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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: \(C=3\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
\(=3\cdot\dfrac{24}{200}=\dfrac{72}{200}=\dfrac{9}{25}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Nếu đề là \(\frac{23.5-23}{4-27}\) thì ok :v
\(\frac{23.5-23}{4-27}=\frac{23.5-23}{-23}=\frac{23.\left(5-1\right)}{-23}=\frac{23.4}{-23}=\frac{92}{-23}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
23-23.5 phần 4-27=23.(1-5)phần -23=23.-4 phần -23=-23.4 phần -23=4
Đáp số 4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{23.5-23}{4-27}\)=\(\frac{23.\left(5-1\right)}{-23}\)=\(\frac{92}{-23}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì ( x + 23 ) . ( x - 8 ) = 0
nên ( x + 23 ) hoặc ( x - 8 ) bằng 0
x + 23 = 0 x - 8 = 0
x = 0 - 23 x = 0 + 8
x = -23 x = 8
Vậy x thuộc { -23;8}
![](https://rs.olm.vn/images/avt/0.png?1311)
+)Theo bài a\(\in\)B(b)
=>|a|\(\in\)B( |b| )
=>|a|\(\in\)B(b)
Chúc bn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=-\left(a-b-c\right)+\left(-c+b+a\right)-\left(a+b\right)\)
\(=-a+b+c-c+b+a-a-b\)
\(=-a+b\)
= \(b-a\)
Vì a > b \(\Rightarrow b-a< 0\) \(\Rightarrow|S|=\left(-1\right).\left(b-a\right)=-b+a=a-b\)
\(\Rightarrow|S|=a-b\)
Kb mình nha!
S = -(a-b-c)+(-c+b+a)-(a+b)
= - a+b+c-c+b+a-a+b
=(-a+a-a)+(b+b-b)+(c-c)
= -a+b+0
=b-a
Vì a>b nên |S| =a-b
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{23.5-23}{-27}\)
\(=\frac{23.\left(5-1\right)}{-27}\)
\(=\frac{23.4}{-27}\)
\(=\frac{92}{-27}\)
\(=\frac{-92}{27}\)
27.125
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