CM:\(\sqrt{10}+\sqrt{17}+1>\sqrt{61}\)

pp: dùng tính chất bắc cầu

có a > b và b > c suy ra a > c

giải:

+Vì\(10>9>0\Rightarrow\sqrt{10}>\sqrt{9}\)

      \(17>16>0\Rightarrow\sqrt{17}>\sqrt{16}\)

\(\Rightarrow\sqrt{10}+\sqrt{17}>\sqrt{9}+\sqrt{16}\)

\(\Leftrightarrow\sqrt{10}+\sqrt{17}+1>3+4+1\)

\(\Leftrightarrow\sqrt{10}+\sqrt{17}+1>8\) (1)

+Vì\(64>61>0\Rightarrow\sqrt{64}>\sqrt{61}\)

\(\Leftrightarrow8>\sqrt{61}\) (2)

Từ (1) và (2) suy ra \(\sqrt{10}+\sqrt{17}+1>\sqrt{61}\) (đpcm)

<không hiểu phần nào thì hỏi mình nhé>

\(\dfrac{x-3}{13}+\dfrac{x-3}{14}=\dfrac{x-3}{15}+\dfrac{x-3}{16}\)

\(\Leftrightarrow\dfrac{x-3}{13}+\dfrac{x-3}{14}-\dfrac{x-3}{15}-\dfrac{x-3}{16}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)

\(\Leftrightarrow x-3=0\) (vì\(\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{15}-\dfrac{1}{16}>0\) (*) )

\(\Leftrightarrow x=3\)

Chứng minh (*):

Vì\(13< 15\Rightarrow\dfrac{1}{13}>\dfrac{1}{15}\Rightarrow\dfrac{1}{13}-\dfrac{1}{15}>0\)

\(14< 16\Rightarrow\dfrac{1}{14}>\dfrac{1}{16}\Rightarrow\dfrac{1}{14}-\dfrac{1}{16}>0\)

Do đó:\(\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{15}-\dfrac{1}{16}>0\)

=> (*) luôn đúng

Vậy x = 3 là giá trị cần tìm.

c)\(0,625+\left(\dfrac{-2}{7}\right)+\dfrac{3}{8}+\left(-\dfrac{5}{7}\right)+1\dfrac{2}{3}\)

\(=\dfrac{5}{8}-\dfrac{2}{7}+\dfrac{3}{8}-\dfrac{5}{7}+\dfrac{5}{3}\)

\(=\left(\dfrac{5}{8}+\dfrac{3}{8}\right)-\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\dfrac{5}{3}\)

\(=1-1+\dfrac{5}{3}\)

\(=\dfrac{5}{3}\)

d)\(\left(-3\right).\left(\dfrac{-38}{21}\right).\left(\dfrac{-7}{6}\right).\left(\dfrac{-3}{19}\right)\)

\(=\dfrac{\left(-3\right).\left(-38\right).\left(-7\right).\left(-3\right)}{21.6.19}\)

\(=\dfrac{\left(-3\right).\left(-2\right).19.\left(-7\right).\left(-3\right)}{3.7.3.2.19}\)

\(=1\)

e)\(\left(\dfrac{11}{18}:\dfrac{22}{9}\right).\dfrac{8}{5}\)

\(=\left(\dfrac{11}{18}.\dfrac{9}{22}\right).\dfrac{8}{5}\)

\(=\dfrac{11.9}{18.22}.\dfrac{8}{5}\)

\(=\dfrac{11.9.2.2.2}{2.9.11.2.5}\)(có thể bỏ qua bước này)

\(=\dfrac{2}{5}\)

g)\(\left[\left(-\dfrac{4}{5}\right).\dfrac{5}{8}\right]:\left(-\dfrac{25}{12}\right)\)

\(=\left[\dfrac{\left(-4\right).5}{5.8}\right].\dfrac{-12}{25}\)

\(=\dfrac{-1}{2}.\dfrac{-12}{25}\)

\(=\dfrac{6}{25}\)

\(x^2+10x+25\)

\(=x^2+2.5x+5^2\)

\(=\left(x+5\right)^2\)