400-396+392-388+...-204+200
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4N
2
H9
HT.Phong (9A5)
CTVHS
6 tháng 2 2023
Các bước giải chi tiết
a) \(\dfrac{392-x}{32}\) + \(\dfrac{390-x}{34}\) + \(\dfrac{388-x}{36}\) = -3
⇔ \(\dfrac{392-x}{32}\)+1+\(\dfrac{390-x}{34}\)+1+\(\dfrac{388-x}{36}\)+1 = 0
⇔\(\dfrac{424-x}{32}\)+\(\dfrac{424-x}{34}\)+\(\dfrac{424-x}{36}\)=0
⇔\(\left(424-x\right)\)\(\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}\right)\)=0
⇔\(424-x\) = 0\(\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}\ne\forall x\right)\)
⇔\(x=424\)
b) \(\dfrac{x-3}{3}\)- \(x\) = \(5-\dfrac{x+1}{4}\)
⇔\(\dfrac{x-3-3x}{3}\) = 5 + \(\dfrac{-\left(x+1\right)}{4}\)
⇔\(\dfrac{-2x-3}{3}\) = 5 + \(\dfrac{-x-1}{4}\)
⇔\(\dfrac{-2x-3}{3}\) = \(\dfrac{20-x-1}{4}\)
⇔\(\dfrac{-2x-3}{3}\) = \(\dfrac{-x+19}{4}\)
⇔ \(4\left(-2x-3\right)\) = \(3\left(-x+19\right)\)
⇔\(-8x-12\) = \(-3x+57\)
⇔\(-8x\) = \(-3x+69\)
⇔\(-5x=69\)
⇔ \(x=-\dfrac{69}{5}\)
6 tháng 2 2023
(392-x)/32+(390-x)/34+(388-x)/36=-3
=>392-x)/32 +1 + (390-x)/34 +1 +(388-x)/36 +1=0
=>(424-x)/32+(424-x)/34+(424-x)/36=0
=>424-x=0(vì 1/32+1/34+1/36 khác 0)
=>x=424
L
0
TL
3
7 tháng 5 2018
B=\(\frac{12}{2^2.4^2}+\frac{20}{4^2.6^2}+......+\frac{388}{96^2.98^2}+\frac{396}{98^2.100^2}\)
=\(\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)
=\(\frac{1}{2^2}-\frac{1}{100^2}\)
=\(\frac{2599}{10000}< \frac{2500}{10000}=\frac{1}{4}\)
=> B<\(\frac{1}{4}\)
A
1
11 tháng 8 2020
204+456=660
123-96=27
37+140=177
160-145=15
129-753= -624
789-298= 491
291+102= 393
183+290= 473
1,000-123= 877
810-723= 87
986+209= 1195
371-289= 82
730-368=362
260+719= 979
102+763= 865
123+456+7= 586
289-200= 89
120+385= 405
392-277= 115
193+240= 433
782+926= 1708
HỌC TỐT ><
Hí hí =))) Mình trả lời đầu tiên đấy =)))
11 tháng 8 2020
204+456=660
123-96=27
37+140=177
160-145=15
129-753=-624
789-298=491
291+102=393
183+290=473
1,000-123=877
810-723=87
986+209=1195
371-289=82
730-368=362
260+719=979
102+763=865
123+456+7=586
289-200=89
120+385=505
392-277=115
193+240=433
782+926=1708
Chúc bạn hok tốt~~
PB
1
MN
22 tháng 1 2020
Ta có :
\(\frac{392-x}{32}+\frac{390-x}{34}+\frac{388-x}{36}+\frac{386-x}{38}+\frac{384-x}{40}=-5\)
\(\Leftrightarrow\left(\frac{392-x}{32}+1\right)+\left(\frac{390-x}{34}+1\right)+\left(\frac{388-x}{36}+1\right)+\left(\frac{386-x}{38}+1\right)+\left(\frac{384-x}{40}\right)=0\)
\(\Leftrightarrow\frac{424-x}{32}+\frac{424-x}{34}+\frac{424-x}{36}+\frac{424-x}{38}+\frac{424-x}{40}=0\)
\(\Leftrightarrow\left(424-x\right)\left(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\right)=0\)
Mà : \(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\ne0\)
\(\Leftrightarrow424-x=0\)
\(\Leftrightarrow x=424\)
Vậy x = 424
30 tháng 4 2017
ta có : \(\dfrac{392-x}{32}+\dfrac{390-x}{34}+\dfrac{388-x}{36}+\dfrac{386-x}{38}\)+\(\dfrac{384-x}{40}=-5\)
\(\Leftrightarrow\)\(\dfrac{392-x}{32}+1+\dfrac{390-x}{34}+1+\dfrac{388-x}{36}+1\)+\(\dfrac{384-x}{40}+1=0\)
\(\Leftrightarrow\)\(\dfrac{424-x}{32}+\dfrac{424-x}{34}+\dfrac{424-x}{36}+\dfrac{424-x}{38}+\dfrac{424-x}{40}=0\)\(\Leftrightarrow\left(424-x\right)\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}+\dfrac{1}{38}+\dfrac{1}{40}\right)=0\)
\(\Leftrightarrow x=424\)(vì \(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}+\dfrac{1}{38}+\dfrac{1}{40}\ne0\))
Vậy tập nghiệm của phương trình là s=\(\left\{424\right\}\)
\(400-396+392-388+...-204+200\)
\(=\left(400-396\right)+\left(392-388\right)+...+\left(208-204\right)+200\)
\(=4+4+...+4+200\)
\(=4\times25+200=300\)
400-396+392-388+...-204+200
=(400-396)+(392-388)+...+(208-204)+200
=4+4+...+4+200
=4x25+200
=300