\(\dfrac{1}{3x}\)+\(^{\dfrac{1}{y}}\)=\(\dfrac{1}{22}\)
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1 tháng 8 2017
\(\dfrac{x-2}{4}=\dfrac{y+1}{5}\)
\(\Rightarrow\dfrac{3\left(x-2\right)}{12}=\dfrac{y+1}{5}\)
\(\Rightarrow\dfrac{3x-6}{12}=\dfrac{y+1}{5}\)
Dựa vào tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{3x-6}{12}=\dfrac{y+1}{5}\)
\(=\dfrac{3x-6-y-1}{12-5}\)
\(=\dfrac{\left(3x-y\right)-\left(6+1\right)}{7}\)
\(=\dfrac{22-7}{7}=\dfrac{15}{7}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{15}{7}.4+2=\dfrac{74}{7}\\y=\dfrac{15}{7}.5-1=\dfrac{68}{7}\end{matrix}\right.\)
1 tháng 8 2017
\(\dfrac{x-2}{4}=\dfrac{y+1}{5}\Leftrightarrow\dfrac{3x-6}{12}=\dfrac{y+1}{5}=\dfrac{3x-y-6-1}{12-5}=\dfrac{15}{7}\)
\(\dfrac{x-2}{4}=\dfrac{15}{7}\Leftrightarrow x=\dfrac{15}{7}.4+2=\dfrac{74}{7}\)
\(y=\dfrac{74}{7}.3-22=\dfrac{68}{7}\)
18 tháng 10 2021
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
em c.ơn trước
Y
9 tháng 2 2023
\(5,\dfrac{4}{x-2}+\dfrac{x}{x+1}-\dfrac{x^2-2}{\left(x-2\right)\left(x+1\right)}=0\left(dkxd:x\ne2;-1\right)\)
\(\Rightarrow4\left(x+1\right)+x\left(x-2\right)-x^2-2=0\)
\(\Rightarrow4x+4+x^2-2x-x^2-2=0\)
\(\Rightarrow2x+2=0\)
\(\Rightarrow x=-1\left(loai\right)\)
Vậy \(S=\varnothing\)
18 tháng 12 2021
Áp dụng BĐT Cauchy-Schwarz:
\(\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\ge\dfrac{16}{3x+3y+2z}\\ \Leftrightarrow\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{2}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\right)\\ \Leftrightarrow\sum\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{4}{x+y}+\dfrac{4}{y+z}+\dfrac{4}{z+x}\right)=\dfrac{4}{16}\cdot6=\dfrac{3}{2}\)
Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)
12 tháng 1 2023
1: Khi x=2 thì \(A=\dfrac{4\cdot2+1}{2-1}=9\)
2: \(=\dfrac{3x+1-2x^2-2x+3x^2-3x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
23 tháng 11 2021
\(a,\dfrac{1}{3x-3y}=\dfrac{x-y}{3\left(x-y\right)^2};\dfrac{1}{x^2-2xy+y^2}=\dfrac{3}{3\left(x-y\right)^2}\\ b,\dfrac{3}{x^2-3x}=\dfrac{6}{2x\left(x-3\right)};\dfrac{5}{2x-6}=\dfrac{5x}{2x\left(x-3\right)}\\ c,\dfrac{x}{x+3}=\dfrac{x^2-3x}{\left(x-3\right)\left(x+3\right)};\dfrac{1}{3-x}=\dfrac{-x-3}{\left(x-3\right)\left(x+3\right)};\dfrac{1}{x^2-9}=\dfrac{1}{\left(x-3\right)\left(x+3\right)}\)
\(d,\dfrac{1}{x^2+xy}=\dfrac{xy-y^2}{xy\left(x+y\right)\left(x-y\right)};\dfrac{1}{xy-y^2}=\dfrac{x^2+xy}{xy\left(x-y\right)\left(x+y\right)};\dfrac{2}{y^2-x^2}=\dfrac{-2xy}{xy\left(x-y\right)\left(x+y\right)}\)