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Câu 8:
Giải:
Ta có: \(a:b=3:4\Rightarrow\frac{a}{3}=\frac{b}{4}\Rightarrow\frac{a^2}{9}=\frac{b^2}{16}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a^2}{9}=\frac{b^2}{16}=\frac{a^2+b^2}{9+16}=\frac{36}{25}\)
+) \(\frac{a^2}{9}=\frac{36}{25}\Rightarrow a^2=\frac{324}{25}\Rightarrow a=\pm\frac{18}{5}\)
+) \(\frac{b^2}{16}=\frac{36}{25}\Rightarrow b^2=\frac{576}{25}\Rightarrow b=\pm\frac{24}{5}\)
Vậy bộ số \(\left(x;y\right)\) là \(\left(\frac{18}{5};\frac{24}{5}\right);\left(\frac{-18}{5};\frac{-24}{5}\right)\)
a) \(15.4=15.\left(2.2\right)=\left(15.2\right).2=30.2=60\)
\(25.12=25.\left(2.6\right)=\left(25.2\right).6=50.6=300\)
\(125.16=125.\left(2.8\right)=\left(125.8\right).2=1000.2=2000\)
b) \(25.12=25.\left(10+2\right)=25.10+25.2=250+50=300\)
\(34.11=34.\left(10+1\right)=34.10+34.1=340+34=374\)
\(47.101=47.\left(100+1\right)=47.100+47.1=4700+47=4747\)
Chúc em học tốt!!!
1. a) 3^2 .2^2 .2^4 = 3^2. 2^(2+4)=3^2. 2^6
b) 10^2. 10^3.10^5= 10^(2+3+5)= 10^10
c) x.x^5=x^(1+5)=x^6
d) a^3 .a^2 ,a^5= a^(3+2+5)= a^10
2. a) 2^3 và 3^2 . Ta có: 2^3 = 8, 3^2=9 => 2^3 < 3^2
b) 2^4 và 4^2. Ta có: 2^4= 16, 4^2=16 => 2^4 = 4^2
c) 2^5 và 5^2. Ta có: 2^5= 32, 5^2=25 => 2^5 > 5^2
d) 2^100 và 100. => 2^100 > 100
3. a) 3^8:3^4= 3^(8-4)=3^4
b) 10^8:10^2=10^(8-6)=10^2
c) a^6: a (a#0) = a^(6-1)=a^5
\(A=10.\left(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+....+\frac{71}{72}+\frac{89}{90}\right)\)
Đặt \(B=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{71}{72}+\frac{89}{90}\)
\(B=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{90}\right)\)
\(B=1+1+1+1+...+1-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{72}+\frac{1}{90}\right)\)
\(B=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(B=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=9-\left(\frac{1}{1}-\frac{1}{10}\right)=9-\frac{9}{10}=\frac{81}{10}=8,1\)
Ta có \(A=10.B=10.B=10.8,1=81\)
Vậy \(A=81\)
\(a,\dfrac{-5}{21}+\dfrac{-6}{42}=\dfrac{-5}{21}+\dfrac{-3}{21}=\dfrac{\left(-5\right)+\left(-3\right)}{21}=\dfrac{-8}{21}\)
\(b,\dfrac{16}{40}+\dfrac{-27}{45}=\dfrac{2}{5}+\dfrac{-3}{5}=\dfrac{2+\left(-3\right)}{5}=\dfrac{-1}{5}\)
\(c,\dfrac{-4}{18}+\dfrac{-12}{30}=\dfrac{-2}{9}+\dfrac{-2}{5}=\dfrac{-10}{45}+\dfrac{-18}{45}=\dfrac{\left(-10\right)+\left(-18\right)}{45}=\dfrac{-28}{45}\)
a)\(\dfrac{-5}{21}\)+\(\dfrac{-6}{42}\)= \(\dfrac{-10}{42}\)+\(\dfrac{-6}{42}\)= \(\dfrac{-16}{42}\)= \(\dfrac{-8}{21}\)
b)\(\dfrac{16}{40}\)+ \(\dfrac{-27}{45}\)= \(\dfrac{2}{5}\)+\(\dfrac{-3}{5}\)= \(\dfrac{-5}{5}\)= (-1)
c) \(\dfrac{-4}{18}\)+\(\dfrac{-12}{30}\)= \(\dfrac{-2}{9}\)+\(\dfrac{-2}{5}\)= \(\dfrac{-10}{45}\)+\(\dfrac{-18}{45}\)= \(\dfrac{-28}{45}\)
Đáp án là C
Ta có: (-30) + (-95) + 40 + 30 = [(-30) + 30] + [(-95) + 40] = 0 + (-55)
Vì -56 < -55 nên C đúng.