2 x 4,8 x 0,5 + 4 x 16 x 0,25 + 20 : 10 x 0,5

=(2 x0,5) x4,8 +(4 x0,25) x16 +2 x0,5

=1 x4,8+16+1

=4,8+16+1

=21,8

12,5 x3,14 x8 +16 x0,125 x1,57

=(12,5 x8 )x3,14 +8 x2x0,125x1,57

=100 x3,14 +(8 x0,125) x2 x1,57

=314 + 2x1,57

=314 +3,14

=317,14

tick nhiệt tình nha nhanh nhất nè

a) 21,8 

b) 317,14 

\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2019}\right)\left(1-\frac{1}{2020}\right)\)

\(B=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{2018}{2019}\cdot\frac{2019}{2020}\)

Số nào xuất hiện 2 lần thì thay thế những số đó bằng số 1.

\(B=\frac{1}{2020}\)

B = \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2019}\right).\left(1-\frac{1}{2020}\right)\)

    = \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2018}{2019}.\frac{2019}{2020}\)

    = \(\frac{1.2.3...2019}{2.3.4..2020}\)(Nếu có 2 thừa số giống nhau lặp lại ở tử số và mẫu số thì rút gọn coi như triệt tiêu hết và không có gì)

   =  \(\frac{1}{2020}\)

\(B=\left(1-\frac{1}{2}\right)....\left(1-\frac{1}{2020}\right)\)

\(B=\frac{1}{2}.\frac{2}{3}....\frac{2019}{2020}\)

\(B=\frac{1.2.3...2019}{2.3.4....2020}\)

\(B=\frac{1}{2020}\)

Vậy B = 1/2020

\(\Leftrightarrow\)2(9x²+6x+1)=(3x+1)(x-2)

\(\Leftrightarrow\)2(3x+1)²-(3x+1)(x-2)=0

\(\Leftrightarrow\)(3x+1)[2(3x+1)-(x-2)]=0

\(\Leftrightarrow\)(3x+1)(6x+2-x+2)=0

\(\Leftrightarrow\)(3x+1)(5x+4)=0

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+1=0\Leftrightarrow3x=-1\Leftrightarrow x=\frac{-1}{3}\\5x+4=0\Leftrightarrow5x=-4\Leftrightarrow x=\frac{-4}{5}\end{cases}}\)​

a) 5^x + 5^x+2 = 650

   5^x + 5^x+2 = 5⁴ + 5²

   5^x +  5^x+2 = 5⁶

 ^{\(\Rightarrow\)}x+x+2 = 6

         2x       = 6-2 

          x        = 2

b) 3x - ( 2x +1) = 2

   3x - 2x -1     = 2 

    3x - 2x         =2+1 

   (3 - 2).x         = 3

     1.x             = 3

       x              = 3

sao ko k  cho mình ?

 +)\(\frac{x}{3}=0\)  \(\Rightarrow x=0\)

+) \(\frac{x}{3}< 0\)  \(\Rightarrow x< 0\)

+)\(1< \frac{x}{3}< 2\)    \(\Rightarrow3< x< 6\)

x/3=0 thì x=0

1)

3*(x+1)-12=15                        4*(x-3)+12=40

=> 3*(x+1)=27                       => 4*(x-3)=28

=> x+1=9                              => x-3=7

=> x=8                                  => x=10

tui nhầm ở câu cuối đó 6 * 19 * 4 + 3 * 81* 8

mình nhầm câu cuối 6*19*4+3*81*8

42-2(x+16) +12 :2=6(x-1)+ (-4)^3 :16

42- 2x- 32 +6 = 6x - 6 - 64: 16

(42 - 32 +6) - 2x = 6x - 6 - 4

16 = 2x = 6x- 10

16+10= 2x+6x

26= 8x 

x= 26/8

x= 13/4

42-2(x+16)+12:2=6(x-1)+(-4)³:16

42-2(x+16)+6=6(x-1)-64:16

42-2x-32+6=6x-6-4

-2x+16=6x-10

6x+2x=16+10

8x=26

x=3,25

       \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left[2\left(3x+1\right)-\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(3x+1\right)\left(6x+2-x+2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(5x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\5x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=-1\\5x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=-\frac{4}{5}\end{cases}}}\)

Vậy phương trình trên có tập nghiệm \(S=\left\{-\frac{1}{3};-\frac{4}{5}\right\}\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2001}{2003}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}\right)=\frac{1}{2}\cdot\frac{2001}{2003}\)

\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)

\(\Rightarrow\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2003}\)

\(\Rightarrow x+1=2003\)

\(\Rightarrow x=2003-1\)

\(\Rightarrow x=2002\)

Vậy x=2002