Tính : [tex]1\tfrac{4}{23} + \frac{5}{21} -\frac{4}{23} + 0,5+ \frac{16}{21}[/tex]
CMR: 10^6 - 5^7 :59
Tìm x ; y: [tex]\frac{5}{x}+ \frac{y}{5}=\frac{1}{8}[/tex]
Cho [tex]a^{2}=b.c[/tex]
CMR: a) [tex]\frac{c}{b}= \frac{a^2+c^2}{b^2+a^2}[/tex]
b) [tex]\frac{4^5.9^4-2.6^9}{2^10.3^8+6^8.20}[/tex]
Tính: $1 \dfrac{4}{23} + \dfrac{5}{21} - \dfrac{4}{23} + 0,5 + \dfrac{16}{21} \\ = 1 + \dfrac{4}{23} + \dfrac{5}{21} - \dfrac{4}{23} + 0,5 + \dfrac{16}{21} \\ = ( 1 + 0,5 ) + ( \dfrac{4}{23} - \dfrac{4}{23} ) + ( \dfrac{5}{21} + \dfrac{16}{21} \\ = 1,5 + 0 + 1 \\ = 2,5$
CMR: $10^6 - 5^7 :59$
$10^6 - 5^7 \\ = ( 2 . 5 )^6 - 5 . 5^6 \\ = 2^6 . 5^6 - 5 . 5^6 \\ = 5^6( 2^6 - 5 ) \\ = 5^6( 64 - 5 ) \\ = 5^6 . 59 \vdots 59 \\ Vậy 10^6 - 5^7 \vdots 59 $
Tìm $x;y {\color{Red} \in \mathbb{Z}}$
$\dfrac{5}{x} + \dfrac{y}{5} = \dfrac{1}{8} \\ \Leftrightarrow \dfrac{5}{x} = \dfrac{1}{8} - \dfrac{y}{5} \\ \Leftrightarrow \dfrac{5}{x} = \dfrac{5}{40} - \dfrac{8y}{40} \\ \Leftrightarrow \dfrac{5}{x} = \dfrac{5 - 8y}{40} \\ \Leftrightarrow x( 5 - 8y ) = 5 . 40 \\ \Leftrightarrow x( 5 - 8y ) = 200 \\ \Rightarrow x \in Ư(200); (5 - 8y) \in Ư(200)
\\\Rightarrow x \in \left \{ -200; -100; -50; -40; -25; -20; -10; -8; -5; -4; -2; -1; 1; 2; 4; 5; 8; 10; 20; 25; 40; 50; 100; 200 \right \}
\\\Rightarrow 5 - 8y \in \left \{ -1; -2; -4; -5; -8; -10; -20; -25; -40; -50; -100; -200; 200; 100; 50; 40; 25; 20; 10; 8; 5; 4; 2; 1 \right \} \\\Rightarrow - 8y \in \left \{ -6; -7; -9; -10; -13; -15; -25; -30; -45; -55; -105; -205; 195; 95; 45; 35; 20; 15; 5; 3; 0; -1; -3; -4\right \}
\\\Rightarrow y \in \left \{ \dfrac{6}{8} ; \dfrac{7}{8}; \dfrac{9}{8}; \dfrac{10}{8}; \dfrac{13}{8} ;\dfrac{15}{8}; \dfrac{25}{8}; \dfrac{30}{8}; \dfrac{45}{8}; \dfrac{5}{8}; \dfrac{105}{8}; \dfrac{205}{8}; \dfrac{-195}{8}; \dfrac{-95}{8}; \dfrac{-45}{8}; \dfrac{-35}{8}; \dfrac{-20}{8}; \dfrac{-15}{8}; \dfrac{-5}{8}; \dfrac{-3}{8}; 0; \dfrac{1}{8}; \dfrac{3}{8}; \dfrac{4}{8} \right \}
\\ Vì\; x, y \in \mathbb{Z} \Rightarrow y = 0 \Rightarrow 5 - 8y = 5 - 8 . 0 = 5 - 0 = 5 \Rightarrow x \frac{200}{5} = 40$
Cho [tex]a^{2}=b.c[/tex]
CMR: a)
$\dfrac{c}{b} = \dfrac{c(c + b)}{b(b + c)} \\\Leftrightarrow \dfrac{c}{b} = \dfrac{b.c + c^2}{b^2 + b.c}\\\Leftrightarrow \dfrac{c}{b}= \dfrac{a^2+c^2}{b^2+a^2}$