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\(\frac{x}{y}=a\Rightarrow x=ay\)
\(\Rightarrow\frac{x+y}{x-y}=\frac{ay+y}{ay-y}=\frac{y\left(a+1\right)}{y\left(a-1\right)}=\frac{a+1}{a-1}\)
\(\frac{a}{b}=2\Rightarrow a=2b;\frac{c}{b}=3\Rightarrow c=3b\Rightarrow c-b=2b\)
\(\Rightarrow a=c-b\)
\(\Rightarrow\frac{a+c}{b+c}=\frac{c-b+b}{b+c}=\frac{b}{b+c}\)
a) \(\frac{3}{2}x-\frac{2}{5}=\frac{1}{3}x-\frac{1}{4}\)
=> \(\frac{3}{2}x-\frac{2}{5}-\frac{1}{3}x+\frac{1}{4}=0\)
=> \(\left(\frac{3}{2}-\frac{1}{3}\right)x+\left(-\frac{2}{5}+\frac{1}{4}\right)=0\)
=> \(\frac{7}{6}x-\frac{3}{20}=0\)
=> \(\frac{7}{6}x=\frac{3}{20}\)
=> \(x=\frac{3}{20}:\frac{7}{6}=\frac{3}{20}\cdot\frac{6}{7}=\frac{9}{70}\)
b) \(2x-\frac{2}{3}=7x+\frac{2}{3}-1\)
=> \(2x-\frac{2}{3}=7x-\frac{1}{3}\)
=> \(2x-\frac{2}{3}-7x+\frac{1}{3}=0\)
=> (2x - 7x) + (-2/3 + 1/3) = 0
=> -5x - 1/3 = 0
=> -5x = 1/3
=> x = -1/15
bài 1 :
a, A = 3|2x - 1| - 5 = 0
có 3|2x - 1| > 0
=> A > -5
xét A = -5 khi
|2x - 1| = 0
=> 2x - 1 = 0
=> 2x = 1
=> x = 1/2
vậy Min A = -5 khi x = 1/2
b, c, d, làm tương tự
Bài 1:
\(a)A=3|2x-1|-5\)
Vì \(|2x-1|\ge0\)\(\forall x\)
\(\Rightarrow3|2x-1|\ge0\) \(\forall x\)
\(\Rightarrow3|2x-1|-5\ge-5\) \(\forall x\)
Dấu "=" xảy ra:
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy \(Min_A=-5\Leftrightarrow x=\frac{1}{2}\)
\(b)x^2+3|y-2|-1\)
Vì \(\hept{\begin{cases}x^2\ge0\forall x\\3|y-2|\ge0\forall y\end{cases}}\)
\(\Rightarrow x^2+3|y-2|-1\ge-1\) \(\forall x,y\)
Dấu '=' xảy ra:
\(\Leftrightarrow\hept{\begin{cases}x^2=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=2\end{cases}}\)
Vậy \(Min_B=-1\Leftrightarrow x=0,y=2\)
\(c)\left(2x^2+1\right)^4-3\)
Vì \(\left(2x^2+1\right)^4\ge0\)\(\forall x\)
\(\Rightarrow\left(2x^2+1\right)^4-3\ge-3\) \(\forall x\)
Dấu "=" xảy ra:
\(\Leftrightarrow2x^2+1=0\)
\(\Leftrightarrow2x^2=-1\)
\(\Leftrightarrow x^2=-\frac{1}{2}\left(voli\right)\)
Vậy không tìm được gt x
\(d)D=|x-\frac{1}{2}|+\left(y+2\right)^2+11\)
Vì \(\hept{\begin{cases}|x-\frac{1}{2}|\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{cases}}\)
\(\Rightarrow|x-\frac{1}{2}|+\left(y+2\right)^2+11\ge11\) \(\forall x,y\)
Dấu '=' xảy ra:
\(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-2\end{cases}}\)
Vậy \(Min_D=11\Leftrightarrow x=\frac{1}{2},y=-2\)
Bài 2:
\(a)A=10-5|x-2|\)
Vì \(|x-2|\ge0\)\(\forall x\)
\(\Rightarrow5|x-2|\ge0\)\(\forall x\)
\(\Rightarrow\)\(10-5|x-2|\le10\) \(\forall x\)
Dấu "=" xảy ra:
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy \(Max_A=10\Leftrightarrow x=2\)
\(b)B=5-|2x-1|^2\)
Vì \(|2x-1|^2\ge0\)\(\forall x\)
\(\Rightarrow5-|2x-1|^2\le5\) \(\forall x\)
Dấu "=" xảy ra:
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy \(Max_B=5\Leftrightarrow x=\frac{1}{2}\)
\(c)C=\frac{1}{|x-2|+3}\)
Vì \(|x-2|\ge0\)\(\forall x\)
\(\Rightarrow|x-2|+3\ge3\) \(\forall x\)
\(\Rightarrow\frac{1}{|x-2|+3}\le\frac{1}{3}\) \(\forall x\)
Dấu "=" xảy ra:
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy \(Max_C=\frac{1}{3}\Leftrightarrow x=2\)
a) \(\frac{x}{y}=\frac{5}{7}\)=>\(\frac{x}{5}=\frac{y}{7}=>\left(\frac{x}{5}\right)^2=\left(\frac{y}{7}\right)^2=\frac{xy}{5.7}\)
=>\(\frac{x^2}{25}=\frac{y^2}{49}=\frac{35}{35}=1\)
=> \(x^2=25;y^2=49\)
=>\(x=\pm5;y=\pm7\)
1. \(AB=-\frac{1}{3}x^2y^2\cdot\left(-6x^3y^4\right)=\left(-\frac{1}{3}\cdot-6\right)\left(x^2x^3\right)\left(y^2y^4\right)=2x^5y^6\)
Bậc = 5 + 6 = 11
2. Thiếu B