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\(\Rightarrow\frac{x+10}{490}+\frac{x+20}{480}+\frac{x+30}{470}+\frac{x+40}{460}+\frac{x+50}{450}+5=0\)
\(\Rightarrow\frac{x+10}{490}+1+\frac{x+20}{480}+1+\frac{x+30}{470}+1+\frac{x+40}{460}+1+\frac{x+50}{450}+1=0\)
\(\Rightarrow\frac{x+500}{490}+\frac{x+500}{480}+\frac{x+500}{470}+\frac{x+500}{460}+\frac{x+500}{450}=0\)
\(\Rightarrow\left(x+500\right).\left(\frac{1}{490}+\frac{1}{480}+\frac{1}{470}+\frac{1}{460}+\frac{1}{450}\right)=0\Rightarrow x+500=0\Rightarrow x=-500\)
a: x/5=32/80
nên x/5=2/5
hay x=2
13/x=26/30
nên 13/x=13/15
hay x=15
-x/7=22/-77
=>x/7=2/7
hay x=2
b: x/9=28/36
=>x/9=7/9
hay x=7
-10/x=50/55
=>-10/x=10/11
hay x=-11
=2/10+3/10+4/10+......+13/10
=\(\frac{2+3+4+......+13}{10}\)
=90/10=9
k cho mình nha
\(\frac{x+5}{95}+\frac{x+10}{90}+\frac{x+15}{85}+\frac{x+20}{80}=-4\)
<=> \(\frac{x+5}{95}+1+\frac{x+10}{90}+1+\frac{x+15}{85}+1+\frac{x+20}{80}+1=0\)
<=> \(\frac{x+100}{95}+\frac{x+100}{90}+\frac{x+100}{85}+\frac{x+100}{80}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{95}+\frac{1}{90}+\frac{1}{85}+\frac{1}{80}\right)=0\)
<=> \(x+100=0\) (do 1/95 + 1/90 + 1/85 + 1/80 khác 0)
<=> \(x=-100\)
Vậy...
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
\(\left(x+50\%\right):\frac{7}{8}=\frac{5}{7}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)=\frac{5}{7}.\frac{7}{8}\)
\(\Rightarrow x+\frac{1}{2}=\frac{5}{8}\)
\(\Rightarrow x=\frac{5}{8}-\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{8}\)
Vậy...
Mình làm tiếp bài của bạn " I have a crazy idea "
b) \(\frac{25-x}{3}=\frac{15}{2}\)
Áp dụng tỉ lệ thức:
\(\left(25-x\right).2=15.3\)
\(\Rightarrow25-x=\frac{15.3}{2}=\frac{45}{2}\Leftrightarrow x=25-\frac{45}{2}=\frac{5}{2}\)
c) \(x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}=1\)
\(\Rightarrow x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=1\)
\(\Rightarrow x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=1\)
\(\Rightarrow x-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\right)=1\)
\(\Rightarrow x-\left(\frac{1}{1}-\frac{1}{7}\right)=1\Leftrightarrow x-\frac{6}{7}=1\Leftrightarrow x=1+\frac{6}{7}=\frac{13}{7}\)
A=\(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{13}-\frac{1}{14}\)=\(\frac{1}{7}-\frac{1}{14}\)=\(\frac{1}{14}\)
B=0
\(\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}\)
\(=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}\)
\(=\frac{1}{7}-\frac{1}{14}=\frac{1}{14}\)
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
\(\frac{x+10}{90}+\frac{x+20}{80}+\frac{x+30}{70}+\frac{x+40}{60}+\frac{x+50}{50}=-5\)
<=> \(\frac{x+10}{90}+1+\frac{x+20}{80}+1+\frac{x+30}{70}+1+\frac{x+40}{60}+1+\frac{x+50}{50}+1=0\)
<=> \(\frac{x+100}{90}+\frac{x+100}{80}+\frac{x+100}{70}+\frac{x+100}{60}+\frac{x+100}{50}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{90}+\frac{1}{80}+\frac{1}{70}+\frac{1}{60}+\frac{1}{50}\right)=0\)
<=> x + 100 = 0
<=> x = -100
Vậy x = -100