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

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

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

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

<=> x + 100 = 0 

<=> x = -100

Vậy x = -100

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

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

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

\(\Leftrightarrow50x=30x\)

\(\Leftrightarrow50x-30x=0\)

\(\Leftrightarrow20x=0\)

\(\Leftrightarrow x=0\)

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 :> 

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

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

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

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

\(x=-280\)
 

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

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 !!!!!!!!

\(\frac{x-20}{x-10}=\frac{x+40}{x+70}=\frac{-20-40}{-10-70}=\frac{6}{8}=\frac{3}{4}.\)

\(\Rightarrow4\cdot\left(x-20\right)=3\cdot\left(x-10\right)\Leftrightarrow4x-80=3x-30\Leftrightarrow x=50.\)

bn dinh thuy linh gioi that, toàn dung ngôn ngữ toán hoc nên bai lam rat ấn tuong

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

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

\(\Rightarrow x^2+70x-20x-1400=x^2+40x-10x-400\)

\(\Rightarrow x^2+70x-20x-x^2-40x+10x=1400-400\)

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

<=> (x-18/74 - 1)+(x-20/72 - 1)+(x-22/70 - 1) = 0

<=> x-92/74 + x-92/72 + x-92/70 = 0

<=> (x-92).(1/74+1/72+1/70) = 0

<=> x-92 = 0 ( vì 1/74 + 1/72 + 1/70 > 0 )

<=> x=92

Vậy S = {92}

Tk mk nha

Ta có :

\(\frac{x-18}{74}+\frac{x-20}{72}+\frac{x-22}{70}=3\)

\(\Leftrightarrow\)\(\left(\frac{x-18}{74}-1\right)+\left(\frac{x-20}{72}-1\right)+\left(\frac{x-22}{70}-1\right)=3-3\)

\(\Leftrightarrow\)\(\frac{x-92}{74}+\frac{x-92}{72}+\frac{x-92}{70}=0\)

\(\Leftrightarrow\)\(\left(x-92\right)\left(\frac{1}{74}+\frac{1}{72}+\frac{1}{70}\right)=0\)

Vì \(\left(\frac{1}{74}+\frac{1}{72}+\frac{1}{70}\right)\ne0\)

\(\Rightarrow\)\(x-92=0\)

\(\Rightarrow\)\(x=92\)

Vậy \(x=92\)

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