Cho x-2y=10
c=\(\frac{x+5}{3x-2y}+\frac{y-5}{bony-x}\)
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DT
1
1 tháng 2 2022
\(=\dfrac{2x+y}{2\left(x+y\right)}-\dfrac{x+2y}{x-y}+\dfrac{5}{x}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{2x^2-2xy+xy-y^2}{2\left(x+y\right)\left(x-y\right)}-\dfrac{2\left(x+2y\right)\left(x-y\right)}{2\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{2x^2-xy-y^2-2\left(x^2+xy-2y^2\right)}{2\left(x-y\right)\left(x+y\right)}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)
\(=\dfrac{2x^2-xy-y^2-2x^2-2xy+4y^2}{2\left(x-y\right)\left(x+y\right)}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)
\(=\dfrac{-3xy+3y^2}{2\left(x-y\right)\left(x+y\right)}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)
\(=\dfrac{-9xy+9y^2-8x}{6\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)
\(=\dfrac{-9x^2y+9xy^2-8x^2+30\left(x^2-y^2\right)}{6x\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{-9x^2y+9xy^2+22x^2-30y^2}{6x\cdot\left(x-y\right)\left(x+y\right)}\)
11 tháng 1 2022
a, \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}x-3,2y-6\in Z\\x-3,2y-6\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\end{matrix}\right.\)
Ta có bảng:
| x-3 | -1 | -5 | 1 | 5 |
| 2y-6 | -5 | -1 | 5 | 1 |
| x | 2 | -2 | 4 | 8 |
| y | \(\dfrac{1}{2}\left(loại\right)\) | \(\dfrac{5}{2}\left(loại\right)\) | \(\dfrac{11}{2}\left(loại\right)\) | \(\dfrac{7}{2}\left(loại\right)\) |
Vậy không có x,y thỏa mãn đề bài
b, tương tự câu a
\(c,xy-5x+2y=7\\ \Rightarrow x\left(y-5\right)+2y-10=-3\\ \Rightarrow x\left(y-5\right)+2\left(y-5\right)=-3\\ \Rightarrow\left(x+2\right)\left(y-5\right)=-3\)
Rồi làm tương tự câu a
\(d,xy-3x-4y=5\\ \Rightarrow x\left(y-3\right)-4y+12=17\\ \Rightarrow x\left(y-3\right)-4\left(y-3\right)=17\\ \Rightarrow\left(x-4\right)\left(y-3\right)=17\)
Rồi làm tương tự câu a
4 tháng 2 2019
\(C=\frac{x+5}{2x+\left(x-2y\right)}+\frac{y-5}{2y-\left(x-2y\right)}\)
\(=\frac{x+5}{2x+10}+\frac{y-5}{2y-10}=\frac{x+5}{2\left(x+5\right)}+\frac{y-5}{2\left(y-5\right)}=\frac{1}{2}+\frac{1}{2}=1\left(x\ne-5,y\ne5\right)\)
BN
4 tháng 2 2019
Trả lời :
Ta có :
C = \(\frac{x+5}{3x-2y}+\frac{y-5}{4y-x}\)
C = \(\frac{2\left(x+5\right)}{2\left(3x-2y\right)}+\frac{2\left(y-5\right)}{2\left(4y-x\right)}\)
C = \(\frac{2x+10}{6x-4y}+\frac{2y-10}{8y-2x}\)
Thay x - 2y = 10 . Ta được :
C = \(\frac{2x+x-2y}{6x-4y}+\frac{2y-x-2y}{8y-2x}\)
C = \(\frac{x\left(2+1\right)-2y}{6x-4y}+\frac{y\left(2+2\right)-x}{8y-2x}\)
C = \(\frac{3x-2y}{6x-4y}+\frac{4y-x}{8y-2x}\)
C = \(\frac{1}{2}+\frac{1}{2}\)
C = \(1\)
Vậy C = 1
Hok tốt
28 tháng 9 2018
\(\hept{\begin{cases}\frac{3x+2}{x+3}+\frac{2y-5}{y-1}=5\left(1\right)\\\frac{3x+5}{x+3}+\frac{2y-4}{y-1}=4\left(2\right)\end{cases}}\)
\(\left(1\right)\Leftrightarrow y=-\frac{3x+2}{7}\)
Thê vào (2) rồi rut gọn ta được
\(3x+11=0\)
\(\Leftrightarrow x=-\frac{11}{3}\)
\(\Rightarrow y=\frac{9}{7}\)
CB
3 tháng 6 2019
Đặt: \(\left\{{}\begin{matrix}x=a\\2y=b\\3z=c\end{matrix}\right.\Rightarrow a+b+c=18\)
