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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:
\(S=\left(100+7\right)\cdot32:2=1712\)
a) Ta có: \(A=\dfrac{7}{12}+\dfrac{5}{12}:6-\dfrac{11}{36}\)
\(=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}\)
\(=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}\)
\(=\dfrac{25}{36}\)
b) Ta có: \(B=\left(\dfrac{4}{5}+\dfrac{1}{2}\right):\left(\dfrac{3}{13}-\dfrac{8}{13}\right)\)
\(=\left(\dfrac{8}{10}+\dfrac{5}{10}\right):\dfrac{-5}{13}\)
\(=\dfrac{13}{10}\cdot\dfrac{13}{-5}\)
\(=-\dfrac{169}{50}\)
c) Ta có: \(C=\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right):\left(\dfrac{5}{12}+1-\dfrac{7}{11}\right)\)
\(=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{115}{132}\cdot\dfrac{132}{103}=\dfrac{115}{103}\)
B2"
`a)3/4+1/2-1/4`
`=3/4-1/4+1/2`
`=1/2+1/2=1`
`b)(-2)/3*5/7+(-2)/3*2/7+5/3`
`=(-2)/3*(5/7+2/7)+5/3`
`=-2/3+5/3=1`
`c)(-5)/9+5/9:(1 2/3-2 1/6)`
`=(-5)/9+5/9:(5/3-13/6)`
`=(-5)/9+5/9:(-3)/6`
`=(-5)/9+5/9*(-2)`
`=5/9*(-1-2)`
`=5/9*(-3)=-5/3`
b3:
`a)x*3/6=2/3`
`=>x*1/2=2/3`
`=>x=4/3`
`b)x/150=5/6*(-7)/25`
`=>x/150=(-7)/(6*5)=-7/30`
`=>x/150=(-35)/150`
`=>x=-35`
`c)1/2x+3/5x=3`
`=>11/10x=3`
`=>x=3*10/11=30/11`
`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
`(-3)/7+15/26+(-2/13+3/7)`
`=(-3)/7+3/7+15/(26)-2/13`
`=15/26-4/26=11/26`
`(-4)/5*11/13+4/5*(-2)/13-1/5`
`=(-4)/5*(11/13+2/13)-1/5`
`=(-4)/5-1/5=-1`
Bài 1:
a) \(\dfrac{4}{5}-\dfrac{7}{6}+\dfrac{-6}{15}=\dfrac{24}{30}-\dfrac{35}{30}+\dfrac{-12}{30}=\dfrac{24-35+-12}{30}=\dfrac{-23}{30}\)
b) \(\dfrac{-5}{9}.\dfrac{7}{13}+\dfrac{6}{13}.\dfrac{-5}{9}+3\dfrac{7}{9}\)
\(=\dfrac{-5}{9}.\left(\dfrac{7}{13}+\dfrac{6}{13}\right)+\dfrac{34}{9}\)
\(=\dfrac{-5}{9}.1+\dfrac{34}{9}\)
\(=\dfrac{-5}{9}+\dfrac{34}{9}\)
\(=\dfrac{29}{9}\)
c) \(6\dfrac{3}{8}-\left(4\dfrac{3}{8}-\dfrac{1}{2}\right)=\dfrac{51}{8}-\dfrac{35}{8}+\dfrac{1}{2}=\left(\dfrac{51}{8}-\dfrac{35}{8}\right)+\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\)
d) \(2\dfrac{1}{3}.1,5-\left(\dfrac{11}{10}+50\%\right):\dfrac{4}{15}\)
\(=\dfrac{7}{3}.1,5-\dfrac{8}{5}:\dfrac{4}{15}\)
\(=\dfrac{7}{2}-6\)
\(=\dfrac{-5}{2}=-2,5\)
a: Tỉ số là 3/2
b: Tỉ số phần trăm là;
40/(30+40+20+20+5)=34,78%
Số tiền Nam mua sách: \(320000\times\dfrac{1}{4}=80000\) (đồng)
Số tiền Nam mua vở: \(90000:\dfrac{2}{3}=135000\) (đồng)
Số tiền Nam mua dụng cụ học tập: \(320000-\left(80000+135000\right)=105000\) (đồng)