Question

1 tìm x

a , ( x + 1 ) + ( x + 4 ) + ( x + 7 ) + ( x + 10) + ..........+ ( x + 28 ) = 155

b, ( x x 0,25 + 1999 ) x 2000 = ( 53 + 1999 ) x 2000

c, 71 + 65 x 4 = \(\frac{x+140}{x}\)x 260

 

Answer 1

( 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

Answer 2

\(a,\left[x+1\right]+\left[x+4\right]+\left[x+7\right]+\left[x+10\right]+...+\left[x+28\right]=155\)

\(\Leftrightarrow x+1+x+4+x+7+x+10+...+x+28=155\)

\(\Leftrightarrow(x+x+x+...+x)+(1+4+7+...+28)=155\)

\(\Leftrightarrow10x+145=155\)

\(\Leftrightarrow10x=155-145\)

\(\Leftrightarrow10x=10\)

\(\Leftrightarrow x=1\)

\(b)(x\times0,25+1999)\times2000=(53+1999)\times2000\)

\(\Leftrightarrow(x\times0,25+1999)\times2000=4104000\)

\(\Leftrightarrow(x\times0,25+1999)=4104000:2000\)

\(\Leftrightarrow(x\times0,25+1999)=2052\)

\(\Leftrightarrow x\times0,25=2052-1999\)

\(\Leftrightarrow x\times0,25=53\)

\(\Leftrightarrow x=53:0,25=212\)

Câu c tự làm

Answer 3

còn câu c nữa giúp mình với đi các bạn.