matrix}3x-6\sqrt{2x-4}=4\sqrt{3y-9}-2y\\6x^3-3x^2y+2xy+4=y^2+4x+6x^2\end{matrix}\right.\)
giải hệ:
a) left{{}begin{matrix}sqrt{x+3y}+sqrt{x+y}2sqrt{x+y}+y-x1end{matrix}right.
b) left{{}begin{matrix}x+y+frac{1}{x}+frac{1}{y}4x^2+y^2+frac{1}{x^2}+frac{1}{y^2}4end{matrix}right.
c) left{{}begin{matrix}left(x-frac{1}{y}right)left(y+frac{1}{x}right)22x^2y+xy^2-4xy2x-yend{matrix}right.
d) left{{}begin{matrix}2x^2+xyy^2-3y+2x^2-y^23end{matrix}right.
e) left{{}begin{matrix}x^2+y^2+z^2+2xy-xz-zy3x^2+y^2-2xy-xz+zy-1end{matrix}right.
f) left{{}begin{matrix}x^2-y^2+5x-y+60x^2+left(x-yright)...
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giải hệ:
a) \(\left\{{}\begin{matrix}\sqrt{x+3y}+\sqrt{x+y}=2\\\sqrt{x+y}+y-x=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\left(x-\frac{1}{y}\right)\left(y+\frac{1}{x}\right)=2\\2x^2y+xy^2-4xy=2x-y\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}2x^2+xy=y^2-3y+2\\x^2-y^2=3\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}x^2+y^2+z^2+2xy-xz-zy=3\\x^2+y^2-2xy-xz+zy=-1\end{matrix}\right.\)
f) \(\left\{{}\begin{matrix}x^2-y^2+5x-y+6=0\\x^2+\left(x-y\right)^2=2+\sqrt{6x+7}+2\sqrt{x+y+1}\end{matrix}\right.\)
Tính
a) \(\frac{x^3+1}{x}.\left(\frac{1}{x+1}+\frac{x-1}{x^2-x+1}\right)\)
b) \(\frac{x^3-3x^2+2x}{3x^2-4x+1}.\left(\frac{x-1}{x}-\frac{2x-6}{x-1}+\frac{x+1}{x-2}\right)\)
c) \(\frac{3x-3y}{2x^2-2xy+2y^2}:\frac{6x^2-12xy+6y^2}{5x^3+5y^3}:\frac{5x}{x-y}\)
a)\(ĐKXĐ:x\ne0;-1\)
Ta có:\(\frac{x^3+1}{x}.\left(\frac{1}{x+1}+\frac{x-1}{x^2-x+1}\right)=\frac{x^3+1}{x}.\frac{\left(x^2-x+1\right)+\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x^3+1}{x}.\frac{x^2-x+1+\left(x^2-1\right)}{x^3+1}=\frac{2x^2-x}{x}=\frac{2x\left(x-1\right)}{x}=2\left(x-1\right)\)
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Giải các hệ phương trình :a) left{{}begin{matrix}left(x-3right)left(2y+5right)left(2x+7right)left(y-1right)left(4x+1right)left(3y-6right)left(6x-1right)left(2y+3right)end{matrix}right.;b) left{{}begin{matrix}left(x+yright)left(x-1right)left(x-yright)left(x+1right)left(2xyright)left(y-xright)left(y+1right)left(y+xright)left(y-2right)-2xyend{matrix}right..
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Giải các hệ phương trình :
a) \(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\);
b) \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)\left(2xy\right)\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\).
