Hãy tính: 1/2 - 1/3; 1/3 - 1/4; 1/4 - 1/5 rồi hay rút ra kết quả tổng quát
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1.
y=f(-1)=3*(-1)-2=-5
y=f(0)=3*0-2=-2
y=f(-2)=3*(-2)-2=-8
y=f(3)=3*3-2=7
Câu 2,3a làm tương tự,chỉ việc thay f(x) thôi.
3b
Khi y=5 =>5=5-2*x=>2*x=0=> x=0
Khi y=3=>3=5-2*x=>2*x=2=>x=1
Khi y=-1=>-1=5-2*x=>2*x=6=>x=3
f(-1)=3.1-2=3-2=1
f(0)=3.0-2=0-2=-2
f(-2)=3.(-2)-2=-6-2=-8
f(3)=3.3-2=9-2=7
\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+20}\)
\(=\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{20\times21}\)
\(=2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{20\times21}\right)\)
\(=2\times\left(\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+...+\frac{21-20}{20\times21}\right)\)
\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\right)\)
\(=2\times\left(\frac{1}{2}-\frac{1}{21}\right)\)
\(=\frac{19}{21}\)
số đó là
5*289+4*289+3*289+2*289+289=4335
tích là
4335*1*2*3*4*5*6*7*8*9*10*11=173039328000
tích 1 là như cũ
tích 2 là 8670
tích 3 là 13005
tích 4 là 17340
còn lại tự tính
A = 4\(\dfrac{1}{3}\) + 3\(\dfrac{3}{4}\) + 2 - \(\dfrac{2}{5}\) + 8 - \(\dfrac{3}{5}\) + 7 - \(\dfrac{1}{4}\) + 6 - \(\dfrac{2}{3}\)
A = \(\dfrac{13}{3}\) + \(\dfrac{15}{4}\) + ( 2 + 8 + 6 + 7) - ( \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) - \(\dfrac{1}{4}\) - \(\dfrac{2}{3}\)
A = \(\dfrac{13}{3}\) + \(\dfrac{15}{4}\) + 23 - 1 - \(\dfrac{1}{4}\) - \(\dfrac{2}{3}\)
A = ( \(\dfrac{13}{3}\) - \(\dfrac{2}{3}\)) + 22 +( \(\dfrac{15}{4}\) - \(\dfrac{1}{4}\) )
A = \(\dfrac{11}{3}\) + 22 + \(\dfrac{14}{4}\)
A = \(\dfrac{11}{3}\) + 22 + \(\dfrac{7}{2}\)
A = \(\dfrac{22}{6}\) + \(\dfrac{132}{6}\) + \(\dfrac{21}{6}\)
A = \(\dfrac{175}{6}\)
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{6}{12}+\frac{4}{12}+\frac{3}{12}\)
\(=\frac{6+4+3}{12}\)
\(=\frac{13}{12}\)
\(=1\frac{1}{12}\)
Bài 3:
Ta có: \(A=\cos^220^0+\cos^240^0+\cos^250^0+\cos^270^0\)
\(=\left(\sin^270^0+\cos^270^0\right)+\left(\sin^250^0+\cos^250^0\right)\)
=1+1
=2
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)
\(=\frac{1}{2}+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)-\frac{1}{5}\)
\(=\frac{1}{2}+0+0-\frac{1}{5}\)
\(=\frac{1}{2}-\frac{1}{5}\)
\(=\frac{5}{10}-\frac{2}{10}=\frac{5-2}{10}=\frac{3}{10}\)