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\(A=\left|4x-3\right|+\left|5y+7,5\right|+10\)
Mà \(\left|4x-3\right|\ge0\)với mọi x
\(\left|5y+7,5\right|\ge0\)với mọi y
\(\Rightarrow A\)có GTNN là 10
Để A có GTNN thì :
\(4x-3=0\) \(5y+7,5=0\)
\(4x=3\) \(5y=-7,5\)
\(x=\frac{3}{4}\) \(y=-1,5\)
\(B=\frac{5,8}{\left|2,5-x\right|+5,8}\)
Mà \(\left|2,5-x\right|\ge0\)
\(\Rightarrow\)GTNN \(\left|2,5-x\right|+5,8=5,8\)
Để B có GTLN \(\Rightarrow2,5-x=0\)
\(\Rightarrow x=2,5\)
Bài 1:
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)
A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)
A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))
A = 1 - 1 - \(\frac{228}{210}\)
A = 0 - \(\frac{128}{210}\)
A = - \(\frac{64}{105}\)
Bài 2:
B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)
B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)
B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))
B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))
B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))
B = 3 - \(\frac{97}{65}\)
B = \(\frac{195}{65}\) - \(\frac{97}{65}\)
B = \(\frac{98}{65}\)
\(\frac{72^2}{24^2}=\frac{3^2.24^2}{24^2}=3^2=9\)
\(\frac{\left(-7,5\right)^2}{2,5^2}=\frac{\left(7,5\right)^2}{\left(2,5\right)^2}=\frac{\left(2,5\right)^2.3^2}{\left(2,5\right)^2}=9\)
\(\frac{15^2}{27}=\frac{3^2.5^2}{27}=\frac{9.25}{27}=\frac{25}{3}\)
\(\frac{72^2}{24^2}=\left(\frac{72}{24}\right)^2=3^2=9\)
\(\frac{\left(-7,5\right)^3}{\left(2,5\right)^3}=\left(\frac{-7,5}{2,5}\right)^3=\left(-3\right)^3=-27\)
\(\frac{15^3}{27}=\frac{15^3}{3^3}=\left(\frac{15}{3}\right)^3=5^3=125\)
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