\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+....+\frac{1}{x\times\left(x+2\right)}=\frac{1005}{2011}\)

\(\Leftrightarrow\frac{1}{2}\times\left(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+.....+\frac{1}{x\times\left(x+2\right)}\right)=\frac{1005}{2011}\)

\(\Leftrightarrow\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\Leftrightarrow\frac{1}{2}\times\left(1-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{1005}{2011}:\frac{1}{2}=\frac{2010}{2011}\)

\(\Leftrightarrow\frac{1}{x+2}=1-\frac{2010}{2011}=\frac{1}{2011}\)

\(\Leftrightarrow x+2=2011\)

\(\Leftrightarrow x=2009\)

Vậy x = 2009

\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{x\left(x+2\right)}=\frac{1005}{2011}\)

\(\Rightarrow\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{1005}{2011}\)

\(\Rightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\Rightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\Rightarrow\frac{x+1}{x+2}=\frac{2010}{2011}\)

\(\Rightarrow x+1=2010\Leftrightarrow x=2009\)

\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+.......+\frac{1}{x.\left(x+2\right)}=\frac{1005}{2011}\) 

<=> \(\frac{1}{1.3}+\frac{1}{3.5}+......+\frac{1}{x.\left(x+2\right)}=\frac{1005}{2011}\)

<=>\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

<=>\(\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

<=> \(1-\frac{1}{x+2}=\frac{1005}{2011}:\frac{1}{2}\)

<=> \(1-\frac{1}{x+2}=\frac{2010}{2011}\) <=> \(\frac{1}{x+2}=1-\frac{2010}{2011}\)

                                                     <=> \(\frac{1}{x+2}=\frac{1}{2011}\) <=>  \(x+2=2011\)<=> \(x=2009\)

vậy \(x=2009\)

\(\frac{\left(-5\right)^2.\left(-5\right)^3.16}{5^4.\left(-2\right)^4}=\frac{\left(-5\right)^5.16}{5^4.16}=\frac{\left(-5\right)^5}{5^4}=-5\)

\(\frac{14^{1005}.5^{1006}}{2^{1007}.35^{1004}}=\frac{14^{1005}.5^{1005}.5}{2^{1004}.2^3.35^{1004}}=\frac{\left(14.5\right)^{1005}.5}{\left(2.35\right)^{1004}.2^3}=\frac{70^{1005}.5}{70^{1004}.2^3}=\frac{70.5}{8}=\frac{350}{8}=\frac{175}{4}\)

Chúc bạn học tốt!

Toán lớp 6 ? 

\(\frac{\left(-5\right)^2\left(-5\right)^3.16}{5^4.\left(-2\right)^4}=\frac{\left(-5\right)^5.2^4}{5^4.2^4}=-5\)

\(\frac{14^{1005}.5^{1006}}{2^{1007}.35^{1004}}=\frac{14^{1005}.5^{1005}.5}{2^{1004}.35^{1004}.2^3}=\frac{\left(14.5\right)^{1005}.5}{\left(2.35\right)^{1004}.8}=\frac{70^{1005}.5}{70^{1004}.8}=\frac{70.5}{8}=\frac{175}{4}\)

C = 48 x 12 + 48 x 3 + 35 x 5 + 13 x 5

C = 48 x (12 + 3) + 5 x (35 + 13)

C = 48 x 15 + 5 x 48

C = 48 x (15 + 5)

C = 48 x 20

C = 960

Đáp số là 960.

Tíck cho mìk vs nhé Kane !

C = 48 x 12 +48 x3 +35 x5 +13 x5

C = 48 x (12 + 3) + 5 x (35 + 13)

C = 48 x 15 + 5 x 48

C = 48 x (15 + 5)

C = 48 x 20

C = 960

C=48(12+3)+5(35+13)

C=48*15+5*48

C=48(15+5)=48*20=960

\(\frac{14^{1005}.5^{1006}}{2^{1007}.35^{1004}}\)

\(=\frac{2^{1005}.7^{1005}.5^{1006}}{2^{1007}.5^{1004}.7^{1004}}\)

\(=\frac{5^2.7}{2^2}=\frac{25.7}{4}=\frac{175}{4}\)

TRẦN TIỂU HY ƠI, BẠN TRÌNH BÀY RA GIÙM MK NHA. MK KO HIỂU LẮM

\(\frac{14^{1005}.5^{1006}}{2^{1007}.35^{1004}}=\frac{2^{1005}.7^{1005}.5^{1006}}{2^{1007}.5^{1004}.7^{1004}}\)

\(=\frac{5^2.7}{2^2}=\frac{25.7}{4}=\frac{175}{4}\)

(X - 1005) x (1 + 3 + 5+  7 + 9 + 11) = 1989 x 1990 x (70 - 35 x 2)

(X - 1005) x (11 + 1)[(11 - 1) : 2 + 1): 2] = 1989 x 1990 x 0

(X - 1005) x 36 = 0

X - 1005 = 0 : 36

X - 1005 = 0

X = 1005

( x - 1005 ) . ( 1 + 3 + 5 + 7 + 9 + 11 ) = 1989 . 1990 . ( 70 - 35 . 2 )

( x - 1005 ) . 36 = 1989 . 1990 . ( 70 - 70 )

( x - 1005 ) . 36 = 1989 . 1990 . 0

( x - 1005 ) . 36 = 0

x - 1005 = 0

x = 1005

(x - 1005) x (1+ 3 + 5 + 7 + 9 + 11) =  1989 x 1990 (70 - 35 x 2)

(x - 1005) x 36 = 1989 x 1990 x 0

(x - 1005) x 36 = 0

X -1005 = 0 : 36

X - 1005 = 0

X = 0

Hok tốt!

https://olm.vn/hoi-dap/question/1038454.html 

Mình vừa làm cách đây 11 phút nhé !

