\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{617}\)
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15 tháng 5 2021
a) Ta có: \(\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{17}\right)\cdot\left(\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{20}\right)\)
\(=\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{17}\right)\cdot\left(\dfrac{5}{20}-\dfrac{4}{20}-\dfrac{1}{20}\right)\)
\(=0\cdot\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{17}\right)=0\)
15 tháng 5 2021
b) Ta có: \(\dfrac{12}{5}\cdot\left(\dfrac{10}{3}-\dfrac{5}{12}\right)\)
\(=\dfrac{12}{5}\cdot\left(\dfrac{40}{12}-\dfrac{5}{12}\right)\)
\(=\dfrac{12}{5}\cdot\dfrac{35}{12}\)
=7
MH
18 tháng 2 2022
\(\left(\dfrac{x}{x+1}+\dfrac{x-1}{x}\right):\left(\dfrac{x}{x+1}-\dfrac{x-1}{x}\right)\) \(\left(đk:x\ne0;-1\right)\)
\(=\dfrac{x^2+\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}:\left(\dfrac{x^2-\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}\right)\)
\(=\dfrac{x^2+x^2-1}{x\left(x+1\right)}.\dfrac{x\left(x+1\right)}{x^2-x^2+1}\)
\(=\dfrac{\left(2x^2-1\right)x\left(x+1\right)}{x\left(x+1\right)}=2x^2-1\)
11 tháng 10 2023
b:
ĐKXĐ: x<>0
\(\dfrac{2}{x}+\dfrac{y}{3}=\dfrac{1}{6}\)
=>\(\dfrac{6+xy}{3x}=\dfrac{1}{6}\)
=>\(6\left(6+xy\right)=3x\)
=>\(x=2\left(6+xy\right)=12+2xy\)
=>\(x\left(1-2y\right)=12\)
mà x,y là các số nguyên
nên \(\left(x;1-2y\right)\in\left\{\left(12;1\right);\left(-12;-1\right);\left(4;3\right);\left(-4;-3\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(12;0\right);\left(-12;1\right);\left(4;-1\right);\left(-4;2\right)\right\}\)
c: ĐKXĐ: y<>-1
\(\dfrac{x}{3}+\dfrac{1}{y+1}=\dfrac{1}{6}\)
=>\(\dfrac{xy+x+3}{3\left(y+1\right)}=\dfrac{1}{6}\)
=>\(\dfrac{2\left(xy+x+3\right)}{6\left(y+1\right)}=\dfrac{y+1}{6\left(y+1\right)}\)
=>\(2xy+2x+6=y+1\)
=>\(2x\left(y+1\right)-\left(y+1\right)=-6\)
=>\(\left(2x-1\right)\left(y+1\right)=-6\)
mà x,y là các số nguyên
nên \(\left(2x-1;y+1\right)\in\left\{\left(1;-6\right);\left(-1;6\right);\left(3;-2\right);\left(-3;2\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(1;-7\right);\left(0;5\right);\left(2;-3\right);\left(-1;1\right)\right\}\)
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{617}\)
\(\Rightarrow\dfrac{x+y}{xy}=\dfrac{1}{617}\)
\(\Rightarrow617\left(x+y\right)=xy\)
\(\Rightarrow617x+617y=xy\)
\(\Rightarrow617x+617y-xy+617^2=617^2\)
\(\Rightarrow\left(y-617\right)\left(x-617\right)=617^2\)
\(\Rightarrow\left[{}\begin{matrix}x-617=617\\y-617=617\end{matrix}\right.\)
\(\Rightarrow x=y=1234\)