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a: \(\widehat{xOm};\widehat{nOm}\)
\(\widehat{nOm};\widehat{yOn}\)
\(\widehat{yOn};\widehat{xOn}\)
\(\widehat{xOm};\widehat{yOm}\)
b: \(\widehat{yOn};\widehat{xOn}\)
\(\widehat{xOm};\widehat{yOm}\)
c: \(\widehat{mOy}=180^0-30^0=150^0\)
\(\widehat{nOx}=180^0-50^0=130^0\)
\(\widehat{mOn}=180^0-80^0=100^0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(0,5\right)^{12}:\left(0,5\right)^{10}=\left(0,5\right)^{12-10}=\left(0,5\right)^2\)
b) \(\sqrt{36}=\pm6\)
c)\(\left(0,75\right)^{22}:\left(0,75\right)^{12}=\left(0,75\right)^{22-12}=\left(0,75\right)^{10}\)
d) \(\sqrt{49}=\pm7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 14:
Để số đó nhân với \(\dfrac{7}{25},\dfrac{9}{35},\dfrac{11}{40}\)là tích các số TN thì số đó phải là BCNN(25,35,40)
25=52
35=5.7
40=23.5
BCNN(25,35,40)=52.23.7=1400
Vậy số đó là 1400
Bài 16:
a) Ta có: \(A=-\dfrac{9}{5}\)
nên \(\dfrac{3x+9}{17-2x}=\dfrac{-9}{5}\)
\(\Leftrightarrow15x+45=-9\left(17-2x\right)\)
\(\Leftrightarrow15x+45+9\left(17-2x\right)=0\)
\(\Leftrightarrow15x+45+153-18x=0\)
\(\Leftrightarrow-3x=-198\)
hay x=66
b) Thay x=3 vào A, ta được:
\(A=\dfrac{3\cdot3+9}{17-2\cdot3}=\dfrac{18}{17-6}=\dfrac{18}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Ta có: \(\dfrac{3}{n\left(n+3\right)}=\dfrac{\left(n+3\right)-n}{n\left(n+3\right)}=\dfrac{n+3}{n\left(n+3\right)}-\dfrac{n}{n\left(n+3\right)}=\dfrac{1}{n}-\dfrac{1}{n+3}\)
Do đó:
\(C=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+..+\dfrac{1}{37}-\dfrac{1}{40}\)
\(=1-\dfrac{1}{40}=\dfrac{39}{40}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(144=2^4\cdot3^2\)
\(120=2^3\cdot3\cdot5\)
\(35=7\cdot5\)
\(\RightarrowƯCLN\left(144,120,35\right)=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-1\right)^5=x^5\)
\(\Leftrightarrow2x-1=x\)
\(\Leftrightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
\(a^3+b^3+...+n^3=\left(a+b+...+n\right)^2\)
\(\Rightarrow1^3+2^3+3^3+...+10^3=\left(1+2+3+...+100\right)^2\)
\(1+2+3+...+100\)
Số số hạng
\(\left(100-1\right):1+1=100\)
Tổng
\(\left(100+1\right)\cdot100:2=5050\)
\(5050^2=25502500\)
Vậy \(1^3+2^3+...+100^3=25502500\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A= \(\dfrac{10.11.\left(1+5.5+7.7\right)}{11.12.\left(1+5.5+7.7\right)}=\dfrac{10}{12}=\dfrac{5}{6}\)
`Answer:`
a. \(\frac{5}{16}-\frac{5}{24}\)
\(=\frac{15}{48}-\frac{10}{48}\)
\(=\frac{5}{48}\)
b. \(-\frac{5}{8}+\frac{12}{7}+\frac{13}{8}+\frac{2}{7}\)
\(=\left(-\frac{5}{8}+\frac{13}{8}\right)+\left(\frac{12}{7}+\frac{2}{7}\right)\)
\(=1+2\)
\(=3\)
c. \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2.\left(1-\frac{1}{100}\right)\)
\(=2.\frac{99}{100}\)
\(=\frac{99}{50}\)