Vì x¹⁰⁰ = x nên x = 1 vì 1¹⁰⁰ =1

1^100=1

Suy ra X=1

x =1 nha

ta có: \(\frac{x}{\frac{y}{z}}=\frac{3}{\frac{4}{5}}\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{x^2}{9}=\frac{y^2}{16}=\frac{z^2}{25}=\frac{2x^2}{18}=\frac{2y^2}{32}=\frac{3z^2}{75}.\)

ADTCDTSBN

...

bn tu lm tiep nha

mà bạn nguyen nguyen ơi, x/3 phải bằng x^2/3*x chứ

không thể bình phương tử và mẫu bạn nha

VD : 2/3 không thể bằng 4/9

a) \(\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-...+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(1-\dfrac{1}{10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\dfrac{9}{10}\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=90-89\)

\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=1\)

\(\Rightarrow x+\dfrac{103}{50}=\dfrac{5}{2}\)

\(\Rightarrow x=\dfrac{11}{25}\)

b) \(x\cdot9,85+x\cdot0,15=0,1\)

\(\Rightarrow x\cdot\left(9,85+0,15\right)=0,1\)

\(\Rightarrow x\cdot10=0,1\)

\(\Rightarrow x=\dfrac{0,1}{10}\)

\(\Rightarrow x=0,01\)

c) \(\dfrac{2}{5}+2022x=\dfrac{4}{10}\)

\(\Rightarrow\dfrac{2}{5}+2022x=\dfrac{2}{5}\)

\(\Rightarrow2022x=\dfrac{2}{5}-\dfrac{2}{5}\)

\(\Rightarrow2022x=0\)

\(\Rightarrow x=\dfrac{0}{2022}\)

\(\Rightarrow x=0\)

a) \(\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\left(1\right)\)

Ta có :

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

\(\left(1\right)\Rightarrow\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right].2=89\)

\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right].2=90-89\)

\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=\dfrac{1}{2}\)

\(\Rightarrow x+\dfrac{206}{100}=\dfrac{5}{2}:\dfrac{1}{2}\)

\(\Rightarrow x+\dfrac{103}{50}=\dfrac{5}{2}.\dfrac{2}{1}\)

\(\Rightarrow x+\dfrac{103}{50}=5\)

\(\Rightarrow x=5-\dfrac{103}{50}\)

\(\Rightarrow x=\dfrac{250}{50}-\dfrac{103}{50}\)

\(\Rightarrow x=\dfrac{147}{50}\)

x ở giữa là sao

C=(1x3+3x5+...+99x101)+(2x4+4x6+...+98x100)

đặt S=1x3+3x5+...+99x101

=>6S=6x(1x3+3x5+...+99x101)

=1x3x(5+1)+3x5x(7-1)+...+97x99x(101-95)+99x101x(103-97)

=1x3x5+1x3x1+3x5x7-1x3x5+....+97x99x101-95x97x99+99x101x103-97x99x101

=1x3x1+99x101x103

=>S=(3+99x101x103):6=171650

=>C=171650+(2x4+4x6+...+98x100)

đặt A=2x4+4x6+...+98x100

=>6A=6x(2x4+4x6+...+98x100)

=>6A=2x4x6+4x6x(8-2)+...+96x98x(100-94)+98x100x(102-96)

=2x4x6+4x6x8-2x4x6+...+96x98x100-94x96x98+98x100x102-96x98x100

=98x100x102

=>A=98x100x102:6=166600

=>C=166600+171650

=>C=338250

B=2x2+4x4+6x6+...+100x100

=2x(4-2)+4x(6-2)+6x(8-2)+...+100x(102-2)

=2x4-4+4x6-8+6x8-12+...+100x102-200

=(2x4+4x6+6x8+...+100x102)-(4+8+12+...+200)

đặt A=2x4+4x6+...+98x100+100x102

=>6A=6x(2x4+4x6+...+98x100+100x102)

=>6A=2x4x6+4x6x(8-2)+...+96x98x(100-94)+98x100x(102-96)+100x102x(104-98)

=2x4x6+4x6x8-2x4x6+...+96x98x100-94x96x98+98x100x102-96x98x100+100x102x104-98x100x102

=100x102x104

=>A=100x102x104:6=176800

=>B=176800-(4+8+12+...+200)

đặt S=4+8+12+..+200

Số số hạng của S là:

(200-4):4+1=50 số

S=(200+4)x50:2=5100

=>B=176800-5100

=>B=171700

k mình đi mình trả lời cho

ko hiểu, nghĩa là sao giúp mình với đấy là bài khảo sát chọn học sinh giỏi đấy.

= 111...1 x (100...0 - 1)

   100 cs 1     100 cs 0

= 111...1 000...0 - 111...1

   100 cs 1 100cs 0    100 cs 1

= 111...10888...89

   99 cs 1  99 cs 8

Kho qua ad

thanks you

a) -12( x - 5 ) + 7( 3 - x ) = 5

-12x + 60 + 21 - 7x  = 5

-19x = 5 -81

-19x = -76

x = 76:19

x= 4

b) 30.( x + 2 ) - 6( x - 5 ) -  24x = 100

30x + 60 - 6x + 30 - 24x = 100

0x = 100 - 60 - 30

0x = 10

=> ko có giá trị x thỏa mãn đề bài

GIẢI
a) \(-12\left(x-5\right)+7\left(3-x\right)=5\Leftrightarrow-12x+60+21-7x=5\)
    \(\Leftrightarrow81-19x=5\Leftrightarrow-19x=5-81=-76\)
     \(\Leftrightarrow19x=76\Rightarrow x=76:19=4\)
Vậy tập nghiệm của pt là: S=[4]

b) \(30\left(x+2\right)-6\left(x-5\right)-24x=100\)
\(\Leftrightarrow30x+60-6x+30-24x=100\)
\(\Leftrightarrow90-0x=100\Rightarrow\)  không có nghiệm nào thỏa mãn.

để mình nghiên cứu đã

333......33333²

100 chữ số 3

33 x 33 = 1089

333 x 333 = 110889

3333 x 3333 = 11108889 

=> có 100 c.số 3 nhân vs nhau thì sẽ dc : 11111111...11( 99 c.số 1)0888888....888( 99 c. số 8)9

k mik nha mn!

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