ếu

Câu 1  \(x.5+112+236=633\Leftrightarrow x.5=285\)

                                                               \(\Leftrightarrow x=57\)

Câu 2  \(\frac{2}{320}+3x=620\Leftrightarrow3x=\frac{99199}{160}\)

                                                \(\Leftrightarrow x=\frac{99199}{480}\)

a)483*8+483+483

=483*8+483*1+483*1

=483*(8+1+1)

=483*10

=4830

b)620*43+620*56+620

=620*43+620*56+620*1

=620*(43+56+1)

=620*100

=62000

a) 483 x 8 + 483 + 483

= 483 x 8 + 483 x 1 + 483 x 1 

= 483 x ( 8 + 1 + 1 )

= 483 x 10

= 4830

b) 620 x 43 + 620 x 56 + 620

= 620 x 43 + 620 x 56 + 620 x 1

= 620 x ( 43 + 56 + 1 )

= 620 x 100

= 62000

    _HT_

\(a,\frac{x+8}{3}+\frac{x+7}{2}=-\frac{x}{5}\)

\(\Leftrightarrow\frac{10\cdot\left(x+8\right)}{30}+\frac{15\left(x+7\right)}{30}=\frac{-6x}{30}\)

\(\rightarrow10x+80+15x+105=-6x\)

\(\Leftrightarrow31x+185=0\)

\(\Leftrightarrow x=-\frac{185}{31}\)

b,\(b,\frac{x-8}{3}+\frac{x-7}{4}=4+\frac{1-x}{5}\)

\(\Leftrightarrow\frac{20\left(x-8\right)}{60}+\frac{15\left(x-7\right)}{60}=\frac{240}{60}+\frac{12\left(1-x\right)}{60}\)

\(\rightarrow20x-160+15x-105=240+12-12x\)

\(\Leftrightarrow47x-517=0\)\(\Leftrightarrow x=11\)

=>2/x(x+2)+2/(x+2)(x+4)+...+2/(x+6)(x+8)=8/105

=>1/x-1/x+2+1/x+2-1/x+4+...+1/x+6-1/x+8=8/105

=>\(\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{105}\)

=>x(x+8)=105

=>x^2+8x-105=0

=>(x+15)(x-7)=0

=>x=7 hoặc x=-15

A. x = 2

B. \(\dfrac{3}{8}=\dfrac{6}{x}\)\(\Leftrightarrow x=\dfrac{6.8}{3}=16\)

C. x = 3

D. \(x=\dfrac{4.6}{8}=3\)

E. \(x=\dfrac{7}{3}\)

G.\(\dfrac{14}{13}=\dfrac{28}{10-x}\)

<=>\(14\left(10-x\right)=364\)

<=> 10 - x = 26 

<=> x = -16 

H. \(3\left(x+2\right)=4\left(x-5\right)\)

<=> 3x + 6  = 4x - 20 

<=> -x = -26

<=> x = 26

K. \(\dfrac{x}{2}=\dfrac{8}{x}\)

<=> \(x^2=16\)

<=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

M. \(\left(x-2\right)^2=100\)

<=> \(\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)

a=2

b=16

c=3

d=3

mik chỉ biết thế này thôi(ko chắc đúng=3)

A, \(\dfrac{x}{5}=\dfrac{2}{5}\Rightarrow x=2\)

B,\(\dfrac{3}{8}=\dfrac{6}{x};\dfrac{3}{8}=\dfrac{3\cdot2}{8\cdot2}=\dfrac{6}{16}\Rightarrow x=16\)

C,\(\dfrac{1}{9}=\dfrac{x}{27};\dfrac{1}{9}=\dfrac{1\cdot3}{9\cdot3}=\dfrac{3}{27}\Rightarrow x=3\)

D,\(\dfrac{4}{x}=\dfrac{8}{6};\dfrac{4}{x}=\dfrac{4\cdot2}{x\cdot2}=\dfrac{8}{2x}\)

   \(\Rightarrow2x=6;6=x:2;x=3\)

\(\dfrac{x^3+8}{x^2+2x+1}.\dfrac{x^2+3x+2}{1-x^2}\left(x\ne\pm1\right)\\ =\dfrac{x^3+2^3}{\left(x+1\right)^2}.\dfrac{\left(x^2+x\right)+\left(2x+2\right)}{1^2-x^2}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{x\left(x+1\right)+2\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{\left(x+2\right)\left(x+1\right)}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{\left(x+2\right)^2\left(x^2-2x+4\right)}{\left(1-x\right)\left(x+1\right)^2}\)

a, x ( x + 1 ) + ( x + 2 ) + ... + ( x + 30 ) = 620

( x + x + ... + x ) + ( 0 + 1 + 2 + ... + 30 ) = 620

Xét dãy số : 0 + 1 + 2 + ... + 30

Số số hạng của dãy số trên là :

( 30 - 0 ) : 1 + 1 = 31 ( số hạng )

Tổng của dãy số trên là :

( 30 + 0 ) . 31 : 2 = 465

Thay vào ta có :

31x + 465 = 620

31x = 620 - 465

31x = 155

=> x = 155 : 31

=> x = 5

Vậy x = 5

b, 2 + 4 + 6 + 8 + ... + 2x = 210

2 ( 1 + 2 + 3 + 4 + ... + x ) = 210

1 + 2 + 3 + 4 + ... + x = 210 : 2

1 + 2 + 3 + 4 + ... + x = 105

( x + 1 ) . x : 2 = 105

=> ( x + 1 ) . x = 105 . 2

=> ( x + 1 ) . x = 210

Mà : x + 1 và x là hai số tự nhiên liên tiếp

Vì : 15 . 14 = 210

=> x = 14

Vậy x = 14

\(x-2=\frac{3}{8}\)

\(x=\frac{3}{8}+2\)

\(x=\frac{3+2\times8}{8}\)

\(x=\frac{19}{8}\)

\(x-2=\frac{3}{8}\)

\(x=\frac{3}{8}+2\)

\(x=\frac{19}{8}\)

x-2=3/8

x=3/8+2

x=19/8