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D

D nha 

bằng 1

Ta có:

$A=\dfrac{1+(1+2)+(1+2+3)+\cdots+(1+2+\cdots+98)}{1\times98+2\times97+3\times96+\cdots+98\times1}$

Tử số:

$1+(1+2)+(1+2+3)+\cdots+(1+2+\cdots+98)$

$=1+2+2^2+?$

Cách dễ nhất là viết:

$1+2+\cdots+k=\dfrac{k(k+1)}2$

Nên tử số:

$=\dfrac{1\cdot2}{2}+\dfrac{2\cdot3}{2}+\cdots+\dfrac{98\cdot99}{2}$

$=\dfrac{1}{2}(1\cdot2+2\cdot3+\cdots+98\cdot99)$

$=\dfrac{1}{2}\cdot\dfrac{98\cdot99\cdot100}{3}$

$=161700$

Mẫu số:

$1\cdot98+2\cdot97+3\cdot96+\cdots+98\cdot1$

$=1(99-1)+2(99-2)+\cdots+98(99-98)$

$=99(1+2+\cdots+98)-(1^2+2^2+\cdots+98^2)$

$=99\cdot\dfrac{98\cdot99}{2}-\dfrac{98\cdot99\cdot197}{6}$

$=161700$

Do đó:

$A=\dfrac{161700}{161700}=1$

Vậy $A=1$.

Ta có\(M=\left[\left(1+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{50}\right)\right].2.3...98\)

\(=\left[\frac{99}{1.98}+\frac{99}{2.97}+...+\frac{99}{49.50}\right].2.3...98=99\left(\frac{1}{1.98}+\frac{1}{2.97}+...+\frac{1}{49.50}\right).2.3...98\)

\(=99\left(\frac{k_1+k_2+...+k_{49}}{1.2.3...98}\right).2.3...98\left(k_1,k_2...k_{49}\varepsilonℕ^∗\right)=99\left(k_1+k_2+...+k_{49}\right)⋮99\Rightarrow M⋮99\left(đpcm\right)\)

Trẻ trâu hỏi cái loz

Con mẹ mày à

lz-12l=98

\(\Rightarrow\)z-12=98 hoac z-12=-98

th1:z-12=98

z=98+12

z=110

th2:z-12=-98

z=-98+12

z=-86

vay z=110 hoac z = -86

lx-98l=0+98

lx-98l=98

\(\Rightarrow\)x-98=98 hoac x-98=-98

th1:x-98=98

x=98+98

x=196

th2:x-98=-98

x=-98+98

x=0

vay x=196 hoac x = 0

chuan do nhe   l i k e    mk nha

98 x a + b x 98 - c x 98 = 98 x (a+b-c) = 98 x 160 = 15680

98 x a + b x 98 - c x 98
=98x(a+b-c)
=98x160
=15680

\(x\times98+x:0,5=2017\)

\(x\times98+x\times2=2017\)

   \(x\times\left(98+2\right)=2017\)

             \(x\times100=2017\)

                          \(x=2017:100\)

                          \(x=20,17\)

\(x\times98+x\div0,5=2017\)

\(x\times98+x\times2=2017\)

\(x\times\left(98+2\right)=2017\)

\(x\times100=2017\)

\(x=2017\div100\)

\(x=20,17\)

\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}=4\)

\(\Leftrightarrow\left(\frac{x-1}{99}+\left(-1\right)\right)+\left(\frac{x-2}{98}+\left(-1\right)\right)+\left(\frac{x-3}{97}+\left(-1\right)\right)+\left(\frac{x-4}{96}+\left(-1\right)\right)=0\)

\(\Leftrightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}=0\)

\(\Leftrightarrow\left(x-1\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)