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![](https://rs.olm.vn/images/avt/0.png?1311)
\(5(\frac{a}{b}+\frac{1}{2})=2\frac{1}{3}\)
\( \iff5.\frac{a}{b}+5.\frac{1}{2}=\frac{7}{3}\)
\(\iff5.\frac{a}{b}+\frac{5}{2}=\frac{7}{3}\)
\(\iff 5.\frac{a}{b}=\frac{7}{3}-\frac{5}{2}\)
\(\iff 5.\frac{a}{b}=\frac{-1}{6}\)
\(\iff\frac{a}{b}=\frac{-1}{6}:5\)
\(\iff\frac{a}{b}=\frac{-1}{6}.\frac{1}{5}\)
\(\iff \frac{a}{b}=\frac{-1}{30}\)
Vậy \(\frac{a}{b}=\frac{-1}{30}\)
~ Hok tốt a~
\(5\left(ab+\frac{1}{2}\right)=2\frac{1}{3}\)
\(5\left(ab+\frac{1}{2}\right)=\frac{7}{3}\)
\(ab+\frac{1}{2}=\frac{7}{3}:5\)
\(ab+\frac{1}{2}=\frac{7}{15}\)
\(ab=\frac{7}{15}-\frac{1}{2}\)
\(ab=\frac{14}{30}-\frac{15}{30}\)
\(ab=-\frac{1}{30}\)
Vậy \(ab=-\frac{1}{30}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(2A=1-\frac{1}{11}\)
\(2A=\frac{10}{11}\)
\(A=\frac{10}{11}.\frac{1}{2}=\frac{5}{11}\)
Vì x2 luôn luôn lơn hơn hoặc bằng 0
Nên x2 + 2015 luôn luôn lơn hơn hoặc bằng 2015
Nên x - 2016 = 0
=> x = 2016 (t/m)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(a^2-b^2\right):\left(a+b\right).\left(a-b\right)\)
\(=\left[5^2-\left(-3\right)^2\right]:\left[5+\left(-3\right)\right].\left[5-\left(-3\right)\right]\)
\(=\left[25-9\right]:2.8\)
\(=14:2.8=7.8=56\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: -a - b - b = -a - b + c
Vậy: (-a-b+c) - (-a-b-c) = (-a-b+c) - (-a-b+c) = (-a-b+c) : 2
b) (-1-1+-2) : 2 = (-2+-2) : 2 = (-4) : 2 = -2
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ Ta có :
\(\left|x-5\right|\ge0\forall x\)
\(\Leftrightarrow\left|x-5\right|+3\ge3\forall x\)
\(\Leftrightarrow A\ge3\)
Dấu "=" xảy ra khi : \(\left|x-5\right|=0\)
\(\Leftrightarrow x=5\)
Vậy \(A_{Min}=3\Leftrightarrow x=5\)
b,c tương tự
a, \(5-\left(\frac{a}{b}+\frac{1}{2}\right)=2\frac{1}{3}\) => \(\frac{a}{b}+\frac{1}{2}=5-2\frac{1}{3}\) => \(\frac{a}{b}+\frac{1}{2}=\frac{8}{3}\) => \(\frac{a}{b}=\frac{8}{3}-\frac{1}{2}\) => \(\frac{a}{b}=\frac{13}{6}\)
b, \((\frac{3}{4}+2\frac{1}{2}):\frac{3}{5-3}=\left(\frac{3}{4}+\frac{5}{4}\right):\frac{3}{5}-1=\frac{9}{4}:\frac{-2}{5}=\frac{-45}{8}\)
a, 5-(\(\frac{a}{b}\)+\(\frac{1}{2}\))=2\(\frac{1}{3}\)
<=>5-\(\frac{a}{b}-\frac{1}{2}\)=\(\frac{7}{3}\)
<=>\(\frac{a}{b}=5-\frac{1}{2}-\frac{7}{3}\)
<=>\(\frac{a}{b}=\frac{13}{6}\)
b,(\(\frac{3}{4}\)+2\(\frac{1}{2}\)):\(\frac{3}{5}\)-3
=(\(\frac{3}{4}\)+\(\frac{5}{2}\)).\(\frac{5}{3}\)-3
=\(\frac{23}{4}\).\(\frac{5}{3}\)-3
=\(\frac{115}{12}\)-3
=\(\frac{115-36}{12}\)
=\(\frac{79}{12}\)