Tìm x, biết:
\(\frac{x+5}{200}+\frac{x+4}{201}=\frac{x+3}{202}+\frac{x+2}{203}\)
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NK
4
JY
1
1 tháng 1 2018
<=> (2-x/201 + 1) + (x/203 - 1) = (1-x/202 + 1) + (1-1)
<=> 203-x/201 + x-203/203 = 203-x/202
<=> 203-x/201 - 203-x/203 - 203-x/202 = 0
<=> (203-x).(1/201-1/203-1/202) = 0
<=> 203-x = 0 ( vì 1/201-1/203-1/202 khác 0 )
<=> x=203
Vậy x=203
k mk nha
BT
22 tháng 2 2017
BT1: 20152014 có tận cùng là 5
20142015=2014.(20142)1007=2014.40561961007=2014.(...6) => Có tận cùng là ...4
=> 20152014-20142015 có tận cùng là ...5-...4=...1
BT2: f(1)=a.1+b=1 (1)
f(2)=a.2+b=4 (2)
Trừ (2) cho (1) => a=3
Thay a=3 vào (1) => b=-2
ĐS: a=3; b=-2
9 tháng 5 2018
a/\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)
=\(\frac{2^3.5^3.7^4}{2^2.5^2.7^4}\)
=2.5
=10
NN
1
14 tháng 9 2016
\(\frac{x+199}{200}+\frac{x+198}{201}+\frac{x+197}{202}=-3\)
\(\frac{x+199}{200}+1+\frac{x+198}{201}+1+\frac{x+197}{202}+1=0\)
\(\frac{x+399}{200}+\frac{x+399}{201}+\frac{x+399}{202}=0\)
\(\left(x+399\right)\left(\frac{1}{200}+\frac{1}{201}+\frac{1}{202}\right)=0\)
Mà \(\left(\frac{1}{200}+\frac{1}{201}+\frac{1}{202}\right)\ne0\)
=> x + 399 = 0
=> x = -399
21 tháng 2 2017
\(\frac{x+1}{203}+\frac{x+2}{202}+\frac{x+3}{201}+\frac{x+4}{200}+\frac{x+5}{199}+5=0\)
\(\Leftrightarrow\frac{x+1}{203}+1+\frac{x+2}{202}+1+\frac{x+3}{201}+1+\frac{x+4}{200}+1+\frac{x+5}{199}+1=0\)
\(\Leftrightarrow\frac{x+204}{203}+\frac{x+204}{202}+\frac{x+204}{201}+\frac{x+204}{200}+\frac{x+204}{199}=0\)
\(\Leftrightarrow\left(x+204\right)\left(\frac{1}{203}+\frac{1}{203}+\frac{1}{201}+\frac{1}{200}+\frac{1}{199}\right)=0\)
\(\Leftrightarrow x+204=0\).Do \(\frac{1}{203}+\frac{1}{203}+\frac{1}{201}+\frac{1}{200}+\frac{1}{199}\ne0\)
\(\Leftrightarrow x=-204\)
IM
21 tháng 2 2017
Ta có :
\(\frac{x+1}{203}+\frac{x+2}{202}+\frac{x+3}{201}+\frac{x+4}{200}+\frac{x+5}{199}+5=0\)
\(\Leftrightarrow\left(\frac{x+1}{203}+1\right)+\left(\frac{x+2}{202}+1\right)+\left(\frac{x+3}{201}+1\right)+\left(\frac{x+4}{200}+1\right)+\left(\frac{x+5}{199}+1\right)=0\)
\(\Leftrightarrow\left(\frac{x+204}{203}\right)+\left(\frac{x+4}{202}\right)+\left(\frac{x+4}{201}\right)+\left(\frac{x+204}{200}\right)+\left(\frac{x+204}{199}\right)=0\)
\(\Leftrightarrow\left(x+204\right)\left(\frac{1}{203}+\frac{1}{202}+\frac{1}{201}+\frac{1}{200}+\frac{1}{199}\right)=0\)
Dễ thấy \(\left(\frac{1}{203}+\frac{1}{202}+\frac{1}{201}+\frac{1}{200}+\frac{1}{199}\right)\ne0\)
=> x + 204 = 0
<=> x = - 204
Vậy pt có nghiệm x = - 204
19 tháng 3 2020
Ta có : \(\frac{x+4}{200}+\frac{x+3}{201}=\frac{x+2}{202}+\frac{x+1}{203}\)
=> \(\frac{x+4}{200}+\frac{x+3}{201}-\frac{x+2}{202}-\frac{x+1}{203}=0\)
=> \(\frac{x+4}{200}+1+\frac{x+3}{201}+1-\frac{x+2}{202}-1-\frac{x+1}{203}-1=0\)
=> \(\frac{x+204}{200}+\frac{x+204}{201}-\frac{x+204}{202}-\frac{x+204}{203}=0\)
=> \(\left(x+204\right)\left(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\right)=0\)
=> \(x+204=0\)
=> \(x=-204\)
Vậy phương trình có tập nghiệm là S = { -204 }
\(\frac{x+5}{200}+\frac{x+4}{201}=\frac{x+3}{202}+\frac{x+2}{203}\)
=> \(\left(1+\frac{x+5}{200}\right)+\left(1+\frac{x+4}{201}\right)=\left(1+\frac{x+3}{202}\right)+\left(1+\frac{x+2}{203}\right)\)
=> \(\frac{x+205}{200}+\frac{x+205}{201}=\frac{x+205}{202}+\frac{x+205}{203}\)
=> \(\frac{x+205}{200}+\frac{x+205}{201}-\frac{x+205}{202}-\frac{x+205}{203}=0\)
=> \(\left(x+205\right).\left(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\right)=0\)
Do \(\frac{1}{200}>\frac{1}{202};\frac{1}{201}>1-\frac{1}{203}\)
=> \(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\ne0\)
=> \(x+205=0\)
=> \(x=-205\)
\(\frac{x+5}{200}+\frac{x+4}{201}=\frac{x+3}{202}+\frac{x+2}{203}\)
\(=>\frac{x+5+200}{200}+\frac{x+4+201}{201}-\frac{x+3+202}{202}-\frac{x+2+203}{203}=0\)
\(=>\frac{x+205}{200}+\frac{x+205}{201}-\frac{x+205}{202}-\frac{x+205}{203}=0\)
\(=>\left(x+205\right).\left(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\right)=0\)
\(Do:\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\ne0\)
\(=>x+205=0\)
\(=>x=-205\)