bài 1:tính bằng cách thuận tiện nhất
A=1/3x6+1/6x9+1/9x12+...+1/96x99+1/99x102
làm đầy đủ mik tick 2.giúp:@@
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\(A=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{93.96}+\frac{3}{96.99}\)
\(A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{93}-\frac{1}{96}+\frac{1}{96}-\frac{1}{99}\)
\(A=1-\frac{1}{99}=\frac{98}{99}\)
Vậy A=\(\frac{98}{99}\)
\(B=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{95.98}\)
\(3B=\)\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{95.98}\)
\(3B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\)
\(3B=\frac{1}{2}-\frac{1}{98}=\frac{24}{49}\)
\(B=\frac{24}{49}:3=\frac{8}{49}\)
Vậy B=\(\frac{8}{49}\)
Dấu "." là dấu nhân.
_Học tốt_
A = \(\dfrac{7}{3\times6}\) + \(\dfrac{7}{6\times9}\) + \(\dfrac{7}{9\times12}\) + \(\dfrac{7}{12\times15}\)+ .....+\(\dfrac{7}{96\times99}\)
A = \(\dfrac{7}{3}\) x ( \(\dfrac{3}{3\times6}\) + \(\dfrac{3}{6\times9}\)+ \(\dfrac{3}{9\times12}\)+ \(\dfrac{3}{12\times15}\)+......+\(\dfrac{3}{96\times99}\))
A = \(\dfrac{7}{3}\) x ( \(\dfrac{1}{3}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{12}\)+ \(\dfrac{1}{12}\) - \(\dfrac{1}{15}\)+....+ \(\dfrac{1}{96}\) - \(\dfrac{1}{99}\))
A = \(\dfrac{7}{3}\) x ( \(\dfrac{1}{3}\)- \(\dfrac{1}{99}\))
A = \(\dfrac{224}{297}\)
\(a,1999-2000+1=\left(1999+1\right)-2000=2000-2000=0\\ b,450:5+550:5=\left(450+550\right):5=1000:5=200\\ c,188\times7+188\times4-188=188\times\left(7+4-1\right)=188\times10=1880\)
a,1999−2000+1=(1999+1)−2000=2000−2000=0b,450:5+550:5=(450+550):5=1000:5=200c,188×7+188×4−188=188×(7+4−1)=188×10=1880
\(28+35+62+95+73+69+27+1\)
\(=\left(28+62\right)+\left(35+95\right)+\left(73+27\right)+\left(69+1\right)\)
\(=90+130+100+70=390\)
(28+62)+(35+95)+(73+27)+(69+1)=
90+130+100+70=390
Đs:390
nhân cả vế với 3 ta có
Ax3=\(\frac{3}{3x6}\)+\(\frac{3}{6x9}\)+.........+\(\frac{3}{99x102}\)
Ax3=\(\frac{1}{3}\)-\(\frac{1}{6}\)+.....+\(\frac{1}{99}\)-\(\frac{1}{102}\)
Ax=\(\frac{1}{3}\)-\(\frac{1}{102}\)
Ax3=\(\frac{11}{34}\)
A=\(\frac{11}{34}\):3
A=\(\frac{11}{102}\)
gạch đi các số lặp lại thì còn phân số 1/3 và 1/102 lấy \(\frac{1}{3}-\frac{1}{102}=\frac{33}{102}\)