Ta có \(\dfrac{2}{1\cdot5}+\dfrac{2}{5\cdot9}+\dfrac{2}{9\cdot13}+...+\dfrac{2}{x\left(x+4\right)}=\dfrac{56}{113}\)

\(\dfrac{1}{2}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{x\left(x+4\right)}\right)=\dfrac{56}{113}\)

\(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{56}{113}:\dfrac{1}{2}\)

\(1-\dfrac{1}{x+4}=\dfrac{112}{113}\)

\(\dfrac{1}{x+4}=1-\dfrac{112}{113}=\dfrac{1}{113}\)

x + 4 = 113 ⇒ x = 109

\(\dfrac{2}{1.5}+\dfrac{2}{5.9}+...+\dfrac{2}{x\left(x+4\right)}=\dfrac{56}{113}\)

Xét: \(A=\dfrac{2}{1.5}+\dfrac{2}{5.9}+...+\dfrac{2}{x\left(x+4\right)}\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{x-4}-\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+4}\right)\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{x+4}\right)\)

Với \(A=\dfrac{56}{113}\) thì

 \(\dfrac{1}{2}.\left(1-\dfrac{1}{x+4}\right)=\dfrac{56}{113}\)

\(\left(1-\dfrac{1}{x+4}\right)=\dfrac{112}{113}\)

\(\dfrac{1}{x+4}=\dfrac{1}{113}\)

\(x=109\)

\(\dfrac{2}{1.5}\) + \(\dfrac{2}{5.9}\)  + \(\dfrac{2}{9.13}\)+...+  \(\dfrac{2}{x\left(x+4\right)}\)   = \(\dfrac{56}{113}\)

(\(\dfrac{2}{1.5}+\dfrac{2}{5.9}+\dfrac{2}{9.13}+....+\dfrac{2}{x\left(x+4\right)}\) ) \(\times\) 2  = \(\dfrac{56}{113}\) \(\times\) 2

\(\dfrac{4}{1.5}\) + \(\dfrac{4}{5.9}\) +\(\dfrac{4}{9.13}\)+...+ \(\dfrac{4}{x\left(x+4\right)}\)                = \(\dfrac{112}{113}\)

\(\dfrac{1}{1}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{13}\) +...+ \(\dfrac{1}{x}\) - \(\dfrac{1}{x+4}\) = \(\dfrac{112}{113}\)

1  - \(\dfrac{1}{x+4}\) = \(\dfrac{112}{113}\)

        \(\dfrac{1}{x+4}\) = 1 - \(\dfrac{112}{113}\)

        \(\dfrac{1}{x+4}\) = \(\dfrac{1}{113}\)

         \(x\) + 4 = 113

         \(x\)        = 113 - 4

         \(x\)        = 109

a) 5.(x-20) = 35

(x-20) = 35:5

x-20 = 7

x = 27

b) (x+125) -301 = 56

x+125 -301 = 56

x - 176 = 56

x = 56 +176

x= 232

c) 215 + (x-21):2 = 235

(x-21):2 = 235 - 215

(x-21):2 = 20

x-21 = 20 .2

x-21 = 40

x = 61

d) (x:23 +45) .67 = 8911

(x:23 +45) = 8911 : 67

x:23+45 = 133

x:23 = 133-45

x:23 = 88

x = 88.23

x = 2024

a) 5.(x - 20) = 35

⇒ x - 20 = 35 : 5

⇒ x - 20 = 7

⇒ x = 7 + 20

⇒ x = 27

b) (x + 125) - 301 = 56

⇒ x + 125 = 56 + 301

⇒ x + 125 = 357

⇒ x = 357 - 125

⇒ x = 232

c) 215 + (x - 21) : 2 = 235

⇒ (x - 21) : 2 = 235 - 215

⇒ (x - 21) : 2 = 20

⇒ x - 21 = 20 x 2 

⇒ x - 21 = 40

⇒ x = 21 + 40

⇒ x = 61

d) (x : 23 + 45).67 = 8911

⇒ x : 23 + 45 = 8911 : 67

⇒ x : 23 + 45 = 133

⇒ x : 23 = 133 - 45

⇒ x : 23 = 88

⇒ x = 88 x 23

⇒ x = 2024

a) \(5.\left(x-20\right)=35\)

