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a) \(-\frac{2x}{5}=\frac{-6}{3}\)
\(\Rightarrow-2x.3=-6.5\)
\(\Rightarrow-2x=\frac{-6.5}{3}\)
\(\Rightarrow-2x=-10\)
\(\Rightarrow x=\frac{-10}{-2}\)
\(\Rightarrow x=5\)
b) \(\frac{x-2}{18}=\frac{-1}{6}\)
\(\Rightarrow\left(x-2\right).6=-1.18\)
\(\Rightarrow x-2=\frac{-1.18}{6}\)
\(\Rightarrow x-2=-3\)
\(\Rightarrow x=-3+2\)
\(\Rightarrow x=-1\)
c) \(\frac{x}{14}=\frac{6}{y}=\frac{-15}{35}\)
\(\Rightarrow\left[\begin{matrix}\frac{x}{14}=\frac{-15}{35}\\\frac{6}{y}=\frac{-15}{35}\end{matrix}\right.\Rightarrow\left[\begin{matrix}x.35=-15.14\\6.35=-15.y\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=\frac{-15.14}{35}\\y=\frac{6.35}{-15}\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=-6\\y=-14\end{matrix}\right.\)
Vậy \(x=-6;y=-14\)
a) \(\frac{-2x}{5}=-\frac{6}{3}\)
\(\Leftrightarrow-6x=-30\)
\(\Leftrightarrow x=5\)
b) \(\frac{x-2}{18}=-\frac{1}{6}\)
\(\Leftrightarrow-6\left(x-2\right)=-18\)
\(\Leftrightarrow x-2=3\)
\(\Leftrightarrow x=5\)
c) \(\frac{x}{14}=\frac{6}{y}=-\frac{15}{35}\)
\(\Leftrightarrow\frac{x}{14}=\frac{6}{y}=-\frac{3}{7}\)
\(\Leftrightarrow\left\{\begin{matrix}x=-6\\y=-14\end{matrix}\right.\)
Cậu có chắc của lớp 6 không ???
Áp dụng Bất đẳng thức Cauchy-Schwarz dạng Engel , có :
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{\left(1+1+1\right)^2}{x+y+z}=\frac{9}{6}=\frac{3}{2}\)
Đẳng thức xảy ra : \(\Leftrightarrow\frac{1}{x}=\frac{1}{y}=\frac{1}{z}=\frac{1}{2}\)
Xét \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(x+y+z\right)=3+\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{x}{z}+\frac{z}{x}\)
Với \(x,y,z\inℕ^∗\)áp dụng bất đẳng thức Cô si \(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x}{y}.\frac{y}{x}}=2\),\(\frac{y}{z}+\frac{z}{y}\ge2\sqrt{\frac{y}{z}.\frac{z}{y}}=2\),\(\frac{x}{z}+\frac{z}{x}\ge2\sqrt{\frac{x}{z}.\frac{z}{x}}=2\)
\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(x+y+z\right)\ge3+2+2+2=9\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}=\frac{9}{6}=\frac{3}{2}\left(x+y+z=6theogt\right)\)
\(\frac{x}{-7}=\frac{5}{-35}\)
\(\frac{x.5}{-35}=\frac{5}{-35}\)
=> x . 5 = 5
x = 5 : 5
x = 1
a, \(\frac{17}{y}=\frac{-7}{11}\)
\(\Rightarrow17\cdot11=-7\cdot y\)
\(\Rightarrow187=-7\cdot y\)
\(\Rightarrow\frac{187}{-7}=y\)
b, \(\frac{-8}{3x-1}=\frac{4}{7}\)
\(\Rightarrow\frac{-8}{3x-1}=\frac{-8}{-14}\)
\(\Rightarrow3x-1=-14\)
\(\Rightarrow3x=-14+1\)
\(\Rightarrow3x=-13\)
\(\Rightarrow x=\frac{-13}{3}\)
c, \(\frac{x}{-3}=\frac{-3}{x}\)
\(\Rightarrow x\cdot x=-3\cdot\left(-3\right)\)
\(\Rightarrow x^2=9\)
\(\Rightarrow x^2=\left(\pm3\right)^2\)
\(\Rightarrow x=\pm3\)
d, \(\frac{-4}{y}=\frac{x}{2}\)
\(\Rightarrow-4\cdot2=x\cdot y\)
\(\Rightarrow-8=x\cdot y\)
\(\Rightarrow x;y\inƯ\left(-8\right)=\left\{-1;1;-2;2;-4;4;-8;8\right\}\)
ta có bảng :
| x | -1 | -8 | -2 | -4 |
| y | 8 | 1 | 4 | 2 |
a)\(\frac{14}{y}\)\(=\) \(\frac{-7}{11}\)
\(\Rightarrow\)\(14\cdot11=y\cdot\left(-7\right)\)
\(y=\)\(\frac{14\cdot11}{-7}\)
\(y=22\)
c) \(\frac{x}{-3}\) = \(\frac{-3}{x}\)
\(\Rightarrow\) \(x\cdot x=\left(-3\right)\cdot\left(-3\right)\)
\(\Rightarrow\)\(x^2=9\)
\(\Rightarrow\)\(x^2=9\)hoặc \(x^2=-9\)
\(TH1:\) \(x^2=9\)
\(\Rightarrow\)\(x=3\)
\(TH2:\)\(x^2=-9\)
\(\Rightarrow\)\(x=-3\)
\(\frac{x}{7}+\frac{1}{14}=\frac{1}{y}\)
\(\frac{x}{7}+\frac{1}{14}=\frac{1}{y}\)
\(\frac{x\times2}{14}+\frac{1}{14}=\frac{1}{y}\)
\(\frac{2x+1}{14}=\frac{1}{y}\)
\(\Rightarrow\left(2x+1\right).y=14\)
Ta có: 14=7.2=-7.(-2)
mà 2x+1 là số lẻ
\(\Rightarrow\orbr{\begin{cases}2x+1=7\\y=2\end{cases}}\)
\(\orbr{\begin{cases}2x+1=-7\\y=-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=6\\y=2\end{cases}}\)
\(\orbr{\begin{cases}2x=-8\\y=-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\y=2\end{cases}}\)
\(\orbr{\begin{cases}x=-4\\y=-2\end{cases}}\)