8: =>2x=210

hay x=105

Câu a

Tick mk nhé

Bn có thể nói cho mk biết quy luật đc ko

C

C

a,\(2,5.16,27.4+7,3=16,27.10+7,3=162,7+7,3=170\)

b,\(8.6.10+4.25.12+3.16.31+2.34.24=48.10+48.25+48.31+48.34\)

\(=48.\left(10+25+31+34\right)=48.100=4800\)

c,\(78.31+78.24+78.17+22.72=78\left(31+24+17\right)+22.72=78.72+22.72\)

\(=72.\left(78+22\right)=72.100=7200\)

d,\(102.67-34.51=102.67-17.102=102.\left(67-17\right)=102.50=5100\)

đéo hiểu thì cút nhé đéo ai khiến m trả lời cả . ngu thì nói ngu đi đừng bày đặt t ghét nhất là cái loại như m đấy ạ 

Bài 2 

a, \(1-\left(12,5+x-4,25\right):21,75=0\)

\(=>21,75-12,5-x+4,25=0\)

\(=>13,5-x=0=>x=13,5\)

b,\(x.101-x+72.99+72=21050\)

\(=>100x+72.100=21050\)

\(=>100\left(x+72\right)=21050\)

\(=>x+72=210,5\)\(=>x=138,5\)

c,\(151-2\left(x-6\right)=2227:17\)

\(=>151-2\left(x-6\right)=131\)

\(=>2\left(x-6\right)=20\)

\(=>2x=32=>x=16\)

d,\(\left(x+4\right)+\left(x+6\right)+\left(x+8\right)+...+\left(x+26\right)=210\)\(=>12x+\left(4+6+8+...+26\right)=210\)

\(=>12x+180=210=>12x=30\)

\(=>x=\frac{30}{12}=\frac{15}{6}=\frac{5}{2}\)

a)    \(\left(-25\right).5.\left(-4\right).2\)

\(=\left[\left(-25\right).\left(-4\right)\right].\left(2.5\right)\)

\(=100.10=1000\)

b)   \(\left(-456\right)-\left(67-456\right)\)

\(=-456-67+456\)

\(=\left(456-456\right)-67\)

\(=67\)

a) ((-25).(-4)).5.2

=100.10

=1000

b) (-456)-67+456

=(-456)+(-67)+456

=0+-(-67)

=-67

c)78.45+-145.78

=78.(45+145)

=78.190

=Tự làm

\(a,\frac{x+1}{65}+\frac{x+2}{64}=\frac{x+3}{63}+\frac{x+4}{62}\)

\(\Rightarrow\left[\frac{x+1}{65}+1\right]+\left[\frac{x+2}{64}+1\right]=\left[\frac{x+3}{63}+1\right]+\left[\frac{x+4}{62}+1\right]\)

\(\Rightarrow\frac{x+1+65}{65}+\frac{x+2+64}{64}=\frac{x+3+63}{63}+\frac{x+4+62}{62}\)

\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}\)

\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}=0\)

\(\Rightarrow\left[x+66\right]\left[\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\right]=0\)

Mà \(\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\ne0\)

\(\Rightarrow x+66=0\)

\(\Rightarrow x=0-66=-66\)

Auto làm nốt câu b

a,  Cộng cả 2 vế với 2 

Ta có \(\frac{x+1}{64}+\frac{x+2}{63}+2=\frac{x+3}{62}+\frac{x+4}{61}+2\)

\(\left(\frac{x+1}{64}+\frac{64}{64}\right)+\left(\frac{x+2}{63}+\frac{63}{63}\right)=\left(\frac{x+3}{62}+\frac{62}{62}\right)+\left(\frac{x+4}{61}+\frac{61}{61}\right)\)

=>  \(\frac{x+65}{64}+\frac{x+65}{63}=\frac{x+65}{62}+\frac{x+65}{61}\)\(\)

=> \(\frac{x+65}{64}+\frac{x+65}{63}-\frac{x+65}{62}-\frac{x+65}{61}=0\)

=> \(\left(x+65\right)\left(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\right)=0\)

Do \(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\ne0\)=> \(x+65=0\)

=> \(x=-65\)

b ,  Lm tương tự như Câu a

Chúc bn hok tốt