2/5 + 2/5 x 3/4 = ?
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\(Đk:\) \(x\ne1,x\ne2,x\ne3\)
\(\Rightarrow\dfrac{x+4}{\left(x-2\right)\left(x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(x-1\right)}=\dfrac{2x+5}{\left(x-3\right)\left(x-1\right)}\)
\(\Rightarrow\dfrac{\left(x+4\right)\cdot\left(x-3\right)+\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)\left(x-3\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(x-3\right)\left(x-1\right)\left(x-2\right)}\)
\(\Rightarrow x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)
\(\Rightarrow0x-14=x-10\)
\(\Rightarrow x=-4\left(tmđk\right)\)
a, \(x-\dfrac{3}{5}=\dfrac{4-14}{7}=\dfrac{-10}{7}\Leftrightarrow x=-\dfrac{10}{7}+\dfrac{3}{5}=-\dfrac{29}{35}\)
b, \(x=\dfrac{5}{9}-\dfrac{2}{9}=\dfrac{3}{9}=\dfrac{1}{3}\)
Ta có: \(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
.....................
\(\frac{1}{2014^2}< \frac{1}{2013.2014}\)
\(\Rightarrow A< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)
Đặt \(B=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=\frac{1}{2}-\frac{1}{2014}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2^2}+\frac{1}{2}=\frac{3}{4}\)
\(\text{Ta có: }n^2>n^2-1=\left(n-1\right)\left(n+1\right)\)
\(\Rightarrow\frac{1}{n^2}< \frac{1}{\left(n-1\right)\left(n+1\right)}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2014^2}< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)+...+\frac{1}{2}\left(\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{2014}-\frac{1}{2015}\right)\)
\(=\frac{1}{2}\left(\frac{3}{2}-\frac{1}{2014}-\frac{1}{2015}\right)\)
\(=\frac{3}{4}-\frac{1}{2}\left(\frac{1}{2014}+\frac{1}{2015}\right)< \frac{3}{4}\)
Vậy .............
Bài 1:
câu a: 4\(\dfrac{4}{9}\) : 2\(\dfrac{2}{3}\) + 3\(\dfrac{1}{6}\)
= \(\dfrac{40}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{19}{6}\)
= \(\dfrac{5}{3}\) + \(\dfrac{19}{6}\)
= \(\dfrac{10}{6}\) + \(\dfrac{19}{6}\)
= \(\dfrac{29}{6}\)
b, (15,25 + 3,75) \(\times\) 4 + ( 20,71 + 5,29)\(\times\) 5
= 19 \(\times\) 4 + 26 \(\times\) 5
= 76 + 130
= 206
c, \(\dfrac{4}{5}\) \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{4}{5}\) \(\times\) \(\dfrac{1}{3}\) - \(\dfrac{4}{5}\) \(\times\) \(\dfrac{1}{4}\)
= \(\dfrac{2}{5}\) + \(\dfrac{4}{15}\) - \(\dfrac{1}{5}\)
= \(\dfrac{6}{15}\) + \(\dfrac{4}{15}\) - \(\dfrac{3}{15}\)
= \(\dfrac{7}{15}\)
d, 1\(\dfrac{5}{7}\) + 7\(\dfrac{3}{6}\) + 2\(\dfrac{2}{7}\) - 4\(\dfrac{3}{6}\)
= (1 + 2 + \(\dfrac{5}{7}\) + \(\dfrac{2}{7}\)) + ( 7 + \(\dfrac{3}{6}\) - 4 - \(\dfrac{3}{6}\))
= 3 + 1 + 3
= 7
a: \(\dfrac{1}{7}\cdot\dfrac{3}{8}+\dfrac{1}{7}\cdot\dfrac{5}{8}+\dfrac{\left(-1\right)^{2023}}{7}\)
\(=\dfrac{1}{7}\left(\dfrac{3}{8}+\dfrac{5}{8}\right)-\dfrac{1}{7}\)
\(=\dfrac{1}{7}-\dfrac{1}{7}=0\)
b: \(-3-\dfrac{16}{23}-\sqrt{\dfrac{4}{49}}-\dfrac{7}{23}+\dfrac{\left(-3\right)^2}{7}\)
\(=-3-\left(\dfrac{16}{23}+\dfrac{7}{23}\right)-\dfrac{2}{7}+\dfrac{9}{7}\)
\(=-3-\dfrac{23}{23}+\dfrac{7}{7}\)
=-3-1+1
=-3
c: \(\dfrac{4^2\cdot0,2^3}{2^6}\)
\(=\dfrac{2^4\cdot0,008}{2^6}=\dfrac{0.008}{4}=0.002\)
\(\dfrac{3}{5}\times\dfrac{4}{7}\times1\dfrac{1}{2}\)
\(=\dfrac{3}{5}\times\dfrac{4}{7}\times\dfrac{3}{2}\)
\(=\dfrac{3\times4\times3}{5\times7\times2}\)
\(=\dfrac{3\times2\times3}{5\times7}\)
\(=\dfrac{18}{35}\)
\(\Leftrightarrow2\sqrt{x-4}=5\left(x\ge4\right)\\ \Leftrightarrow\sqrt{x-4}=\dfrac{5}{2}\\ \Leftrightarrow x-4=\dfrac{25}{4}\\ \Leftrightarrow x=\dfrac{41}{4}\left(tm\right)\)
2/5 + 2/5 x 3/4
=2/5 x (1 + 3/4)
=2/5 x 7/4
=7/10
7/10 nha cậu