\(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)

\(\dfrac{1}{3}x+\dfrac{2}{5}x-\dfrac{2}{5}=0\\ \dfrac{11}{15}x=\dfrac{2}{5}\\ x=\dfrac{6}{11}\)

a. \(-0,8-\dfrac{7}{13}\cdot\dfrac{26}{39}=-\dfrac{4}{5}-\dfrac{14}{39}=-\dfrac{156}{195}-\dfrac{70}{195}=-\dfrac{226}{195}\)

Để B có giá trị lớn nhất thì \(2\left(n-1\right)^2+3\) phải nhỏ nhất

Mà \(2\left(n-1\right)^2+3\ge3\forall x\)

\(\Rightarrow MAX_B=\dfrac{1}{3}\) khi \(n-1=0\Leftrightarrow n=1\)

Áp dụng bất đẳng thức tam giác ta có

\(BC< AB+AC=5+1=6\)

\(BC>AB-AC=5-1=4\)

\(\Rightarrow4< BC< 6\)

Mà BC là số nguyên \(\Rightarrow BC=5\Rightarrow\Delta ABC\)cân tại B

\(Q\left(x\right)=0\)

\(\Leftrightarrow x-5x^2=0\\ \Leftrightarrow x\left(1-5x\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x=0\\1-5x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)

a) \(\dfrac{-23}{3}=7\dfrac{-2}{3}\)

b) \(123,45=\dfrac{12345}{100}\cdot100\%=123,45\%\)

a) \(\dfrac{8}{5}-\dfrac{9}{5}=\dfrac{8-9}{5}=\dfrac{-1}{5}\)

b) \(\dfrac{5}{2}+\dfrac{2}{3}=\dfrac{15}{6}+\dfrac{4}{6}=\dfrac{15+4}{6}=\dfrac{19}{6}\)

c) \(\dfrac{-5}{9}\cdot\dfrac{2}{11}=\dfrac{-5\cdot2}{9\cdot11}=\dfrac{-10}{99}\)

d) \(\dfrac{-2}{9}:\dfrac{1}{3}=\dfrac{-2}{9}\cdot3=\dfrac{-2}{3}\)

e) \(\dfrac{3}{8}-\dfrac{1}{4}+\dfrac{5}{12}=\dfrac{9}{24}-\dfrac{6}{24}+\dfrac{10}{24}=\dfrac{9-6+10}{24}=\dfrac{13}{24}\)

f) \(\dfrac{-4}{3}\cdot\dfrac{5}{4}:\dfrac{7}{3}=\dfrac{-4}{3}\cdot\dfrac{5}{4}\cdot\dfrac{3}{7}=\dfrac{-4\cdot5\cdot3}{3\cdot4\cdot7}=\dfrac{-5}{7}\)

\(-\left(-23\right)+\left(-36\right)+\left|-57\right|-\left(-29\right)-35=23-36+57+29-3=38\)

\(\dfrac{5}{4}+\dfrac{9}{11}\cdot\dfrac{11}{12}=\dfrac{5}{4}+\dfrac{9}{12}=\dfrac{5}{4}+\dfrac{3}{4}=\dfrac{8}{4}=2\)

\(\dfrac{8}{5}+0,25\cdot\dfrac{16}{15}-\dfrac{13}{15}=\dfrac{8}{5}+\dfrac{1}{4}\cdot\dfrac{16}{15}-\dfrac{13}{15}=\dfrac{8}{5}+\dfrac{4}{15}-\dfrac{13}{15}=\dfrac{8}{5}-\dfrac{3}{5}=\dfrac{5}{5}=1\)