\(\left[x-7\right]\times\left[xy+1\right]=9\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy\right)^3+7\left(xy+x+y+1\right)=31\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3+\left(xy\right)^3+7\left(xy+x+y\right)=30\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\)
\(\Rightarrow\left\{{}\begin{matrix}uv=2\\u^3+v^3+7\left(u+v\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3-3uv\left(u+v\right)+7\left(u+v\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3+\left(u+v\right)-30=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\u+v=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=2\\v=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\) \(\Leftrightarrow\left(x;y\right)=\left(1;1\right)\)
2.
ĐKXĐ: \(0\le x\le\dfrac{3}{2}\)
\(\Leftrightarrow9x\left(3-2x\right)+81+54\sqrt{x\left(3-2x\right)}=49x+25\left(3-2x\right)+70\sqrt{x\left(3-2x\right)}\)
\(\Leftrightarrow9x^2-14x-3+8\sqrt{x\left(3-2x\right)}=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)-4\left(3-x-2\sqrt{x\left(3-2x\right)}\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2-\dfrac{36\left(x-1\right)^2}{3-x+2\sqrt{x\left(3-2x\right)}}=0\)
\(\Leftrightarrow9\left(x-1\right)^2\left(1-\dfrac{4}{3-x+2\sqrt{x\left(3-2x\right)}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\3-x+2\sqrt{x\left(3-2x\right)}=4\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\sqrt{x\left(3-2x\right)}=x+1\)
\(\Leftrightarrow4x\left(3-2x\right)=x^2+2x+1\)
\(\Leftrightarrow9x^2-10x+1=0\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)
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(1+7/9).(1+7/20).(1+7/33).(1+7/48)......(1+7/180)
=16/9.27/20.40/33.55/48........187/180
=2.8/1.9 . 3.9/2.10 . 4.10/3.11 . 5.11/4.12 ........ 11.17/18.10
=(2.3.4.5.......11).(8.9.10......17)/(1.2.3.4.....18).(9.10.11.12......18)
=11.8/1.18=88/18=44/9
![](https://rs.olm.vn/images/avt/0.png?1311)
Đk: \(\left\{{}\begin{matrix}y\ge0\\x>1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\sqrt{9\left(x-1\right)y}=y\left(2+\sqrt{\dfrac{y}{x-1}}\right)\left(1\right)\\y^2+xy-5x+7=0\left(2\right)\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{\left(x-1\right)y}\left(a\ge0\right)\\b=\sqrt{\dfrac{y}{x-1}}\left(b\ge0\right)\end{matrix}\right.\)
\(\left(1\right)\Rightarrow3a=ab\left(2+b\right)\)
Với \(a=0\Rightarrow\sqrt{\left(x-1\right)y}=0\Rightarrow y=0\) (vì \(x\ne1\)).
Thay \(y=0\) vào (2) ta được:
\(2^2+x.2-5x+7=0\)
\(\Leftrightarrow x=\dfrac{11}{3}\left(nhận\right)\)
Với \(a\ne0\Rightarrow3=b\left(2+b\right)\)
\(\Leftrightarrow b^2+2b-3=0\)
\(\Leftrightarrow b^2-b+3b-3=0\)
\(\Leftrightarrow b\left(b-1\right)+3\left(b-1\right)=0\)
\(\Leftrightarrow\left(b-1\right)\left(b+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b=1\left(nhận\right)\\b=-3\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{\dfrac{y}{x-1}}=1\Rightarrow x=y+1\)
Thay vào (2) ta được:
\(y^2+\left(y+1\right)y-5\left(y+1\right)+7=0\)
\(\Leftrightarrow y^2+y^2+y-5y-5+7=0\)
\(\Leftrightarrow2y^2-4y+2=0\)
\(\Leftrightarrow2\left(y-1\right)^2=0\)
\(\Leftrightarrow y=1\Rightarrow x=1+1=2\)
Vậy hệ phương trình đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(\dfrac{11}{3};0\right),\left(2;1\right)\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x+1) * (x2 +x+1) * (x-1) * (x2-x+1) = 7
[(x+1) * (x2 +x+1) ]*[(x-1) * (x2-x+1)]= 7 [Áp dụng hằng đẳng thức a3+b3=(a+b)*(a2+ab+b2)]
(x3+13) * (x3-13) = 7
x3 * x3 - x3 * 13 + x3 * 13 - 13 *13 =7
(x3)2 - 1 = 7
(x3)2 =7+1
(x3)2 =8
suy ra x = 3 căn 2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\)\(-\frac{3}{5}\cdot x=\frac{1}{4}+0,75\)
\(-\frac{3}{5}\cdot x=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)
\(x=1\div\left(-\frac{3}{5}\right)\)
\(x=-\frac{5}{3}\)
\(b,\)\(\left(\frac{1}{7}-\frac{1}{3}\right)\cdot x=\frac{28}{5}\times\left(\frac{1}{4}-\frac{1}{7}\right)\)
\(\left(\frac{3}{21}-\frac{7}{21}\right)\cdot x=\frac{28}{5}\cdot\left(\frac{7}{28}-\frac{4}{28}\right)\)
\(-\frac{4}{21}\cdot x=\frac{28}{5}\cdot\frac{3}{28}\)
\(-\frac{4}{21}\cdot x=\frac{3}{5}\)
\(x=\frac{3}{5}\div\left(-\frac{4}{21}\right)\)
\(x=-\frac{63}{20}\)
\(c,\)\(\frac{5}{7}\cdot x=\frac{9}{8}-0,125\)
\(\frac{5}{7}\cdot x=\frac{9}{8}-\frac{1}{8}\)
\(\frac{5}{7}\cdot x=1\)
\(x=1\div\frac{5}{7}\)
\(x=\frac{7}{5}\)
\(d,\)\(\left(\frac{2}{11}+\frac{1}{3}\right)\cdot x=\left(\frac{1}{7}-\frac{1}{8}\right)\cdot36\)
\(\left(\frac{6}{33}+\frac{11}{33}\right)\cdot x=\left(\frac{8}{56}-\frac{7}{56}\right)\cdot36\)
\(\frac{17}{33}\cdot x=\frac{1}{56}\cdot36\)
\(\frac{17}{33}\cdot x=\frac{9}{14}\)
\(x=\frac{9}{14}\div\frac{17}{33}\)
\(x=\frac{9}{14}\cdot\frac{33}{17}=\frac{297}{238}\)