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1) \(\frac{2x-y}{x+y}=\frac{2}{3}\)
\(\Leftrightarrow\left(2x-y\right).3=2\left(x+y\right)\)
\(\Leftrightarrow6x-3y=2x+2y\)
\(\Leftrightarrow6x-2x=2y+3y\)
\(\Rightarrow4x=5y\)
\(\Rightarrow\frac{x}{y}=\frac{5}{4}\)
2) \(\begin{cases}\frac{b}{a}=2\\\frac{c}{b}=3\end{cases}\) \(\Rightarrow\begin{cases}b=2a\\c=3b\end{cases}\)
Khi đó: \(\frac{a+b}{b+c}=\frac{a+2a}{2a+3}\)
\(=\frac{a+2a}{2a+3.2a}\)
\(=\frac{3a}{8a}=\frac{3}{8}\)
Chúc bạn học tốt
Ta có: \(\frac{2x-y}{x+y}\)=\(\frac{2}{3}\)
=> (2x - y).3 = (x+y) .2
6x - 3y = 2x + 2y
6x - 2x = 3y + 2y
4x = 5y
=> \(\frac{x}{5}\)=\(\frac{y}{4}\)
Vậy tỉ số \(\frac{x}{y}\)=\(\frac{5}{4}\)
\(\frac{2x-y}{x+y}=\frac{2}{3}\)
\(\Rightarrow3\left(2x-y\right)=2\left(x+y\right)\)
\(\Rightarrow6x-3y=2x+2y\)
\(\Rightarrow6x-2x=2y+3y\)
\(\Rightarrow4x=5y\)
\(\Rightarrow\frac{x}{y}=\frac{5}{4}\)
Vậy \(\frac{x}{y}=\frac{5}{4}\)
Ta có : \(\frac{2x-y}{x+y}=\frac{2}{3}\Leftrightarrow3\left(2x-y\right)=2\left(x+y\right)\Leftrightarrow6x-3y=2x+2y\Leftrightarrow4x=5y\Leftrightarrow\frac{x}{y}=\frac{5}{4}\)
Vì \(\frac{2x-y}{x+y}=\frac{2}{3}=>\left(2x-y\right).3=\left(x+y\right).2=>6x-3y=2x+2y\)
\(=>6x-2x=2y-\left(-3y\right)=>6x-2x=2y+3y=>4x=5y=>\frac{x}{y}=\frac{5}{4}\)
Vậy tỉ số x/y=5/4
\(\frac{2x-y}{x+y}=\frac{2}{3}\Rightarrow3\left(2x-y\right)=2\left(x+y\right)\)
\(\Rightarrow6x-3y=2x+2y\)
\(\Rightarrow6x-2x=3y+2y\)
\(\Rightarrow4x=5y\)
\(\Rightarrow\frac{x}{y}=\frac{5}{4}\)
T ủng hộ mk nhé bạn ^...^ ^_^
\(\frac{2x-y}{x+y}=\frac{2}{3}\)
\(\Rightarrow3\left(2x-y\right)=2\left(x+y\right)\)
\(\Rightarrow6x-3y=2x+2y\)
\(\Rightarrow4x=5y\)
\(\Rightarrow\frac{x}{y}=\frac{4}{5}\)
Vậy.................
\(\frac{2x-y}{x+y}=\frac{2}{3}\Rightarrow2\left(x+y\right)=3\left(2x-y\right)\)
\(\Rightarrow6x-3y=2x+2y\)
\(\Rightarrow6x-2x=2y+3y\)
\(\Rightarrow\frac{x}{y}=\frac{5}{4}\)
Ta có: \(\frac{2x-y}{x+y}=\frac{2}{3}\)
\(\Rightarrow3.\left(2x-y\right)=2.\left(x+y\right)\)
\(\Rightarrow6x-3y=2x+2y\)
\(\Rightarrow6x-2x=3y+2y\)
\(\Rightarrow4x=5y\)
\(\Rightarrow\frac{x}{y}=\frac{5}{4}\)
Vậy tỉ số \(\frac{x}{y}=\frac{5}{4}\)
Theo bài ra ta có:
2x-y/x+y=2/3
=>3(2x-y)=2(x+y)
6x-3y=2x+2y
=>(6x-3y)-(2x+2y)=0
6x-3y-2x-2y=0
4x-y=0
=>4x=y
=>x/y=1/4
Theo đầu bài ta có:
\(\frac{2x-y}{x+y}=\frac{2}{3}\)
\(\Rightarrow3\left(2x-y\right)=2\left(x+y\right)\)
\(\Rightarrow6x-3y=2x+2y\)
\(\Rightarrow6x-2x=2y+3y\)
\(\Rightarrow4x=5y\)
\(\Rightarrow\frac{x}{y}=\frac{5}{4}\)
a)Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x-1\right|+\left|3+x\right|=\left|1-x\right|+\left|3+x\right|\ge\left|1-x+3+x\right|=4\)
\(\Rightarrow VT\ge VP."="\Leftrightarrow-3\le x\le1\)
b) \(\hept{\begin{cases}\left|2x+3\right|+\left|2x-1\right|=\left|2x+3\right|+\left|1-2x\right|\ge4\\\frac{8}{2\left(y-5\right)^2+2}\le4\end{cases}}\Leftrightarrow VT\ge VP."="\Leftrightarrow\hept{\begin{cases}-\frac{3}{2}\le x\le\frac{1}{2}\\y=5\end{cases}}\)
c Tương tự b
2) \(\frac{1}{x}+\frac{1}{y}=5\Leftrightarrow x+y-5xy=0\Leftrightarrow5x+5y-25xy=0\Leftrightarrow5x\left(1-5y\right)-\left(1-5y\right)=-1\)
\(\Leftrightarrow\left(5x-1\right)\left(1-5y\right)=-1\)
Xét ước
(2x-y)/(x+y)=2/3
nên 3(2x-y)=2(x+y)
6x-3y=2x+2y
6x-3y-2x-2y=0
4x-5y=0
4x=5y
x/y=5/4
\(\frac{2x-y}{x+y}=\frac{2}{3}\)
Nên \(3\left(2x-y\right)=2\left(x+y\right)\)
\(\Rightarrow6x-3y=2x+2y\)
\(\Rightarrow6x-2x=3y+2y\)
\(\Rightarrow4x=5y\)
Vậy ta có \(\frac{x}{y}=\frac{5}{4}\)