Tính giá trị các biểu thức:
A=(-1)+(-5)+(-9)+...+(-101)
B=\(\dfrac{-5}{17}\).\(\dfrac{8}{19}\)+\(\dfrac{-12}{17}\). \(\dfrac{8}{19}\) - \(\dfrac{11}{19}\)
C=\(\dfrac{10}{1.6}\)+\(\dfrac{10}{6.11}\)+\(\dfrac{10}{11.16}\)+...+\(\dfrac{10}{2016.2021}\)
`#lv`
`A=(-1)+(-5)+(-9)+...+(-101)`
`=-(1+5+9+...+101)`
Số số hạng là :
`[101-(-1)]:4+1=26(` số hạng `)`
Tổng là :
`[(-101)+(-1)]xx26:2=-1326`
Vậy `A=-1326`
__
`B=-5/17 . 8/19 + (-12)/17 . 8/19 - 11/19`
`=((-5)/17+(-12)/17).8/19-11/19`
`=-1.8/19-11/19`
`=-8/19-11/19`
`=-8/19+(-11)/19`
`=-19/19`
`=-1`
__
`C=10/1.6 + 10/6.11 + 10/11.16 + ... + 10/2016.2021`
`=2.(1-1/6+1/6-1/11+...+1/2016-1/2021)`
`=2(1-1/2021)`
`=2. (2021/2021-1/2021)`
`=2. 2020/2021`
`=4040/2021`
Tính giá trị các biểu thức sau một cách hợp lí :
\(A=\dfrac{7}{19}.\dfrac{8}{11}+\dfrac{7}{19}.\dfrac{3}{11}+\dfrac{12}{19}\)
\(B=\dfrac{5}{9}.\dfrac{7}{13}+\dfrac{5}{9}.\dfrac{9}{13}-\dfrac{5}{9}.\dfrac{3}{13}\)
\(C=\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right).\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.
Giá trị của tích \(\dfrac{1}{11}\) . \(\dfrac{1}{12}\)bằng giá trị của biểu thức nào sau đây?
A.\(\dfrac{1}{12}\)-\(\dfrac{1}{11}\) B.\(\dfrac{1}{23}\) C.\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\) D.\(\dfrac{1}{11}\)-\(\dfrac{1}{12}\)
Bài 1: Thực hiện phép tính:
1) \(\dfrac{-17}{30}-\dfrac{11}{-15}+\dfrac{-7}{12}\)
2) \(\dfrac{-5}{9}+\dfrac{5}{9}:\left(1\dfrac{2}{3}-2\dfrac{1}{12}\right)\)
3) \(\dfrac{-7}{25}.\dfrac{11}{13}+\dfrac{-7}{25}.\dfrac{2}{13}-\dfrac{18}{25}\)
1) âm năm phần 12
2) âm mười bảy phần 9
3) -1
Đây là đáp án còn làm bài từ làm nhé
\(\dfrac{1}{8}-\dfrac{1}{2}+(\dfrac{-11}{12}+1)\)
\(\dfrac{3}{5}-\dfrac{-7}{10}+\dfrac{-13}{10}\)
`1/8 -1/2 + (-11/12 + 1)`
`=1/8 -1/2 + (-11/12 +12/12)`
`=1/8 -1/2 + 1/12`
`= 1/8- 4/8+1/12`
`= -3/8 + 1/12`
`=-7/24`
`---------`
`3/5 -(-7/10) + (-13/10)`
`= 3/5 + 7/10 + (-13/10)`
`= 6/10 + 7/10 + (-13/10)`
`= 13/10 +(-13/10)`
`= 0/10=0`
\(\dfrac{1}{8}-\dfrac{1}{2}+\left(\dfrac{-11}{12}+1\right)\\ =\dfrac{-3}{8}+\dfrac{1}{12}\\ =\dfrac{-7}{24}\\ \dfrac{3}{5}-\dfrac{-7}{10}+\left(-\dfrac{13}{10}\right)\\ =\dfrac{13}{10}-\dfrac{13}{10}\\ =0\)
\(\dfrac{1}{8}-\dfrac{1}{2}+\left(\dfrac{-11}{12}+1\right)=\dfrac{-3}{8}+\dfrac{1}{12}=\dfrac{-7}{24}\)
\(\dfrac{3}{5}-\dfrac{-7}{10}+\dfrac{-13}{10}=\dfrac{6}{10}-\dfrac{-7}{10}+\dfrac{-13}{10}=\dfrac{13}{10}+\dfrac{-13}{10}=\dfrac{0}{10}=0\)
Tính hợp lý nếu có thể:
a) \(\dfrac{2}{9}+\dfrac{-3}{10}+\dfrac{-7}{10}\)
b) \(\dfrac{-11}{6}+\dfrac{2}{5}+\dfrac{-1}{6}\)
c) \(\dfrac{27}{13}-\dfrac{106}{111}+\dfrac{-5}{111}\)
