\(x=-280\)
 

HD dùng PP Quy đồng tử (không quy đồng Mẫu)

\(\left(\frac{x+10}{270}+10\right)+\left(\frac{x+20}{260}+10\right)=\left(\frac{x+30}{250}+10\right)+\left(\frac{x+40}{240}+10\right)\\ \)

\(\left(x+280\right)\left(....\right)=0\)chú ý (...) thường khác không nếu bằng =0=> đúng với mọi x

nếu khác không=> x=-280

\(\frac{x+10}{90}+\frac{x+20}{80}+\frac{x+30}{70}+\frac{x+40}{60}+\frac{x+50}{50}=-5\)

<=> \(\frac{x+10}{90}+1+\frac{x+20}{80}+1+\frac{x+30}{70}+1+\frac{x+40}{60}+1+\frac{x+50}{50}+1=0\)

<=> \(\frac{x+100}{90}+\frac{x+100}{80}+\frac{x+100}{70}+\frac{x+100}{60}+\frac{x+100}{50}=0\)

<=> \(\left(x+100\right)\left(\frac{1}{90}+\frac{1}{80}+\frac{1}{70}+\frac{1}{60}+\frac{1}{50}\right)=0\)

<=> x + 100 = 0 

<=> x = -100

Vậy x = -100

a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\) 

\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)

\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)

Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)

Nên x + 1 = 0

=> x = -1

còn b vs c thì sao ạ

b) Sai đề à bạn đề \(\frac{x+2}{42}+\frac{x+4}{22}=\frac{x+5}{23}+\frac{x+3}{43}\)  hả đề này mk làm đc 

\(\Rightarrow\frac{x+10}{490}+\frac{x+20}{480}+\frac{x+30}{470}+\frac{x+40}{460}+\frac{x+50}{450}+5=0\)

\(\Rightarrow\frac{x+10}{490}+1+\frac{x+20}{480}+1+\frac{x+30}{470}+1+\frac{x+40}{460}+1+\frac{x+50}{450}+1=0\)

\(\Rightarrow\frac{x+500}{490}+\frac{x+500}{480}+\frac{x+500}{470}+\frac{x+500}{460}+\frac{x+500}{450}=0\)

\(\Rightarrow\left(x+500\right).\left(\frac{1}{490}+\frac{1}{480}+\frac{1}{470}+\frac{1}{460}+\frac{1}{450}\right)=0\Rightarrow x+500=0\Rightarrow x=-500\)

Khó Lazy kute

Biết khó mới hỏi ! Câu dễ thì tớ gửi cho zui thui !!!!!!!!

\(\text{Ta có: }\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+.....+\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.....+\frac{1}{\left(x+2\right)}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)

\(\Rightarrow\frac{1}{\left(x+5\right)}=\frac{1}{2}-\frac{3}{20}\)

\(\frac{x-20}{x-10}=\frac{x+40}{x+70}\)

\(\Leftrightarrow\left(x-20\right)\left(x+70\right)=\left(x-10\right)\left(x+40\right)\)

\(\Leftrightarrow x^2+70x-20x-140=x^2+40x-10x-140\)

\(\Leftrightarrow50x=30x\)

\(\Leftrightarrow50x-30x=0\)

\(\Leftrightarrow20x=0\)

\(\Leftrightarrow x=0\)

Chết , sửa chút !!! Nhân ngu wá

\(\frac{x-20}{x-10}=\frac{x+40}{x+70}\)

\(\Leftrightarrow\left(x-20\right)\left(x+70\right)=\left(x-10\right)\left(x+40\right)\)

\(\Leftrightarrow x^2+70x-20x-140=x^2+40x-10x-400\)

\(\Leftrightarrow20x+260=0\)

\(\Leftrightarrow x=-13\)

Incursion_03 :> 

\(\frac{x-20}{x-10}=\frac{x+40}{x+70}\)

\(\Rightarrow\left(x-20\right).\left(x+70\right)=\left(x-10\right).\left(x+40\right)\)

\(\Rightarrow x^2+50x-1400=x^2+30-400\)

\(\Rightarrow20x=1000\Rightarrow x=50\)

mink cần gấp

\(\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{128}=\frac{1}{x-2}\)

\(\Leftrightarrow\frac{1}{10\cdot1}+\frac{1}{10\cdot2}+\frac{1}{10\cdot3}+\frac{1}{10\cdot4}+...+\frac{1}{10\cdot128}=\frac{1}{x-2}\)

\(\Leftrightarrow\frac{1}{10}\cdot\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)=\frac{1}{x-2}\)

Đặt \(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\)

\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^6}\)

\(2A-A=2-\frac{1}{2^7}\)

Thay vào biểu thức ta có :

\(\frac{1}{10}\cdot\left(2-\frac{1}{2^7}\right)=\frac{1}{x-2}\)

\(\Leftrightarrow\frac{1}{10}\cdot\frac{255}{128}=\frac{1}{x-2}\Leftrightarrow\frac{51}{256}=\frac{1}{x-2}\)

\(\Leftrightarrow51x-102=256\)

\(51x=358\Rightarrow x=\frac{358}{51}\)

Vậy ..................................