\(y.9,2+y:0,25-y.3,2=\frac{3}{4}.\left(1\frac{1}{2}-\frac{6}{8}\right)\)
Những câu hỏi liên quan
e) y x 9,2 + y : 0,25 - y x 3,2 = \(\frac{3}{4}x\left(1\frac{1}{2}-\frac{6}{8}\right)\) g) 71 + 65 x 4 = \(\frac{X+140}{X}+260\)
tìm x,y biết:
a)\(\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\le0\)
b)\(x=6:\frac{1}{3}-0,8:\frac{1,5}{\frac{3}{2}.0,4.\frac{50}{1:\frac{1}{2}}}+\frac{1}{4}+\frac{1+\frac{1}{2}.\frac{1}{0,25}}{6-\frac{46}{1+2,2.10}}\)
c)\(y=77^{-1}.7^4.11^2.\left(1.11\right)^4.\left(7^2\right)^{-8}.\left(7^{-8}\right)^{-3}.\frac{1}{\left(-11\right)^{-3}}\)
a) Vì (2x - 5)2000 và (3y + 4)2002 đều có số mũ là chẵn => (2x - 5)2000 \(\ge\) 0; (3y + 4)2002 \(\ge\) 0
Mà tổng trên lại \(\le\) 0
=> (2x - 5)2000 = (3y + 4)2002 = 0
=> 2x - 5 = 3y + 4 = 0
=> x = 2,5; y = \(\frac{-4}{3}\)
b) x = 18 - 0,8 : \(\frac{1,5}{\frac{3}{2}.\frac{4}{10}.\frac{50}{2}}\)+ \(\frac{1}{4}\)+ \(\frac{1+0,5.4}{6-\frac{46}{23}}\)
= 18 - \(\frac{8}{10}:\frac{1,5}{15}+\frac{1}{4}+\frac{3}{4}\)
= \(18-8+1=11\)
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a) x = 2,5; y = -4/3
Câu b với c nhìn chóng mặt quá, không dám đụng vào
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khó quá à , ai giải được thì nhắn tin cho nhắn nha .
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hệ phương trình
1, left{{}begin{matrix}frac{1}{x+y}+frac{1}{x-y}frac{5}{8}frac{1}{x+y}-frac{1}{x-y}-frac{3}{8}end{matrix}right.
2, left{{}begin{matrix}frac{4}{2x-3y}+frac{5}{3x+y}2frac{3}{3x+y}-frac{5}{2x-3y}21end{matrix}right.
3, left{{}begin{matrix}frac{7}{x-y+2}+frac{5}{x+y-1}frac{9}{2}frac{3}{x-y+2}+frac{2}{x+y-1}4end{matrix}right.
4, left{{}begin{matrix}frac{3}{x}+frac{5}{y}-frac{3}{2}frac{5}{x}-frac{2}{y}frac{8}{3}end{matrix}right.
5 , left{{}begin{matrix}frac{2}{x+y-1}-frac{4}{x-y+1...
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hệ phương trình
1, \(\left\{{}\begin{matrix}\frac{1}{x+y}+\frac{1}{x-y}=\frac{5}{8}\\\frac{1}{x+y}-\frac{1}{x-y}=-\frac{3}{8}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{4}{2x-3y}+\frac{5}{3x+y}=2\\\frac{3}{3x+y}-\frac{5}{2x-3y}=21\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\frac{7}{x-y+2}+\frac{5}{x+y-1}=\frac{9}{2}\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\frac{3}{x}+\frac{5}{y}=-\frac{3}{2}\\\frac{5}{x}-\frac{2}{y}=\frac{8}{3}\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}\frac{2}{x+y-1}-\frac{4}{x-y+1}=-\frac{14}{5}\\\frac{3}{x+y-1}+\frac{2}{x-y+1}=-\frac{13}{5}\end{matrix}\right.\)
6 , \(\left\{{}\frac{\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}}{2\left(x-3\right)-3\left(y+20=-16\right)}}\)
7\(\left\{{}\begin{matrix}\left(x+3\right)\left(y+5\right)=\left(x+1\right)\left(y+8\right)\\\left(2x-3\right)\left(5y+7\right)=2\left(5x-6\right)\left(y+1\right)\end{matrix}\right.\)
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 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
CMR: \(\frac{1}{\left(x+y\right)^3}.\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^4}.\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}.\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{x^3y^3}\)
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\(\frac{1}{\left(x+y\right)^2}\cdot\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^{\text{4}}}\cdot\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}\cdot\left(\frac{1}{x}+\frac{1}{y}\right)\)
Giúp vs cần gấp
Thiếu điều kiện xy = 1; x+y khác 0 nhá bn
Bài này tương tự câu 1 ở đây
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Rút gọn:
\(\frac{1}{\left(x+y\right)^3}\cdot\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^4}\cdot\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}\cdot\left(\frac{1}{x}+\frac{1}{y}\right)\)
hệ phương trình
1 ,left{{}begin{matrix}frac{2x-3}{2y-5}frac{3x+1}{3y-4}2left(x-3right)-3left(y+2right)-16end{matrix}right.
2, left{{}begin{matrix}frac{x}{y}frac{3}{2}3x-2y5end{matrix}right.
3, left{{}begin{matrix}frac{x^2-y-6}{x}x-2x+3y8end{matrix}right.
4, left{{}begin{matrix}frac{x}{y}frac{2}{3}x+y10end{matrix}right.
5, left{{}begin{matrix}frac{y^2+2x-8}{y}y-3x+y10end{matrix}right.
6 , left{{}begin{matrix}frac{x+1}{y-1}53left(2x-2right)-4left(3x+4right)5end{matrix}right.
7, left{{}begi...
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hệ phương trình
1 ,\(\left\{{}\begin{matrix}\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}\\2\left(x-3\right)-3\left(y+2\right)=-16\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{x}{y}=\frac{3}{2}\\3x-2y=5\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\frac{x^2-y-6}{x}=x-2\\x+3y=8\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\frac{x}{y}=\frac{2}{3}\\x+y=10\end{matrix}\right.\)
5, \(\left\{{}\begin{matrix}\frac{y^2+2x-8}{y}=y-3\\x+y=10\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}\frac{x+1}{y-1}=5\\3\left(2x-2\right)-4\left(3x+4\right)=5\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}2x+y=4\\\left|x-2y\right|=3\end{matrix}\right.\)
8 , \(\left\{{}\begin{matrix}\frac{2x}{x+1}+\frac{y}{y+1}=3\\\frac{x}{x+1}-\frac{3y}{y+1}=-1\end{matrix}\right.\)
9 , \(\left\{{}\begin{matrix}y-\left|x\right|=1\\2x-y=1\end{matrix}\right.\)
10 , \(\left\{{}\begin{matrix}\sqrt{x+3y}=\sqrt{3x-1}\\5x-y=9\end{matrix}\right.\)
Giải các hpt:
1)\(\left\{{}\begin{matrix}\frac{10}{x-1}+\frac{1}{y+2}=1\\\frac{25}{x-1}+\frac{3}{y+2}=2\end{matrix}\right.\)
2)\(\left\{{}\begin{matrix}4\left|x+y\right|-3\left|x-y\right|=8\\3\left|x+y\right|-5\left|x-y\right|=6\end{matrix}\right.\)