Đặt \(A=\frac{5}{1.7}+\frac{5}{7.13}+\frac{5}{13.19}+....+\frac{5}{x.\left(x+6\right)}\)

\(\Rightarrow A=\frac{5}{6}.\left(\frac{6}{1.7}+\frac{6}{7.13}+...+\frac{6}{x.\left(x+6\right)}\right)\)

\(\Rightarrow A=\frac{5}{6}.\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+....+\frac{1}{x}-\frac{1}{x+6}\right)\)

\(\Rightarrow A=\frac{5}{6}.\left(1-\frac{1}{x+6}\right)\)

\(\Rightarrow\frac{5}{6}.\frac{x+5}{x+6}=\frac{10075}{12096}\)

Làm nốt nha

\(\frac{5}{1.7}+\frac{5}{7.13}+...+\frac{5}{x.\left(x+6\right)}=\frac{10075}{12096}\)

\(\Rightarrow\frac{5}{6}.\left(\frac{6}{1.7}+\frac{6}{7.13}+...+\frac{6}{x.\left(x+6\right)}\right)=\frac{10075}{12096}\)

\(\Rightarrow\frac{5}{6}.\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{x}-\frac{1}{x+6}\right)=\frac{10075}{12096}\)

\(\Rightarrow\frac{5}{6}.\left(1-\frac{1}{x+6}\right)=\frac{10075}{12096}\)

\(\Rightarrow1-\frac{1}{x+6}=\frac{10075}{12096}:\frac{5}{6}\)

\(\Rightarrow1-\frac{1}{x+6}=\frac{10075}{12096}.\frac{6}{5}\)

\(\Rightarrow1-\frac{1}{x+6}=\frac{2015}{2016}\)

\(\Rightarrow\frac{1}{x+6}=1-\frac{2015}{2016}\)

\(\Rightarrow\frac{1}{x+6}=\frac{1}{2016}\)

\(\Rightarrow x+6=2016\)

\(\Rightarrow x=2016-6\)

\(\Rightarrow x=2010\)

Chúc bạn học tốt !!! 

1.7 5 + 7.13 5 + ... + x. x + 6 5 = 12096 10075 ⇒ 6 5 . 1.7 6 + 7.13 6 + ... + x. x + 6 6 = 12096 10075 ⇒ 6 5 . 1 − 7 1 + 7 1 − 13 1 + ... + x 1 − x + 6 1 = 12096 10075 ⇒ 6 5 . 1 − x + 6 1 = 12096 10075 ⇒1 − x + 6 1 = 12096 10075 : 6 5 ⇒1 − x + 6 1 = 12096 10075 . 5 6 ⇒1 − x + 6 1 = 2016 2015 ⇒ x + 6 1 = 1 − 2016 2015 ⇒ x + 6 1 = 2016 1 ⇒x + 6 = 2016 ⇒x = 2016 − 6 ⇒x = 2010\(1.7 5 + 7.13 5 + ... + x. x + 6 5 = 12096 10075 ⇒ 6 5 . 1.7 6 + 7.13 6 + ... + x. x + 6 6 = 12096 10075 ⇒ 6 5 . 1 − 7 1 + 7 1 − 13 1 + ... + x 1 − x + 6 1 = 12096 10075 ⇒ 6 5 . 1 − x + 6 1 = 12096 10075 ⇒1 − x + 6 1 = 12096 10075 : 6 5 ⇒1 − x + 6 1 = 12096 10075 . 5 6 ⇒1 − x + 6 1 = 2016 2015 ⇒ x + 6 1 = 1 − 2016 2015 ⇒ x + 6 1 = 2016 1 ⇒x + 6 = 2016 ⇒x = 2016 − 6 ⇒x = 2010\)

50375

ai bít thì trả lời giúp tớ với đang gấp quá trời nè 3;;

50375 nhé em

huhuhuhu help me cứi tui

\(2)1+1+2+2^2+\cdot\cdot\cdot+2^n=2^{101}\)

\(\Rightarrow1+2+2^2+\cdot\cdot\cdot+2^n=2^{101}-1\)

\(\Rightarrow2+2^2+2^3+\cdot\cdot\cdot+2^{n+1}=2^{102}-2\)

\(\Rightarrow\left(2+2^2+\cdot\cdot\cdot+2^{n+1}\right)-\left(1+2+\cdot\cdot\cdot+2^n\right)=\left(2^{102}-2\right)-\left(2^{101}-1\right)\)

\(\Rightarrow2^{n+1}-1=2^{101}-1\)

\(\Rightarrow2^{n+1}=2^{101}\)

\(\Rightarrow n+1=101\)

\(\Rightarrow n=100\)

\(1)a+\left(a+2\right)+\left(a+4\right)+\left(a+6\right)+\left(a+8\right)=10075(a⋮̸2)\)

\(\Rightarrow5a+\left(2+4+6+8\right)=10075\)

\(\Rightarrow5a=10075-20\)

\(\Rightarrow5a=10055\)

\(\Rightarrow a=2011\)

bài 2 nữa mà bạn

\(\left(x+5\right)^3=\left(x+5\right)^2\)

`(x+5)^3 -(x+5)^2 =0`

`(x+5)^2 (x+5-1)=0`

`(x+5)^2 (x+4)=0`

\(=>\left[{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)

a. A=1000-|x+5| < 1000

=> GTLN của A là 1000

<=> x + 5 = 0

<=> x = -5

b. B = |x-3| + 5 > 5

=> GTNN của B là 5

<=> x - 3 = 0

<=> x = 3

a, x= -5

b, x= -3

a) A = 1000 - |x + 5| \(\le\)1000

Vậy GTLN của A = 1000 khi

|x + 5|  = 0 => x=  -5

b)B =  |x - 3| + 5 \(\ge\) 5 

Vậy GTNN của B = 5 khi

|x - 3| = 0 => x = 3 

\(a,\dfrac{1}{5}-\dfrac{1}{5}:x=\dfrac{3}{5}\\ \Rightarrow\dfrac{1}{5}:x=\dfrac{1}{5}-\dfrac{3}{5}\\ \Rightarrow\dfrac{1}{5}:x=\dfrac{-2}{5}\\ \Rightarrow x=\dfrac{1}{5}:\dfrac{-2}{5}\\ \Rightarrow x=\dfrac{-1}{2}\)

câu b thiếu đề

a: =>1/5:x=-2/5

hay x=-1/2

<=> x + x/5 = 5 - 9/2

<=> 6/5x = 1/2

<=> x = 5/12 

\(x+\frac{9}{2}=5-\frac{x}{5}\)

\(x+\frac{x}{5}=5-\frac{9}{2}\)

\(x\left(1+\frac{1}{5}\right)=\frac{10}{2}-\frac{9}{2}\)

\(x\cdot\left(\frac{5}{5}+\frac{1}{5}\right)=\frac{1}{2}\)

\(x\cdot\frac{6}{5}=\frac{1}{2}\)

\(x=\frac{1}{2}:\frac{6}{5}\)

\(x=\frac{5}{12}\)

\(x+\frac{9}{2}=5-\frac{x}{5}\)

\(x=5-\frac{9}{2}-\frac{x}{5}\)

\(x=\frac{1}{2}-\frac{x}{5}\)

\(\frac{1}{2}=x-\frac{x}{5}=\frac{4x}{5}\)

\(\frac{5}{10}=\frac{8x}{10}\)

\(=>x=\frac{5}{8}\)