đề bài là tìm x hả bn

\(A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-...+\frac{1}{3^{2017}}-\frac{1}{3^{2018}}\)

\(\Rightarrow3A=1-\frac{1}{3}+\frac{1}{3^2}-...+\frac{1}{3^{2016}}-\frac{1}{3^{2017}}\)

\(4A=\)\(1-\frac{1}{3^{2018}}\)

\(A=\frac{1}{4}-\frac{1}{3^{2018}.4}\)

bạn ơi còn nx ko bạn hay đáp án là vầy

bài này nha

961⁴²  < 10 * 23⁹³

Đi 2, 3 thôi là đc.

Bài 3:

Bạn tham khảo

Nguồn : h.o.c24.vn

Ta có : \(\frac{a+b}{b+c}=\frac{c+d}{d+a}\)

=> (a + b)(d + a) = (b + c)(c + d) 

=> ad + a² + bd + ab = bc + bd + c² + cd 

=> ad + a² + ab = bc + c² + cd 

=> ad + a² + ab - bc - c² - cd = 0 

=> (ad - cd) + (a² - c²) + (ab - bc) = 0

=> d(a - c) + (a - c)(a + c) + b(a - c) = 0 

=> (a - c)(d + a + c + b) = 0

=> \(\orbr{\begin{cases}a-c=0\\d+a+c+b=0\end{cases}}\Rightarrow\orbr{\begin{cases}a=c\\a+b+c+d=0\end{cases}}\)(đpcm) 

\(\frac{5}{x+1}=\frac{x+1}{20}\)ĐK : \(x\ne-1\)

\(\Rightarrow\left(x+1\right)^2=100=10^2\)

TH1 : \(x+1=10\Leftrightarrow x=9\)

TH2 : \(x+1=-10\Leftrightarrow x=-11\)

như này được chưa ? 

5/x+1 = x+1/20

x= -11 

x= 9

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2n}\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2n}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2n}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2n}-\left(1+\frac{1}{2}+...+\frac{1}{n}\right)\)

\(=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\)

Bạn đăng tách ra cho mn cùng giúp nhé 

Bài 6 :

a, \(\frac{3x+5y}{x-2y}=\frac{1}{4}\Rightarrow12x+20y=x-2y\)ĐK : \(x\ne2y\)

\(\Leftrightarrow11x=-22y\Leftrightarrow\frac{x}{y}=-\frac{22}{11}=-2\)

b, Với \(a\ne3;b\ne-6\)

\(\frac{a-3}{a+3}=\frac{b-6}{b+6}\Rightarrow\left(a-3\right)\left(b+6\right)=\left(b-6\right)\left(a+3\right)\)

\(\Leftrightarrow ab+6a-3b-18=ab+3b-6a-18\)

\(\Leftrightarrow-6b+12a=0\Leftrightarrow6b=12a\Leftrightarrow\frac{a}{b}=\frac{6}{12}=\frac{1}{2}\)( đpcm )

Bạn đăng tách ra cho mn cùng giúp nhé 

Bài 3 :

a, \(\frac{x+6}{8}+\frac{x+8}{6}+\frac{x+1}{13}+3=0\)

\(\Leftrightarrow\frac{x+6}{8}+1+\frac{x+8}{6}+1+\frac{x+1}{13}+1=0\)

\(\Leftrightarrow\frac{x+14}{8}+\frac{x+14}{6}+\frac{x+14}{13}=0\)

\(\Leftrightarrow\left(x+14\right)\left(\frac{1}{8}+\frac{1}{6}+\frac{1}{13}\ne0\right)=0\Leftrightarrow x=-14\)

b, \(\frac{x-5}{10}+\frac{x-7}{8}+\frac{x-1}{14}=3\)

\(\Leftrightarrow\frac{x-5}{10}+\frac{x-7}{8}+\frac{x-1}{14}-3=0\)

\(\Leftrightarrow\frac{x-5}{10}-1+\frac{x-7}{8}-1+\frac{x-1}{14}-1=0\)

\(\Leftrightarrow\frac{x-15}{10}+\frac{x-15}{8}+\frac{x-15}{14}=0\)

\(\Leftrightarrow\left(x-15\right)\left(\frac{1}{10}+\frac{1}{8}+\frac{1}{14}\ne0\right)=0\Leftrightarrow x=15\)

vâng lần sau mik sửa cảm ơn bạn góp ý