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ko chép lại đầu bài nha
=(53x+800):7020=(49x+147):6000
=(53x+800).6000=70
x=145,0369515
lưu ý:mik làm theo cách suy luận nhé @@@
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
nhân cả 2 vế của đẳng thức với 1/2 ta được
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{x\left(x+1\right)}=\frac{2014}{2015}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}=\frac{2014}{2015}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-......+\frac{1}{x}-\frac{1}{x+1}=\frac{2014}{2015}\)
\(=\frac{1}{2}-\frac{1}{x+1}=\frac{2014}{2015}\)
\(=>\frac{1}{x+1}=\frac{1}{2}-\frac{2014}{2015}\)
\(\frac{1}{x+1}=-\frac{2013}{4030}\)
hay \(1:\left(x+1\right)=-\frac{2013}{4030}\)
\(x+1=-\frac{4030}{2013}\)
\(=>x=-\frac{6043}{2013}\)
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{97.100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{97}-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\frac{99}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{33}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow x=\frac{0,33\times100}{0,33}=100\)
Ta có: \(x-\frac{20}{11\cdot13}-\frac{20}{13\cdot15}-...-\frac{20}{53\cdot55}=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\left(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+...+\frac{2}{53\cdot55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\frac{4}{55}=\frac{3}{11}\)
\(\Leftrightarrow x-\frac{8}{11}=\frac{3}{11}\)
\(\Leftrightarrow x=\frac{3}{11}+\frac{8}{11}\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)thỏa mãn đề.
a) (x + 5) / 95 + (x +10)/90 + (x + 15)/85 + (x + 20)/80 = -4
<=> (x + 5)/95 + (x + 5)/90 + 5/90 + (x + 5)/85 + 10/85+ (x + 5)/80 + 15/80 = -4
<=> (x + 5)(1/95+1/90+1/85+1/80) =-4 -5/90-10/85-15/85
<=> (x + 5)(1/95+1/90+1/85+1/80)= -1-(1 + 5/90 )-(1 + 10/85) - (1 + 15/80)
<=>(x + 5)(1/95+1/90+1/85+1/80) = -1 - 95/90 - 95/85 - 95/80
<=>(x + 5)(1/95+1/90+1/85+1/80) = -95 (1/95+1/90+1/85+1/80)
<=> x + 5 = -95 => x = -100
\(\Leftrightarrow\frac{x+10}{270}+1+\frac{x+20}{260}+1=\frac{x+30}{250}+1+\frac{x+40}{240}+1\)
\(\Leftrightarrow\frac{x+280}{270}+\frac{x+280}{260}=\frac{x+280}{250}+\frac{x+280}{240}\)
\(\Leftrightarrow\left(x+280\right)\left(\frac{1}{270}+\frac{1}{260}\right)=\left(x+280\right)\left(\frac{1}{250}+\frac{1}{240}\right)\)
\(\Leftrightarrow\left(x+280\right)\left(\frac{1}{270}+\frac{1}{260}-\frac{1}{250}-\frac{1}{240}\right)=0\)
=>x=-280
đề vế phải sai rồi