Tìm x :
a) \(\times.\left(0,25+1999\right).2000=\left(53.1999\right).2000\)
b) 71 + 65 x 4 = \(\frac{x+140}{x}+260\)
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\(a.\left(x\times0,25+1999\right)\times2000=2052\times2000\\ \left(x\times0,25+1999\right)\times2000=4104000\\ x\times0,25+1999=4104000:2000\\ x\times0,25+1999=2052\\ x\times0,25=2052-1999\\ x\times0,25=53\\ x=53:0,25\\ x=212\)
( X x 0,25+ 1999) x2000= ( 53+1999) x2000
( X x 0,25+ 1999) = ( 53+1999)x2000:2000
( X x 0,25+1999) =( 53+1999)
( X x 0,25+1999) = 2052
X+0,25 = 2052 - 1999
X+ 0,25 = 53
X = 53:0,25
X = 212
71+65 x4 = x+140/x + 260
x+140/x + 260= 71+65x4
x+140/x + 260=71 + 260
x+140/x=71
x/x +140/x=71
1+ 140/x=71
140/x=71-1
140/x=70
x=140/70
x=2
k nha linh
a)(x.0,25+1999)x2000=(53+1999)x2000
(x.0,25+1999)=53+1999
x.0,25=53
x=53x4
x=210
a)(x.0,25+1999)x2000=(53+1999)x2000
(x.0,25+1999)=53+1999
x.0,25=53
x=53x4
x=210
k mik đi
( x + 1 ) + ( x + 4 ) + ( x + 7 ) + ( x + 10) + ..........+ ( x + 28 ) = 155
10x + ( 1 + 4 + 7 + ... + 28 ) = 155
10x + 145 = 155
10x = 155 - 145
10x = 10
x = 10 : 10
Vậy x = 1
\(a,\left[x+1\right]+\left[x+4\right]+\left[x+7\right]+\left[x+10\right]+...+\left[x+28\right]=155\)
\(\Leftrightarrow x+1+x+4+x+7+x+10+...+x+28=155\)
\(\Leftrightarrow(x+x+x+...+x)+(1+4+7+...+28)=155\)
\(\Leftrightarrow10x+145=155\)
\(\Leftrightarrow10x=155-145\)
\(\Leftrightarrow10x=10\)
\(\Leftrightarrow x=1\)
\(b)(x\times0,25+1999)\times2000=(53+1999)\times2000\)
\(\Leftrightarrow(x\times0,25+1999)\times2000=4104000\)
\(\Leftrightarrow(x\times0,25+1999)=4104000:2000\)
\(\Leftrightarrow(x\times0,25+1999)=2052\)
\(\Leftrightarrow x\times0,25=2052-1999\)
\(\Leftrightarrow x\times0,25=53\)
\(\Leftrightarrow x=53:0,25=212\)
Câu c tự làm
a, ( x . 0,25+1999).2000=(53+1999).2000
=> 500x + 1999 . 2000 = 53 . 2000 + 1999 . 2000
=> 500x = 53 . 2000 + 1999 . 2000 - 1999 .2000
=> 500x = 53 . 2000
=> x = 53 . 2000 : 500
=> x = 53 . 4
=> x = 212