x - 1 + x - 3 + x - 5 + x - 7 + .... + x - 201 = 202

Ta có : ( x + x + x + ... + x ) - ( 1 + 3 + 5 + 7 + .... + 201 ) = 202 

Ta thấy 1 + 3 + 5 + 7 + .... + 201 là dãy cách đều 2 đơn vị . Số số hạng là : ( 201 - 1 ) : 2 + 1 = 101 ( số ) và tương ứng với 101 lần số x . Tổng của dãy là : ( 201 + 1 ) x 101 : 2 = 10201

Thay vào ta được : 101 . x - 10201 = 202

                             101 . x = 202 + 10201

                              101 . x = 10403

                                       x = 10403 : 101

                                       x = 103

Vậy x = 103

103 mới tra google

x . 2 + x . 5 + x . 8 = x . 9 + x . 7 + x . 1/3 + 201

x . ( 2 + 5 + 8 ) = x . ( 9 + 7 + 1/3 ) +201

x . 15 = x . 49/3 + 201

15x = 49/3x + 201

15x - 201 = 49/3x

giảm mỗi vế đi 15x , ta có :

-201 = 4/3x

x = (-201) : 4/3

x = -150,75

Vậy x = -150,75

\(x\times2+x\times5+x\times8=x\times9+x\times7+x\times13+201\)

\(x\times\left(2+5+8\right)=x\times\left(9+7+13\right)+201\)

\(x\times15-x\times29=201\)

\(x\times\left(-14\right)=201\)

\(x=201:\left(-14\right)=\)Số dài quá bạn ơi!

Mk chưa hiểu chỗ [ - 4 ] là gì

\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)

\(\Rightarrow\frac{x+2}{2002}+1+\frac{x+5}{1999}+1+\frac{x+201}{1803}+1=0\)

\(\Rightarrow\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)

\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)

Dễ thấy \(\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)>0\)nên x + 2004 = 0

Vậy x = -2004

\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)

\(\Leftrightarrow\frac{x+2}{2002}+1+\frac{x+5}{1999}+1+\frac{x+201}{1803}+1=-3+1+1+1\)

\(\Leftrightarrow\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)

\(\Leftrightarrow x+2004=0\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\ne0\right)\)

<=> x=-2004

a,\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)

\(< =>\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+5}{1999}+1\right)+\left(\frac{x+201}{1803}+1\right)=0\)

\(< =>\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)

\(< =>\left(x+2004\right).\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)

Do \(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\ne0\)

\(=>x+2004=0\)

\(=>x=-2004\)

\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)

\(\Leftrightarrow\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+5}{1999}+1\right)+\left(\frac{x+201}{1803}+1\right)=0\)

\(\Leftrightarrow\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)

\(\Leftrightarrow x=-2004\)

\(\frac{x+1}{99}+\frac{x+3}{97}+\frac{x+5}{95}=\frac{x+9}{91}+\frac{x+8}{92}+\frac{x+7}{93}\)

\(\Leftrightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)=\left(\frac{x+9}{91}+1\right)+\left(\frac{x+8}{92}+1\right)+\left(\frac{x+7}{93}+1\right)\)

\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}=\frac{x+100}{91}+\frac{x+100}{92}+\frac{x+100}{93}\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}-\frac{1}{91}-\frac{1}{92}-\frac{1}{93}\right)=0\)

Để ý thấy cụm đằng sau < 0 nên x=-100

Tính :

1-2+3-4+5-6+...+201-202

=(1-2)+(3-4)+(5-6)+...+(201-202)  (có 101 cặp)

=(-1)+(-1)+(-1)+...+(-1)

=(-1).101=-101.

Tìm x là số nguyên :

30-(x+1)=-17

x+1=30-(-17)

x+1=47

x=47-1

x=46

Vậy x=46.

x(|15|-10)=5³

x(15-10)=5³

x.5=5³

x=5³:5

x=5²=25

Vậy x=25

2x+5=7-x-5

2x+x+5=7-5

3x+5=2

3x=2-5

3x=-3

x=(-3):3

x=-1

Vậy x=-1.

P(x)= 1+x+x.3+x.5+...+x.199+x.201

       = 1+x(1+3+5+...+199+201)

       = 1+x.10201

Thay x=1 vào đa thức ta được: 1+1.10201=1+10201=10202

Thay x=-1 vào đa thức ta được: 1+(-1).10201= 1+(-10201)=-10200  

Ta có: \(P\left(x\right)=1+x+3x+5x+...+199x+201x\)

\(=1+x\left(1+3+5+...+199+201\right)\)

\(=10201x+1\)

\(P\left(1\right)=10201\cdot1+1=10202\)

\(P\left(-1\right)=10201\cdot\left(-1\right)+1=-10201+1=-10200\)