Gukmin
Giới thiệu về bản thân
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)\(\left[\left(-\dfrac{4}{5}\right).\dfrac{5}{8}\right]:\left(-\dfrac{25}{12}\right)\)
\(=\left[\dfrac{\left(-4\right).5}{5.8}\right].\dfrac{-12}{25}\)
\(=\dfrac{-1}{2}.\dfrac{-12}{25}\)
\(=\dfrac{6}{25}\)
\(x^2+10x+25\)
\(=x^2+2.5x+5^2\)
\(=\left(x+5\right)^2\)