x=20     

Lí do = 0 nè:

\(\frac{x-39}{19}+\frac{x-49}{29}=\frac{x-5}{39}+\frac{x-69}{49}\)

=> \(\frac{x-39}{19}+\frac{x-49}{29}=0+\left(\frac{x-5}{39}+\frac{x-69}{49}\right)\)

=> \(\frac{x-39}{19}+\frac{x-49}{29}-\left(\frac{x-5}{39}+\frac{x-69}{49}\right)=0\)

=> \(\frac{x-39}{19}+\frac{x-49}{29}-\frac{x-5}{39}-\frac{x-69}{49}=0\)

Câu 2:

a: x-158=32

=>x=158+32

=>x=190

b: \(x\cdot24=264\)

=>\(x=\dfrac{264}{24}\)

=>x=11

c: \(6x+9=3^7:3^4\)

=>\(6x+9=3^3\)

=>6x+9=27

=>6x=18

=>x=18/6=3

Câu 1:

a: \(86\cdot19+14\cdot19\)

\(=19\left(86+14\right)\)

\(=19\cdot100=1900\)

b: \(4\cdot\left(-5\right)^2-104\cdot\left(-5\right)^2\)

\(=4\cdot25-104\cdot25\)

\(=25\left(4-104\right)=-100\cdot25=-2500\)

c: \(7\cdot\left(-2\right)\cdot8\left(-5\right)\)

\(=7\cdot2\cdot8\cdot5\)

\(=56\cdot10=560\)

d: \(59-\left[59+\left(-76\right)\right]\)

\(=59-59+76\)

=76

a,x/8=2/x

=>x²=16

=>x=-4;4

vậy x=-4;4

b,\(\frac{x-39}{19}+\frac{x-49}{29}=\frac{x-59}{39}+\frac{x-69}{49}\)

\(\Rightarrow\left(\frac{x-39}{19}+1\right)+\left(\frac{x-49}{29}+1\right)=\left(\frac{x-59}{39}+1\right)+\left(\frac{x-69}{49}+1\right)\)

\(\Rightarrow\frac{x-20}{19}+\frac{x-20}{29}=\frac{x-20}{39}+\frac{x-20}{49}\)

\(\Rightarrow\frac{x-20}{19}+\frac{x-20}{29}-\frac{x-20}{39}-\frac{x-20}{49}=0\)

\(\Rightarrow\left(x-20\right)\left(\frac{1}{19}+\frac{1}{29}-\frac{1}{39}-\frac{1}{49}\right)=0\)

vì \(\frac{1}{19}+\frac{1}{29}-\frac{1}{39}-\frac{1}{49}\ne0\Rightarrow x-20=0\Rightarrow x=20\)

vậy x=20

\(\frac{x}{8}=\frac{2}{x}\Rightarrow x^2=8.2\)

\(x^2=16\Rightarrow x^2=\left(\pm4\right)^2\Rightarrow x=4;x=-4\)

Học tốt

\(x=20\)\(nha\)

 lấy chữ số tận cùng ta có: 9x9=81 mà 1 x 1=1

mà 6 x 6 x6x6x6 = có chữ số tận cùng là 6

mà 1 x 6 =6

vậy có chữ số tận cùng là 6

6 bạn ạ

1) 39×51+39×49=39x(51+49)=39x100=3900

2. phân tích 100 thành 50 r lm như câu 1

1. = 39 x (51 + 49)

= 39 x 100

= 3900

2. = 50 x 2 x 31 + 100 x 19

= 100 x 31 + 100 x 19

= 100 x (31 + 19)

= 100 x 50

= 5000

=>67-x=28

hay x=39

=>67-x = 7/19 x 76

67-x=28

x=67-28

x=39

\(19(67-x)=532=>1273-19x=532=>19x=67=>x=39\)

67-x/76 = 7/19

=> x/76 = 1266/19

=> x= 5064

\(\frac{x-39}{19}+\frac{x-49}{29}=\frac{x-59}{39}+\frac{x-69}{49}\)

Công mỗi vế cho 2 ta được:

\(\frac{x-39}{19}+\frac{x-49}{29}+2=\frac{x-59}{39}+\frac{x-69}{49}+2\)

\(\frac{x-39}{19}+1+\frac{x-49}{29}+1=\frac{x-59}{39}+1+\frac{x-69}{49}+1\)

\(\frac{x-39}{19}+\frac{19}{19}+\frac{x-49}{29}+\frac{29}{29}=\frac{x-59}{39}+\frac{39}{39}+\frac{x-69}{49}+\frac{49}{49}\)

\(\frac{x-20}{19}+\frac{x-20}{29}=\frac{x-20}{39}+\frac{x-20}{49}\)

\(\frac{x-20}{19}+\frac{x-20}{29}-\frac{x-20}{39}-\frac{x-20}{49}=0\)

\(\left(x-20\right).\frac{1}{19}+\left(x-20\right).\frac{1}{29}-\left(x-20\right).\frac{1}{39}-\left(x-20\right).\frac{1}{49}=0\)

Đặt thừa số chung (x-20) ra ngoài

\(\left(x-20\right).\left(\frac{1}{19}+\frac{1}{29}-\frac{1}{39}-\frac{1}{49}\right)=0\)

\(\text{Vì }\frac{1}{19}+\frac{1}{29}-\frac{1}{39}-\frac{1}{49}\ne0\text{ nên }x-20=0\Rightarrow x=20\)