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1:
a: BC=8-3=5cm
b: MN=MC+CN=1/2(CA+CB)
=1/2*AB=4cm
2:
a: Có 2 tia là OA và OB
b: AB=OB+OA=11cm
c: AC=BC=11/2=5,5cm
a,\(\dfrac{2}{3}\)-\(\dfrac{5}{7}\).\(\dfrac{14}{25}\)
=\(\dfrac{2}{3}\)-\(\dfrac{2}{5}\)
=\(\dfrac{4}{15}\)
b,\(\dfrac{-2}{5}\).\(\dfrac{5}{8}\)+\(\dfrac{5}{8}\).\(\dfrac{3}{5}\)
=\(\dfrac{5}{8}\).(\(\dfrac{-2}{5}\)+\(\dfrac{3}{5}\))
=\(\dfrac{5}{8}\).\(\dfrac{1}{5}\)
=\(\dfrac{1}{8}\)
Giải:
a) \(\dfrac{2}{3}-\dfrac{5}{7}.\dfrac{14}{25}=\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{4}{15}\)
b) \(\dfrac{-2}{5}.\dfrac{5}{8}+\dfrac{5}{8}.\dfrac{3}{5}\)
\(=\dfrac{5}{8}.\left(\dfrac{-2}{5}+\dfrac{3}{5}\right)\)
\(=\dfrac{5}{8}.\dfrac{1}{5}\)
\(=\dfrac{1}{8}\)
c) \(\left(\dfrac{1}{2}\right)^2-1\dfrac{1}{2}+0,5.\dfrac{12}{5}+5\%\)
\(=\dfrac{1}{4}-\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{12}{5}+\dfrac{1}{20}\)
\(=\dfrac{-5}{4}+\dfrac{6}{5}+\dfrac{1}{20}\)
\(=0\)
Chúc bạn học tốt!
`B=(2015+2016+2017)/(2016+2017+2018)`
`=2015/(2016+2017+2018)+2016/(2016+2017+2018)+2017/(2016+2017+2018)`
Vì `2015/(2016+2017+2018)<2015/2016`
`2016/(2016+2017+2018)<2016/2017`
`2017/(2016+2017+2018)<2017/2018`
`=>B<A`
Bài 5
B= \(\dfrac{2015}{2016+2017+2018}\)+\(\dfrac{2016}{2016+2017+2018}\)+\(\dfrac{2017}{2016+2017+2018}\)
Ta có:\(\dfrac{2015}{2016}\)>\(\dfrac{2015}{2016+2017+2018}\),\(\dfrac{2016}{2017}\)>\(\dfrac{2016}{2016+2017+2018}\),\(\dfrac{2017}{2018}\)>\(\dfrac{2017}{2016+2017+2018}\)
⇒A>B
Đặt A = 3-3^2+3^3-...-3^100
3A=3^2-3^3+3^4-...-3^101
3A+A=3-3^101
4A=3-3^101
A=(3-3^101):4
Bài 8:
a: \(A=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\cdot\dfrac{6}{70}=\dfrac{6}{10}=\dfrac{3}{5}\)
b: \(B=2\left(\dfrac{1}{15}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{21}+...+\dfrac{1}{87}-\dfrac{1}{90}\right)\)
\(=2\left(\dfrac{1}{15}-\dfrac{1}{90}\right)\)
\(=2\cdot\dfrac{5}{90}=\dfrac{10}{90}=\dfrac{1}{9}\)
c: \(C=3\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
\(=3\cdot\dfrac{24}{200}=\dfrac{72}{200}=\dfrac{9}{25}\)
Dạng 3:
Bài 1:
a) Số lượng số hạng là:
\(\left(999-1\right):1+1=999\) (số hạng)
Tổng dãy là:
\(A=\left(999+1\right)\cdot999:2=499500\)
b) Số lượng số hạng là:
\(\left(100-7\right):3+1=32\) (số hạng)
Tổng dãy là:
\(S=\left(100+7\right)\cdot32:2=1712\)