\(x-20=35:5\)

\(x-20=7\)

\(x=7+20\Rightarrow x=27\)

b) \(\left(x+125\right)-301=56\)

\(x+125=357\)

\(x=357-125\Rightarrow x=232\)

c) \(215+\left(x-21\right):2=235\)

\(\left(x-21\right):2=20\)

\(x-21=40\Rightarrow x=40+21\Rightarrow x=61\)

d) \(\left(x:23+45\right).67=8911\)

\(x:23+45=133\)

\(x:23=88\Rightarrow x=88.23\Rightarrow x=2024\)

\(\frac{x}{5}-\frac{3}{7}=\frac{10}{13}\)

\(\Rightarrow\frac{91x}{455}-\frac{195}{455}=\frac{350}{455}\)

\(\Rightarrow91x-195=350\)

\(\Rightarrow91x=545\)

\(\Rightarrow x=\frac{545}{91}\)

\(\frac{-3x}{4}+\frac{1}{3}=\frac{11}{5}\)

\(\Rightarrow\frac{-45x}{60}+\frac{20}{60}=\frac{132}{60}\)

\(\Rightarrow-45x+20=132\)

\(\Rightarrow-45x=112\)

\(\Rightarrow x=-\frac{112}{45}\)

-3x/4+1/3=11/5

=>-45x/60+20/60=132/60

=>-45x+20=132

=>-45x=112

=>x=-112/45

a) \(5+3^{x+1}=86\)

\(=>3^{x+1}=86-5\)

\(=>3^{x+1}=81=3^4\)

\(=>x+1=4\) ( cùng cơ số )

\(=>x=4-1\)

\(=>x=3\)

b) \(15:\left(x+2\right)=\left(3^3+3\right):10\)

\(=>15:\left(x+2\right)=\left(27+3\right):10\)

\(=>15:\left(x+2\right)=30:10=3\)

\(=>x+2=15:3\)

\(=>x+2=5\)

\(=>x=5-2\)

\(=>x=3\)

c) \(\left(9x+2\right).4=80\)

\(=>9x+2=80:4\)

\(=>9x+2=20\)

\(=>9x=20-2\)

\(=>9x=18\)

\(=>x=18:9\)

\(=>x=2\)

d) \(\left(245-x\right)+7^2=14\)

\(=>\left(245-x\right)+14=14\)

\(=>245-x=14-14\)

\(=>245-x=0\)

\(=>x=245-0\)

\(=>x=245\)

?

1) A=(-125)(8x-8y)

A=(-125).8(x-y)

A=(-1000)(x-y)

Thay vào đó ta có :

A=(-1000).[(-43)-17]

A=(-1000).(-60)

A=60000

Hi SVĐ Mỹ Đình

<=> abcabc = abcx(1000+1) = abc x 1001

ta có: ax bcd x abc = abcabc

<=> a x bcd x abc = abc x 1001

<=> a x bcd = 1001

đây là tích giữa số có 1 chữ số và số có 3 chữ số nên ta tìm được a = 7 ( vì 1-> 9 chỉ có 1001 mới chia hết cho 7) => bcd = 143

vậy a = 7 ; b = 1 ; c = 4 ; d = 3

vậy abcd = 7143

a x bcd x abc = abcabc

Ta có: abc x 1001 = abcabc, mà abc x a x bcd = abcabc 

=> a x bcd = 1001 => a thuộc Ư(1001) = {1;7;11;...}

a là số 1 chữ số nên a là 1 hoặc 7

TH1: a = 1 => bcd = 1001 (loại)

Th2: a = 7 => bcd = 143 (đúng)

Vậy abcd = 7143

a) 5^x  . 5¹⁵ = 5²¹

=> 5^x = 5²¹ : 5¹⁵

=> 5^x = 5⁶

=> x = 6

a) 5^x  . 5¹⁵ = 5²¹

    5^x         = 5²¹ : 5¹⁵

    5^x         = 5⁶

=> x = 6

b) 7^x : 49 = 75

    7^x       = 75 . 49

     7^x      = 3675

7^x : 49 = 7⁵

=> 7^x = 7⁵ . 49

=> 7^x = 7⁵ . 7²

=> 7^x = 7⁷

=> x = 7

Cho mik hỏi tí z có gạch ngang ở giữa là j thế