tìm x biết
\(\frac{x+10}{270}+\frac{x+20}{260}=\frac{x+30}{250}+\frac{x+40}{240}\)
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NN
2
16 tháng 9 2016
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 @@@
NN
tìm x giúp m nha đúng mình tíck
\(\frac{x+10}{270}+\frac{x+20}{260}=\frac{x+3}{250}+\frac{x+3}{240}\)
2
16 tháng 9 2016
\(\Leftrightarrow\frac{x+10}{270}+1+\frac{x+20}{260}+1=\frac{x+30}{250}+1+\frac{x+40}{240}+1\)
\(\Leftrightarrow\frac{x+280}{270}+\frac{x+280}{260}=\frac{x+280}{250}+\frac{x+280}{240}\)
\(\Leftrightarrow\left(x+280\right)\left(\frac{1}{270}+\frac{1}{260}\right)=\left(x+280\right)\left(\frac{1}{250}+\frac{1}{240}\right)\)
\(\Leftrightarrow\left(x+280\right)\left(\frac{1}{270}+\frac{1}{260}-\frac{1}{250}-\frac{1}{240}\right)=0\)
=>x=-280
NN
0
3 tháng 7 2017
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
NN
1
27 tháng 5 2020
\(\frac{x+10}{90}+\frac{x+20}{80}+\frac{x+30}{70}+\frac{x+40}{60}+\frac{x+50}{50}=-5\)
<=> \(\frac{x+10}{90}+1+\frac{x+20}{80}+1+\frac{x+30}{70}+1+\frac{x+40}{60}+1+\frac{x+50}{50}+1=0\)
<=> \(\frac{x+100}{90}+\frac{x+100}{80}+\frac{x+100}{70}+\frac{x+100}{60}+\frac{x+100}{50}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{90}+\frac{1}{80}+\frac{1}{70}+\frac{1}{60}+\frac{1}{50}\right)=0\)
<=> x + 100 = 0
<=> x = -100
Vậy x = -100
MT
6 tháng 1 2016
\(\Rightarrow\frac{x+10}{490}+\frac{x+20}{480}+\frac{x+30}{470}+\frac{x+40}{460}+\frac{x+50}{450}+5=0\)
\(\Rightarrow\frac{x+10}{490}+1+\frac{x+20}{480}+1+\frac{x+30}{470}+1+\frac{x+40}{460}+1+\frac{x+50}{450}+1=0\)
\(\Rightarrow\frac{x+500}{490}+\frac{x+500}{480}+\frac{x+500}{470}+\frac{x+500}{460}+\frac{x+500}{450}=0\)
\(\Rightarrow\left(x+500\right).\left(\frac{1}{490}+\frac{1}{480}+\frac{1}{470}+\frac{1}{460}+\frac{1}{450}\right)=0\Rightarrow x+500=0\Rightarrow x=-500\)
DT
0
MR
3
1 tháng 1 2020
áp dụng DSTCBN:
Ta có:
\(\frac{40}{x-30}=\frac{20}{y-15}=\frac{28}{z-21}\Leftrightarrow\frac{x-30}{40}=\frac{y-15}{20}=\frac{z-21}{28}\)
\(\Rightarrow\frac{x-30}{10}=\frac{y-15}{5}=\frac{z-21}{7}\)
\(\frac{\Rightarrow x}{10}-\frac{30}{10}=\frac{y}{5}-\frac{15}{5}=\frac{z}{7}-\frac{21}{7}\)
\(\frac{\Rightarrow x}{10}-3=\frac{y}{3}-3=\frac{z}{7}-3\)
\(\frac{\Rightarrow x}{10}=\frac{y}{5}=\frac{z}{7}\)
\(\frac{x}{10}=\frac{y}{5}=\frac{z}{7}=t=\hept{\begin{cases}x=10t\\y=5t\\z=7t\end{cases}}\)
\(xyz=22400\Leftrightarrow350t^3=22400\Leftrightarrow t^3=64\Rightarrow t=4\)
\(\Rightarrow\hept{\begin{cases}x=40\\y=20\\z=28\end{cases}}\)
HT
1 tháng 1 2020
\(\text{Ta có:}\)\(\frac{40}{x-30}=\frac{20}{y-15}=\frac{28}{z-21}\)
\(\Leftrightarrow\frac{x-30}{40}=\frac{y-15}{40}=\frac{z-21}{28}\)
\(\Leftrightarrow\frac{x}{40}-\frac{30}{40}=\frac{y}{40}-\frac{15}{40}=\frac{z}{28}-\frac{21}{28}\)
\(\Leftrightarrow\frac{x}{40}-\frac{3}{4}=\frac{y}{20}-\frac{3}{4}=\frac{z}{28}-\frac{3}{4}\)\
\(\Leftrightarrow\frac{x}{40}=\frac{y}{20}=\frac{z}{28}\)
\(\text{đặt:}\)\(\frac{x}{40}=\frac{y}{20}=\frac{z}{28}=k\)
\(\Rightarrow x=40k\)
\(\Rightarrow y=20k\)
\(\Rightarrow z=28k\)
\(\text{Theo đề ta có :}\)\(x.y.z=22400\Rightarrow40k.20k.28k=22400\)
\(\Rightarrow22400.k^3=22400\)
\(\Rightarrow k^3=1\)
\(\Rightarrow k=\pm1\)
\(\text{Với k=1 thì :}\)\(\hept{\begin{cases}x=40\\y=20\\z=28\end{cases}}\)
\(\text{Với k=-1 thì :}\)\(\hept{\begin{cases}x=-40\\y=-20\\z=-28\end{cases}}\)
\(x=-280\)
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