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\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}+\frac{x-5}{95}=5\)
\(\Rightarrow\left(\frac{x-1}{99}-1\right)+\left(\frac{x-2}{98}-1\right)+\left(\frac{x-3}{97}-1\right)+\left(\frac{x-4}{96}-1\right)+\left(\frac{x-5}{95}-1\right)\)\(=5-1-1-1-1-1\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}+\frac{x-100}{95}=0\)
\(\Rightarrow\left(x-100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}\right)=0\)
Mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x=100
Chúc bạn học tốt
Ta có \(1\frac{1}{5}x+\frac{2}{3}x=-\frac{56}{125}\)
<=> \(\frac{6}{5}x+\frac{2}{3}x=-\frac{56}{125}\)
<=> \(\frac{28}{15}x=-\frac{56}{125}\)
<=> \(x=-\frac{2}{15}\)
b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
<=> \(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=0\)
<=> \(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
<=> x + 100 = 0
<=> x = -100
\(\frac{x+1}{99}+\frac{x+3}{97}+\frac{x+5}{95}=\frac{x+2}{98}+\frac{x+4}{96}+\frac{x+6}{94}\)
\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)=\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)+\left(\frac{x+6}{94}+1\right)\)
\(\left(\frac{x+1}{99}+\frac{99}{99}\right)+\left(\frac{x+3}{97}+\frac{97}{97}\right)+\left(\frac{x+5}{95}+\frac{95}{95}\right)=\left(\frac{x+2}{98}+\frac{98}{98}\right)+\left(\frac{x+4}{96}+\frac{96}{96}\right)+\left(\frac{\left(x+6\right)}{94}+\frac{94}{94}\right)\)
\(\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}=\frac{x+100}{92}+\frac{x+100}{94}+\frac{x+100}{96}\)
\(\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}-\frac{x+100}{92}-\frac{x+100}{94}-\frac{x+100}{96}=0\)
\(\left(x+100\right).\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{92}-\frac{1}{94}-\frac{1}{96}\right)=0\)
\(Mà\) \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{92}-\frac{1}{94}-\frac{1}{96}\ne0\)
Nên x+ 100 = 0
x = 0 - 100 = -100
Vậy x= -100
cộng 1 vào mỗi tỉ số,ta được:
\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)=\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)+\left(\frac{x+6}{94}+1\right)\)\(\Rightarrow\frac{x+1+99}{99}+\frac{x+3+97}{97}+\frac{x+5+95}{95}=\frac{x+2+98}{98}+\frac{x+4+96}{96}+\frac{x+6+94}{94}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}=\frac{x+100}{98}+\frac{x+100}{96}+\frac{x+100}{94}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}-\frac{x+100}{98}-\frac{x+100}{96}-\frac{x+100}{94}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{98}-\frac{1}{96}-\frac{1}{94}\right)\)
Vì \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{98}-\frac{1}{96}-\frac{1}{94}\ne0\)
=>x+100=0
=>x=-100
Vậy x=-100
2x - 4(5 - x) = 2x + 16
=> 4(5 - x) = 2x - (2x + 16)
=> 4(5 - x) = 2x - 2x - 16
=> 4(5 - x) = -16
=> 5 - x = -4
=> x = 9
Vậy...
