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=1x2x3+2x3x3+...+98x99x3
=1x2x3+2x3x(4-1)+...+98x99x(100-97)
=1x2x3+2x3x4-1x2x3+...+98x99x100-97x98x99
=98x99x100
=970200
Đặt X=phép tính trên
Ta có X=X x 1/2 :1/2
X=(1/6+1/12+...+1/6480):1/2
X=(1/2x3+1/3x4+...+1/80x81):1/2
X=(1/2-1/3+1/3-1/4+...+1/80-1/81):1/2
X=(1/2-1/81):1/2
Đến đây bạn tự tính nhé!!!
Đặt: A=...
\(\frac{A}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{6480}\)
\(\frac{A}{2}=\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+...+\frac{1}{80x81}\)
\(\frac{A}{2}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{80}-\frac{1}{81}\)
\(\frac{A}{2}=\frac{1}{2}-\frac{1}{81}=\frac{79}{162}\) => A=\(\frac{79}{81}\)
a) Xét \(\frac{1999.2000-2}{1998.1999+3997}=\frac{1999.\left(1998+2\right)-2}{1998.1999+3997}=\frac{1999.1998+1999.2-2}{1998.1999+3997}=\frac{1999.1998+3996}{1998.1998+3997}\)
=> A < B
a)
\(=\dfrac{4.4.25+4.11.4.25}{29.2.48+71.2.48}=\dfrac{400+400.11}{96.29+71.96}=\dfrac{400\left(1+11\right)}{96\left(29+71\right)}=\dfrac{400.12}{96.100}=\dfrac{2.2.100.12}{12.2.2.2.100}=\dfrac{1}{2}\)
a)\(\frac{864.48-432.96}{864.48+432}\)
=\(\frac{864.48-432.2.48}{864.48+432}\)
=\(\frac{864.48-864.48}{864.48+432}\)
=\(\frac{0}{864.48+432}\)
=0
864x48-432x96/864x48+432=2x432x48-432x48x2/864x48+432=0/864x48+432=0.
**** mình nha
\(\frac{317.452-201}{451.317+116}=\frac{317.\left(451+1\right)-201}{451.317+116}=\frac{317.451+317-201}{451.317+116}\)\(=\frac{451.317+\left(317-201\right)}{451.317+116}=\frac{451.317+116}{451.317+116}\)\(=1\)
Vậy: \(\frac{317.452-201}{451.317+116}=1\)
[317 x 452 - 201][451x317+116]=[317x(451+1)-201][451x317+116]
=[317x451+317-201][451x317+116]
=[317x451-116][451x317-116]=1
nhớ nhé!
\(\frac{864.48-432.96}{864.48.432}=\frac{864.48-432.2.48}{864.48.432}=\frac{864.48-864.48}{864.48.432}\)
\(=\frac{0}{864.48.432}=0\)
ta có tử số= 864.48-432.96=432.2.48-432.96=432.96-432.96=0
vậy phân thức đã cho có giá trị bằng 0