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

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

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

\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

\(\Rightarrow x-1=0\) ( vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\) )

Vậy x = 1

b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=3\)

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

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)

\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\ne0\)

=> x + 100 = 0

=> x           = -100

c) \(\frac{x-1}{99}+\frac{x-2}{49}+\frac{x-4}{32}=6\)

\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{49}-2+\frac{x-4}{32}-3=6-6\)

\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{49}+\frac{x-100}{32}=0\)

\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\ne0\)

=> x - 100 = 0

=> x           = 100

Chúc bạn học tốt

có người khác trả lời trước rồi nên chị ko trả lời đâu nhé em trai

2S=1x2x3+2x3x3+3x4x3+...+98x99x3+99x100x3

3S=1x2x3+2x3x(4-1)+3x4x(5-2)+...+98x99x(100-97)+99x100x(101-98)

3S=1x2x3-1x2x3+2x3x4-2x3x4+3x4x5-...-97x98x99+98x99x100-98x99x100+99x100x101=99x100x101

S=33x100x101=333300

3xS = 1x 2x 3 + 2x3x3 + 3x4x3 + ...+ 98x99x3 + 99x100x3

= 1x2x3 + 2x3x(4-1) + 3x4x(5-2) +...+98x99x(100-97) + 99x100x(101-98)

= 1x2x3 -1x2x3 + 2x4x4 -2x3x4 + 3x4x5 +...- 97x98x99 +98x99x100  -98x99x199 + 99x100x101

= 99x100x101 = 999900 

=> S = 999900 : 3 =333300

=333300 nha

mong bn k mik

\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)

\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)

\(=\frac{100}{2}=50\)

Thanks nhìu

\(S=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\\ 3S=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot3\cdot4+...+3\cdot99\cdot100\\ 3S=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+99\cdot100\cdot\left(101-98\right)\\ 3S=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+....+99\cdot100\cdot101-98\cdot99\cdot100\\ 3S=99\cdot100\cdot101\\ S=\dfrac{99\cdot100\cdot101}{3}=33\cdot100\cdot101=3300\cdot101=333300\)

t.i.c.k câu này mình làm cho

tl đi rùi tk cho

S = 1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 5 + 4 x 5 x 6 + … + 97 x 98 x 99 

=) S = ( 2+3 ) * ( 2+3+4) * ........ * ( 97 + 98 +99 ) ( 1 khi nhân không có giá trị nên không cộng )

=) S = 5 * 6 * ........ * 294

vậy S  = 5 * 6 * ........ * 294 ( với * là dấu nhân )

x+1/99+x+2/98+x+3/97+x+4/96=0

=> 4x+(1/99+2/98+3/97+4/96)=0

=> x=-0,025775918

\(\left(x-7\right).\left(x+3\right)< 0\)

TH1: \(\hept{\begin{cases}x-7< 0=>x< 0+7=>x< 7\\x+3>0=>x>0-3=>x>-3\end{cases}}\)

                    => x thuộc {-2;-1;0;1;2;3;4;5;6}

TH2: \(\hept{\begin{cases}x-7>0=>x>0+7=>x>7\\x+3< 0=>x< 0-3=>x< -3\end{cases}}\)

                   => x thuộc rỗng 

(x - 7) . (x + 3) < 0

Trường hợp 1 : x - 7 > 0 và x + 3 < 0

x - 7 > 0 => x > 7

x + 3 < 0 => x < -3

=> 7 < x < -3 (vô lý nên loại)

Trường hợp 2 : x - 7 < 0 và x + 3 > 0

x - 7 < 0 => x < 7

x + 3 > 0 => x > -3

=> -3 < x < 7 (thỏa mãn)

Vậy x thuộc {-2 ; -1 ; 0 ; 1 ; 2 ; 3 ; 4 ; 5 ; 6}

ko biết