\(\dfrac{\left(44.52.60\right)}{11.13.15}=\dfrac{44}{11}\cdot\dfrac{52}{13}\cdot\dfrac{60}{15}=4.4.4=64\)
b)Tận cùng=5 hoặc 0 nhưng mình ngại viết lắm,thông cảm nha
A = ( 100 + 98 + 96 + ... + 2 ) - ( 97 + 95 + ... + 1 )
Xét vế trái :
SSH là : ( 100 - 2 ) : 2 + 1 = 50 ( số )
Tổng là : ( 100 + 2 ) . 50 : 2 = 2550
Xét vế phải :
SSH là : ( 97 - 1 ) : 2 + 1 = 50 ( số )
Tổng là : ( 97 + 1 ) . 50 : 2 = 2450
=> A = 2550 - 2450
=> A = 100
a) \(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}=4\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-3}{97}-1+\frac{x-3}{96}-1=4-4\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
\(\Rightarrow x-1=0\) ( vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\) )
Vậy x = 1
b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=3\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=3-3\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)
\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\ne0\)
=> x + 100 = 0
=> x = -100
c) \(\frac{x-1}{99}+\frac{x-2}{49}+\frac{x-4}{32}=6\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{49}-2+\frac{x-4}{32}-3=6-6\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{49}+\frac{x-100}{32}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\ne0\)
=> x - 100 = 0
=> x = 100
Chúc bạn học tốt
có người khác trả lời trước rồi nên chị ko trả lời đâu nhé em trai
a) \(x-\frac{1}{12}+x-\frac{1}{20}+x-\frac{1}{30}+x-\frac{1}{42}+x-\frac{1}{56}+x-\frac{1}{72}=224\)
\(\left(x+x+x+x+x+x\right)-\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=224\)
\(6x-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=224\)
\(6x-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=224\)
\(6x-\left(\frac{1}{3}-\frac{1}{9}\right)=224\)
\(6x-\frac{2}{9}=224\)
\(6x=224+\frac{2}{9}\)
\(6x=\frac{2018}{9}\)
\(\Rightarrow x=\frac{2018}{9}:6=\frac{1009}{27}\)
b) ( 2x - 1 ) 2 = \(\frac{1}{4}\)
( 2x - 1 ) 2 = \(\left(\frac{1}{2}\right)^2\)
\(\Rightarrow\)2x - 1 = \(\frac{1}{2}\)
\(\Rightarrow\)2x = \(\frac{1}{2}+1\)
\(\Rightarrow\)2x = \(\frac{3}{2}\)
\(\Rightarrow\)x = \(\frac{3}{2}:2\)
\(\Rightarrow\)x = \(\frac{3}{4}\)
2.
Ta có : 3300 = ( 33 ) 100 = 27100
5200 = ( 52 ) 100 = 25100
Vì 27100 > 25100 nên 3300 > 5200
3.
150 - ( 100 - 99 + 98 - 97 + 96 - 95 + ... + 4 - 3 + 2 - 1 )
= 150 - [ (100 - 99 ) + ( 98 - 97 ) + ( 96 - 95 ) + ... + ( 4 - 3 ) + ( 2 - 1 ) ]
= 150 - ( 1 + 1 + 1 + ... + 1 + 1 )
= 150 - 50
= 100
4.
ta có :
9x + 5y + 4 .( 2x + 3y )
= 9x + 5y + 8x + 12y
= ( 9x + 8x ) + ( 5y + 12y )
= 17x + 17y
= 17 ( x + y ) \(⋮\)17
Vì 9x + 5y \(⋮\)17 \(\Rightarrow\)4 . ( 2x + 3y ) \(⋮\)17
Mà ( 4 ; 17 ) = 1
\(\Rightarrow\)2x + 3y \(⋮\)17
bài 1
a) \(\frac{x-1}{12}+\frac{x-1}{20}+\frac{x-1}{30}+\frac{x-1}{42}+\frac{x-1}{56}+\frac{x-1}{72}=224\)
\(\left(x-1\right).\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=224\)
\(\left(x-1\right).\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=224\)
\(\left(x-1\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=24\)
\(\left(x-1\right).\left(\frac{1}{3}-\frac{1}{9}\right)=\left(x-1\right)\cdot\frac{2}{9}=224\)
\(\Rightarrow\left(x-1\right)=224:\frac{2}{9}=1008\Rightarrow x=1008+1=1009\)
\(\frac{2x+1}{4-x}=\frac{x-1}{5}\)
\(\Rightarrow\)\(5.\left(2x+1\right)=\left(4-x\right).\left(x-1\right)\)
\(\Rightarrow\)\(10x+5=2x-4\)
\(\Rightarrow\)\(10x-2x=-4-5\)
\(\Rightarrow\)\(8x=-9\)
\(\Rightarrow\)\(x=-\frac{9}{8}\)