Có: BDT
\(\Leftrightarrow\sum_{cyc}\left(\frac{b+c+5}{a+1}\right)\ge\frac{51}{7}\)
\(\Leftrightarrow\sum_{cyc}\left(\frac{a+b+c-a+5}{a+1}\right)\ge\frac{51}{7}\)(1)
Đặt tiếp tục: \(\left\{{}\begin{matrix}m=a+1\\n=b+1\\p=c+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=m-1\\b=n-1\\c=p-1\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sum_{cyc}\left(\frac{24-m}{m}\right)\ge\frac{51}{7}\)
\(\Leftrightarrow\sum_{cyc}\left(\frac{24}{m}-1\right)\ge\frac{51}{7}\)
\(\Leftrightarrow24\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\ge\frac{72}{7}\)
\(\Leftrightarrow\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\ge\frac{3}{7}\)
\(\Leftrightarrow\left(m+n+p\right)\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\ge21\cdot\frac{3}{7}=9\)
\(\left(\frac{m}{n}-2+\frac{n}{m}\right)+\left(\frac{p}{m}-2+\frac{m}{p}\right)+\left(\frac{n}{p}-2+\frac{p}{n}\right)\ge0\)
\(\Leftrightarrow\frac{\left(m-n\right)^2}{mn}+\frac{\left(p-m\right)^2}{pm}+\frac{\left(n-p\right)^2}{pn}\ge0\)(đúng)
CB
4 tháng 6 2019
Đặt: \(\left\{{}\begin{matrix}x=a\\2y=b\\3z=c\end{matrix}\right.\)
BĐT
\(\Leftrightarrow\frac{b+c+5}{a+1}+\frac{a+c+5}{b+1}+\frac{a+b+5}{c+1}\ge\frac{51}{7}\)
\(\Leftrightarrow\frac{a+b+c-a+5}{a+1}+\frac{a+c+b-b+5}{b+1}+\frac{a+b+c-c+5}{c+1}\ge\frac{51}{7}\)
\(\Leftrightarrow\frac{24-\left(a+1\right)}{a+1}+\frac{24-\left(b+1\right)}{b+1}+\frac{24-\left(c+1\right)}{c+1}\ge\frac{51}{7}\)(1)
Đặt tiếp: \(\left\{{}\begin{matrix}a+1=m\\b+1=n\\c+1=p\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=m-1\\b=n-1\\c=p-1\end{matrix}\right.\)
(1)\(\Leftrightarrow\frac{24-m}{m}+\frac{24-n}{n}+\frac{24-p}{p}\ge\frac{51}{7}\)
\(\Leftrightarrow24\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)-3\ge\frac{51}{7}\)
\(\Leftrightarrow24\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\ge\frac{72}{7}\)
\(\Leftrightarrow\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\ge\frac{3}{7}\)
\(\Leftrightarrow\left(m+n+p\right)\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\ge\frac{3}{7}\left(m+n+p\right)\)( do m+n+p>0)
\(\Leftrightarrow3+\frac{m}{n}+\frac{n}{m}+\frac{p}{n}+\frac{n}{p}+\frac{m}{p}+\frac{p}{m}\ge\frac{3}{7}\left[\left(a+b+c\right)+3\right]\)
\(\Leftrightarrow\frac{m}{n}+\frac{n}{m}+\frac{p}{n}+\frac{n}{p}+\frac{p}{m}+\frac{m}{p}-6\ge0\)
Tới đây chắc bn làm đc rồi
18 tháng 9 2016
Đặt \(\hept{\begin{cases}a=x\\b=2y\\c=3z\end{cases}}\) => a + b + c = 18
\(P=\frac{2y+3z+5}{1+x}+\frac{3z+x+5}{1+2y}+\frac{x+2y+5}{1+3z}=\frac{b+c+5}{a+1}+\frac{a+c+5}{b+1}+\frac{a+b+5}{c+1}\)
Lại đặt \(\hept{\begin{cases}m=a+1\\n=b+1\\p=c+1\end{cases}}\Rightarrow\hept{\begin{cases}a=m-1\\b=n-1\\c=p-1\end{cases}}\)
Ta có : \(\frac{b+c+5}{a+1}+\frac{a+c+5}{b+1}+\frac{a+c+5}{c+1}=\frac{24-m}{m}+\frac{24-n}{n}+\frac{24-p}{p}\)
\(=24\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)-3\ge\frac{24.9}{m+n+p}-3=\frac{24.9}{\left(a+1\right)+\left(b+1\right)+\left(b+1\right)}-3\)
\(=\frac{24.9}{18+3}-3=\frac{51}{7}\)