Ta có : a²⁰¹⁰ + b²⁰¹⁰ + c²⁰¹⁰ = a¹⁰⁰⁵b¹⁰⁰⁵ + b¹⁰⁰⁵c¹⁰⁰⁵ + c¹⁰⁰⁵a¹⁰⁰⁵ 

<=> 2a²⁰¹⁰ + 2b²⁰¹⁰ + 2c²⁰¹⁰ = 2a¹⁰⁰⁵b¹⁰⁰⁵ + 2b¹⁰⁰⁵c¹⁰⁰⁵ + 2c¹⁰⁰⁵a¹⁰⁰⁵ 

<=> 2a²⁰¹⁰ + 2b²⁰¹⁰ + 2c²⁰¹⁰  - 2a¹⁰⁰⁵b¹⁰⁰⁵ - 2b¹⁰⁰⁵c¹⁰⁰⁵ - 2c¹⁰⁰⁵a¹⁰⁰⁵ = 0

<=> (a²⁰¹⁰ - 2a¹⁰⁰⁵b¹⁰⁰⁵ + b²⁰¹⁰) + (b²⁰¹⁰ - 2b¹⁰⁰⁵c¹⁰⁰⁵ + c²⁰¹⁰) + (c²⁰¹⁰ - 2c¹⁰⁰⁵a¹⁰⁰⁵  +  a²⁰¹⁰​) = 0

<=> (a¹⁰⁰⁵ - b¹⁰⁰⁵)² + (b¹⁰⁰⁵ - c¹⁰⁰⁵)² + (c¹⁰⁰⁵ - a¹⁰⁰⁵ ​)² = 0

=> a¹⁰⁰⁵ - b¹⁰⁰⁵ = b¹⁰⁰⁵ - c¹⁰⁰⁵ = c¹⁰⁰⁵ - a¹⁰⁰⁵ ​ = 0

=> a = b = c 

Vậy (a - b)²⁰ + (b - c)¹¹ + (c - a)²⁰¹⁰ = (a - a)²⁰ + (a - a)¹¹ + (a - a)²⁰¹⁰ = 0 + 0 + 0 = 0 .

Ta có : a²⁰¹⁰ + b²⁰¹⁰ + c²⁰¹⁰ = a¹⁰⁰⁵b¹⁰⁰⁵ + b¹⁰⁰⁵c¹⁰⁰⁵ + c¹⁰⁰⁵a¹⁰⁰⁵ 

<=> 2a²⁰¹⁰ + 2b²⁰¹⁰ + 2c²⁰¹⁰ = 2a¹⁰⁰⁵b¹⁰⁰⁵ + 2b¹⁰⁰⁵c¹⁰⁰⁵ + 2c¹⁰⁰⁵a¹⁰⁰⁵ 

<=> 2a²⁰¹⁰ + 2b²⁰¹⁰ + 2c²⁰¹⁰  - 2a¹⁰⁰⁵b¹⁰⁰⁵ - 2b¹⁰⁰⁵c¹⁰⁰⁵ - 2c¹⁰⁰⁵a¹⁰⁰⁵ = 0

<=> (a²⁰¹⁰ - 2a¹⁰⁰⁵b¹⁰⁰⁵ + b²⁰¹⁰) + (b²⁰¹⁰ - 2b¹⁰⁰⁵c¹⁰⁰⁵ + c²⁰¹⁰) + (c²⁰¹⁰ - 2c¹⁰⁰⁵a¹⁰⁰⁵  +  a²⁰¹⁰​) = 0

<=> (a¹⁰⁰⁵ - b¹⁰⁰⁵)² + (b¹⁰⁰⁵ - c¹⁰⁰⁵)² + (c¹⁰⁰⁵ - a¹⁰⁰⁵ ​)² = 0

=> a¹⁰⁰⁵ - b¹⁰⁰⁵ = b¹⁰⁰⁵ - c¹⁰⁰⁵ = c¹⁰⁰⁵ - a¹⁰⁰⁵ ​ = 0

=> a = b = c 

Vậy (a - b)²⁰ + (b - c)¹¹ + (c - a)²⁰¹⁰

 = (a - a)²⁰ + (a - a)¹¹ + (a - a)²⁰¹⁰ 

= 0 + 0 + 0

= 0 .

=> ĐPCM

135 - 3 ( x + 1 ) = 30

3 ( x + 1 ) = 105

x + 1 = 35

x = 34

\(1005^2\cdot1005^x=1005^7\)

\(1005^{2+x}=1005^7\)

=> x + 2 = 7

=> x = 5

Vậy,........

\(135-3.\left(x+1\right)=30\)

\(\Rightarrow3.\left(x+1\right)=105\)

\(\Rightarrow x+1=35\)

\(\Rightarrow x=34\)

\(1005^2.1005^x=1005^7\)

\(\Rightarrow1005^x=1005^7\div1005^2\)

\(\Rightarrow1005^x=1005^{7-2}\)

\(\Rightarrow1005^x=1005^2\)

\(\Rightarrow x=2\)

1005^2.1005^x=1005^7

<=>1005^2+x=1005^7

<=>2+x=7

<=>x=5