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\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}.\dfrac{5}{1}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{1}{3}.2+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{1}{3}.\left(\dfrac{2}{5}+\dfrac{3}{5}-2\right)\)=\(\dfrac{1}{3}.\left(-1\right)=\dfrac{-1}{3}\)
\(=4+\dfrac{1}{6}-2-\dfrac{1}{3}+3+\dfrac{7}{12}-5-\dfrac{3}{4}\)
\(=\dfrac{1}{6}-\dfrac{1}{3}+\dfrac{7}{12}-\dfrac{3}{4}=\dfrac{2}{12}-\dfrac{4}{12}+\dfrac{7}{12}-\dfrac{9}{12}=0\)
d: =-2/7-3/11+2/7=-3/11
e: =2+3/7+1+4/7-17/7
=4-17/7=11/7
f: =-2/3*4/5+1/5*11/9
=-8/15+11/45
=-24/45+11/45=-13/45
h: =(-3,1+3,7):0,2=0,6:0,2=3
b\()\)
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 3/4
Tương tự như vậy với câu a\()\)
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 1/2
a: =-21/36-3/36=-24/36=-2/3
b: =43/12*1/2+5/24=43/24+5/24=2
c: =8/9+1/9=1
e: =1-1/4+1/4-1/7+...+1/97-1/100
=1-1/100=99/100
\(P=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
\(\Rightarrow\dfrac{1}{2}P=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{101}}\)
\(\Rightarrow\dfrac{1}{2}P-P=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{101}}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{100}}\)
\(\Rightarrow-\dfrac{1}{2}P=\dfrac{1}{2^{101}}-\dfrac{1}{2^2}\)
\(\Rightarrow P=\left(\dfrac{1}{2^{101}}-\dfrac{1}{2^2}\right):\left(-\dfrac{1}{2}\right)\)
2A=1-1/2+1/2^2-...+1/2^98-1/2^99
=>3A=1-1/2^100
=>\(A=\dfrac{2^{100}-1}{3\cdot2^{100}}\)
\(M=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\cdot\left(\dfrac{1}{2}+1\right)\cdot\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{101}{200}\)