d) \(\dfrac{12}{11}-\dfrac{-7}{19}+\dfrac{12}{19}\)
a, \(=\dfrac{2}{9}-\dfrac{10}{10}=\dfrac{2}{9}-1=-\dfrac{7}{9}\)
b, \(=-\dfrac{12}{6}+\dfrac{2}{5}=-2+\dfrac{2}{5}=-\dfrac{8}{5}\)
c, \(=\dfrac{27}{13}-1=\dfrac{14}{13}\)
d, \(=\dfrac{12}{11}+\dfrac{7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+1=\dfrac{23}{11}\)
a) \(\dfrac{2}{9}+\dfrac{-3}{10}+\dfrac{-7}{10}\)
\(=\dfrac{2}{9}+\dfrac{-10}{10}=\dfrac{2}{9}-1=-\dfrac{7}{9}\)
b) \(\dfrac{-11}{6}+\dfrac{2}{5}+\dfrac{-1}{6}\)
\(=-2+\dfrac{2}{5}=-\dfrac{8}{5}\)
a,\(=\dfrac{2}{9}+\left(\dfrac{-3}{10}+\dfrac{-7}{10}\right)=\dfrac{2}{9}+\left(-1\right)=\dfrac{-7}{9}\)
b,\(=\left(\dfrac{-11}{6}+\dfrac{-1}{6}\right)+\dfrac{2}{5}=-2+\dfrac{2}{5}=\dfrac{-8}{5}\)
c,\(=\dfrac{27}{13}-\left(\dfrac{106}{111}+\dfrac{-5}{111}\right)=\dfrac{27}{13}-1=\dfrac{14}{13}\)
Tính rồi rút gọn (theo mẫu):
| Mẫu: \(\dfrac{9}{10}-\dfrac{4}{10}=\dfrac{9-4}{10}=\dfrac{5}{10}=\dfrac{1}{2}\) |
a) \(\dfrac{15}{8}-\dfrac{13}{8}\) b) \(\dfrac{7}{15}-\dfrac{2}{15}\) c) \(\dfrac{11}{12}-\dfrac{2}{12}\) d) \(\dfrac{19}{7}-\dfrac{5}{7}\)
a: \(\dfrac{15}{8}-\dfrac{13}{8}=\dfrac{15-13}{8}=\dfrac{2}{8}=\dfrac{1}{4}\)
b: \(\dfrac{7}{15}-\dfrac{2}{15}=\dfrac{7-2}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)
c: \(\dfrac{11}{12}-\dfrac{2}{12}=\dfrac{11-2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)
d: \(\dfrac{19}{7}-\dfrac{5}{7}=\dfrac{19-5}{7}=\dfrac{14}{7}=2\)
\(\dfrac{x+1}{10}\)+\(\dfrac{x+1}{11}\)+\(\dfrac{x+1}{12}\)=\(\dfrac{x+1}{13}\)+\(\dfrac{x+1}{14}\)
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)\(\left(x+1\right)\times\dfrac{1}{10}+\left(x+1\right)\times\dfrac{1}{11}+\left(x+1\right)\times\dfrac{1}{12}-\left(x+1\right)\times\dfrac{1}{13}-\left(x+1\right)\times\dfrac{1}{14}=0\)
\(\left(x+1\right)\times\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
Vì \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}>0\)
=> \(x+1=0\)
\(x=0-1\)
\(x=-1\)
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\\ \Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\\ \Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\\ \Rightarrow x+1=0\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\\ \Rightarrow x=-1\)
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
Lời giải:
$\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}$
$\Rightarrow (x+1)(\frac{1}{10}+\frac{1}{11}+\frac{1}{12})=(x+1)(\frac{1}{13}+\frac{1}{14})$
$\Rightarrow (x+1)(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14})=0$
Hiển nhiên $\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0$
$\Rightarrow x+1=0$
$\Rightarrow x=-1$
