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MH
16 tháng 9 2021
a) \(-y^2+\dfrac{1}{9}\)
\(=-\left(y^2-\left(\dfrac{1}{3}\right)^2\right)\)
\(=-\left(y+\dfrac{1}{3}\right)\left(y-\dfrac{1}{3}\right)\)
b) \(4^4-256\)
\(=4^4-4^4\)
\(=0\)
B
1
HQ
Hà Quang Minh
Giáo viên
12 tháng 1 2024
Xét tứ giác EFGH có:
\(\) \(\widehat E + \widehat F + \widehat G + \widehat H = {360^o}\)(định lí tổng các góc trong một tứ giác).
Hay \({90^o} + \widehat F + {90^o} + {55^o} = {360^o}\)
Suy ra \(\widehat F\)+235°=360°
Do đó \(\widehat F\)=360°−235°=125°
Vậy \(\widehat F\)=125o
27 tháng 9 2016
a, 4n+4-3.4n+2
=4n+2(42-3)
=4n+2.16 chia hết cho 16 => ................
b tương tự đặt 7n làm nhân tử chung
PT
17 tháng 10 2017
1. \(\dfrac{1}{x-1}-\dfrac{1}{x+1}\)
\(=\dfrac{1.\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}-\dfrac{x-1}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x+1+\left(-x+1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x+1-x+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{1}{x^2-1}\)
2. \(\dfrac{x}{x^2-1}-\dfrac{1}{x-1}\)
\(=\dfrac{x}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x}{\left(x+1\right)\left(x-1\right)}+\dfrac{-\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x+\left(-x-1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{-1}{x^2-1}\)
3. \(\dfrac{1}{x\left(x-y\right)}-\dfrac{1}{x\left(x-y\right)}\)
\(=\dfrac{1}{y\left(x-y\right)}+\dfrac{-1}{x\left(x-y\right)}\)
\(=\dfrac{1x}{y\left(x-y\right)x}+\dfrac{-1y}{x\left(x-y\right)y}\)
\(=\dfrac{x}{xy\left(x-y\right)}+\dfrac{-y}{xy\left(x-y\right)}\)
\(=\dfrac{x-y}{xy\left(x-y\right)}=\dfrac{1}{xy}\)
4. \(\dfrac{1}{x}-\dfrac{1}{x-1}\)
\(=\dfrac{1\left(x-1\right)}{x\left(x-1\right)}-\dfrac{1x}{\left(x-1\right)x}\)
\(=\dfrac{x-1}{x\left(x-1\right)}+\dfrac{-x}{x\left(x-1\right)}\)
\(=\dfrac{\left(x-1\right)-x}{x\left(x-1\right)}\)
\(=\dfrac{-1}{x\left(x-1\right)}\)
5. \(\dfrac{1}{x}-\dfrac{1}{x+1}\)
\(=\dfrac{1\left(x+1\right)}{x\left(x+1\right)}-\dfrac{1x}{\left(x+1\right)x}\)
\(=\dfrac{x+1}{x\left(x+1\right)}+\dfrac{-x}{x\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}\)
6. \(\dfrac{1}{2x^2-10x}-\dfrac{1}{x-5}\)
\(=\dfrac{1}{2x\left(x-5\right)}-\dfrac{1}{x-5}\)
\(=\dfrac{1}{2x\left(x-5\right)}-\dfrac{1.2x}{2x\left(x-5\right)}\)
\(=\dfrac{1}{2x\left(x-5\right)}+\dfrac{-2x}{2x\left(x-5\right)}\)
\(=\dfrac{1-2x}{2x\left(x-5\right)}\)
7. \(\dfrac{x-1}{x^2-1}.\dfrac{x+1}{x+3}\)
\(=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x^2-1\right)\left(x+3\right)}\)
\(=\dfrac{x^2-1}{\left(x^2-1\right)\left(x+3\right)}\)
8. \(\dfrac{2}{2x^2+10x}.\dfrac{x+5}{3x}\)
\(=\dfrac{2x\left(x+5\right)}{2x^2+10x.3x}\)
\(=\dfrac{2\left(x+5\right)}{2x\left(x+5\right)3x}\)
\(=\dfrac{2}{6x^2}=\dfrac{1}{3x^2}\)
3 tháng 7 2016
Ta có:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2014}+\frac{1}{2015}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+\frac{1}{2015}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)=\frac{1}{1008}+\frac{1}{1009}+....+\frac{1}{2015}\)
Mà \(P=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}\)
\(\Leftrightarrow S-P=0\) \(\Rightarrow\left(S-P\right)^{2016}=0\)
\(=\dfrac{1}{4}\left(\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{51.55}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{51}-\dfrac{1}{55}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{55}\right)=\dfrac{1}{4}\times\dfrac{52}{165}=\dfrac{13}{165}\)