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

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

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

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

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

Vì 1/99 + 1/98 - 1/97 - 1/96 khác 0

=> x + 100 = 0 => x = -100

b) \(\frac{x-3}{47}+\frac{x-2}{48}=\frac{x-1}{49}+1\)

\(\Rightarrow\frac{x-3}{47}-1+\frac{x-2}{48}-1=\frac{x-1}{49}+1-2\)

\(\Rightarrow\frac{x-50}{47}+\frac{x-50}{48}-\frac{x-50}{49}=0\)

\(\Rightarrow\left(x-50\right)\left(\frac{1}{47}+\frac{1}{48}-\frac{1}{49}\right)=0\)

Vì 1/47 + 1/48 - 1/49 khác 0

Nên x -50 = 0 => x = 50

\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)

\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)

\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)

\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)

\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)

\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)

\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)

\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)

a)  A=x^2+4x+4=(x+2)^2.

Giờ ta tính giá trị của đa thức A với x=98:

A=(98+2)^2=100^2=10000

b)  B=x^3+9x^2+27x+27=(x+3)^3.

Thế x=-103 => (-103+3)^3=-1000000

c) Tách C = a⋅b−a⋅c+2⋅c−2⋅b rồi kết hợp lại thành C=(a−2)⋅b+(2−a)⋅c.

Thế a,b,c vào được vậy 

C=(2−2)⋅1.007+(2−2)⋅(−0.006) =0

d) Bài này khó quá mà tui nghĩ là đưa mấy cặp (2023^2-2022^2) thành dạng a^2-b^2=(a-b)(a+b) á

d: D=(2023^2-2022^2)+(2021^2-2020^2)+...+(3^2-2^2)+(1^2-0^2)

=2023+2022+...+3+2+1+0

=2023*2024/2=2047276

Bài 1:

A =  1996 x 1997 x 1998 x 1999 + 2021 x 2022 x 2023 x 2024

A = (1996 x 1997) x (1998  x 1999) + (2021 x 2022) x (2023 x 2024)

A = \(\overline{..2}\) x \(\overline{..2}\) + \(\overline{..2}\) x \(\overline{..2}\)

A = \(\overline{..4}\) + \(\overline{..4}\)

A = \(\overline{..8}\)

,làm ơn giúp mik với ah

\(\left(1+\dfrac{2}{3}\right).\left(1+\dfrac{2}{4}\right).\left(1+\dfrac{2}{5}\right)....\left(1+\dfrac{2}{2020}\right).\left(1+\dfrac{2}{2021}\right)\)

= \(\dfrac{5}{3}.\dfrac{6}{4}.\dfrac{7}{5}.\dfrac{8}{6}.\dfrac{9}{7}....\dfrac{2022}{2020}.\dfrac{2023}{2021}\)

= \(\dfrac{1}{3}.\dfrac{1}{4}.2022.2023\)

= \(\dfrac{337.2023}{2}\)

= \(\dfrac{\text{681751}}{2}\)

\(\Rightarrow2x+\frac{146}{4753}=2x+\frac{58}{825}\)

\(\Rightarrow\frac{146}{4753}=\frac{58}{825}\)( vô lí)

Vậy phương trình vo nghiệm

a, -1+3 - 5 + 7 - ...... +97 - 99

[ - 1+ 3] - [ 5 + 7] -  .... - [ 95 + 97] - 99

  [2 - 12] - ..... - [184 - 192] - 99

còn lại tự giải

nnnnnnnnnnnnnnnnnnnnnnnoooooooooooooooooo.......

b

-100

c

-50 

chức bạn học tốt

1) -12.(x-5) + 7.(3-x)=5

   -12x+ 60+21-7x =5

    -12x-7x = 5-60-21

    -19x=-76

     x=-76:(-19)

     x=4

2) (x-2).(x+4) =0

   \(\Rightarrow\)x-2=0 hoặc x+4=0

x-2=0                    x+4=0

x=0+2                    x=0-4     

x=2                         x=-4

Vậy x=2 hoặc x=-4

3) (x-2).(x+15) =0

 \(\Rightarrow\)x-2=0 hoặc x+15=0

 x-2=0                  x+15=0

x=0+2                   x=0-15 

 x=2                       x=-15

1)\(-12.\left(x-5\right)+7.\cdot\left(3-x\right)=5\)

\(-12x+60+21-7x=5\)

\(-19x+81=5\)

\(-19x=5-81\)

-\(-19x=-76\)

\(x=-76:-19\)

\(x=4\)

2) Ta có 2 trường hợp

TH1: x-2=0 =>x=2

TH2: x+4=0 => x=-4

Vậy \(x\in\left(-4;2\right)\)

3) Ta có

TH1: x-2=0=>x=2

TH2: x+15=0=>x=-15

Vậy \(x\in\left(-15;2\right)\)

2)   (x - 2).(x + 4) = 0

\(\hept{\begin{cases}x-2=0\\x+4=0\end{cases}\Rightarrow\hept{\begin{cases}x=0+2\\x=0-4\end{cases}\Rightarrow}\hept{\begin{cases}x=2\\x=-4\end{cases}}}\)

Vậy x = 2 và (-4)

3)   (x-2).(x + 15) = 0

\(\hept{\begin{cases}x-2=0\\x+15=0\end{cases}\Rightarrow}\hept{\begin{cases}x=0+2\\x=0-15\end{cases}\Rightarrow}\hept{\begin{cases}x=2\\x=-15\end{cases}}\)

Vậy x = 2 và (-15)