5(x+35) = 515

=> x + 35 = 103

=> x = 103 - 35

=> x = 68

vậy_

\(5.\left(x+35\right)=515\)

\(\Rightarrow x+35=515:5\)

\(\Rightarrow x+35=103\)

\(\Rightarrow x=103-35\)

\(\Rightarrow x=68\)

Mấy câu còn lại tương tự nhé !

\(x=68\)

a.x-4300-(5250:1050.250)=4250

=>x-4300-1250=4250

=>x=4250+1250+4300

=>x=9800

b.x-6-(48-24.2:6-3)=100

=>x-6-(48-8-3)=100

=>x-6-37=100

=>x=100+37+6

=>x=143

c.20-[7.(x-3)+4]=2

=>7.(x-3)+4=20-2

=>7.(x-3)+4=18

=>7.(x-3)=18-4

=>7.(x-3)=14

=>x-3=14:7

=>x-3=2

=>x=2+3

=>x=5

d. [(6x-39):3].28=5628

=>(6x-39):3=5628:28

=>(6x-39):3=201

=>6x-39=201.3

=>6x-39=603

=>6x=603+39

=>6x=642

=>x=642:6

=>x=107

                                  a) x - 4300 - 1250 = 4250                                                b) x - 6 : 2 - (48 - 8 - 3) = 0

                                      x - 4300 = 4250 + 1250                                                   x - 6 : 2 - 37 = 0

                                      x - 4300 = 5500                                                               x - 3 - 37 = 0

                                      x = 5500 - 4300                                                               x - 3 = 37

                                      x = 1200                                                                          x = 37 - 3

                                                                                                                              x = 34

a: \(\Leftrightarrow5x-42=251\)

=>5x=293

hay x=293/5

b: \(\Leftrightarrow20-x=20\)

hay x=0

c: \(\Leftrightarrow x-4300-\dfrac{1}{50}=4250\)

\(\Leftrightarrow x=\dfrac{427501}{50}\)

d: =>(x+200):4=460-340=120

=>x+200=480

hay x=280

e: =>5+15(x+1)=500-480=20

=>15(x+1)=15

=>x+1=1

hay x=0

\(5.\left(x+35\right)=515\)

     \(x+35=515:5\)

     \(x+35=103\)

     \(x=103-35\)

 \(\Rightarrow x=68\)

\(\left(x+40\right).15=75.12\)

\(\Rightarrow\left(x+40\right)=5.12\)[ rút gọn theo t/c đẳng thức ]

\(x+40=60\)

\(x=20\)

trả lời giùm mình mấy câu kia với

a.  x - 4300 - (5250 : 1050 : 250) =4240

 x - 4300 - (5 : 250) =4240

 x - 4300 - 0,02 =4240

 x - 4300 = 4240 + 0,02

 x - 4300 =4240,02

x=4240,02+4300

x=854,02

b.  460 + 85 .4 : (x + 200) =4

a.  x - 4300 - (5250 : 1050 : 250) = 4240

=> x - 4300 - 1/50 = 4240

        x - 4300 = 4240 + 1/50 

lk ùi ko lm nữa !! ==" 

Ta có: \(\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{1}{x+2}-\frac{1}{x+5}\);  \(\frac{5}{\left(x+5\right)\left(x+10\right)}=\frac{1}{x+5}-\frac{1}{x+10}\)

\(\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{1}{x+10}-\frac{1}{x+17}\); 

=> Phương trình tương đương:

\(\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)<=> \(\frac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

<=> \(\frac{15}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

=> x=15

Đáp số: x=15

\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}-\frac{1}{x+8}-\frac{1}{x+16}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+16}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{\left(x+16\right)-\left(x+2\right)}{\left(x+2\right)\left(x+16\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow x+16-x-2=x\)

\(\Rightarrow x=14\)

\(\left(\frac{x}{2}\right)^2+\left(\frac{x}{3}\right)^2+\left(\frac{x}{4}\right)^2=\left(\frac{x}{5}\right)^2+\left(\frac{x}{6}\right)^2+\left(\frac{x}{7}\right)^2\)

\(\frac{x^2}{2^2}+\frac{x^2}{3^2}+\frac{x^2}{4^2}=\frac{x^2}{5^2}+\frac{x^2}{6^2}+\frac{x^2}{7^2}\)

\(\frac{x^2}{2^2}+\frac{x^2}{3^2}+\frac{x^2}{4^2}-\frac{x^2}{5^2}-\frac{x^2}{6^2}-\frac{x^2}{7^2}=0\)

\(x^2.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}-\frac{1}{5^2}-\frac{1}{6^2}-\frac{1}{7^2}\right)=0\)

vì \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}-\frac{1}{5^2}-\frac{1}{6^2}-\frac{1}{7^2}\ne0\)nên \(x^2=0\)

\(\Rightarrow x=0\)

x^2(1/4+1/9+1/16-1/25-1/36/1/49)=0
mà (1/2+1/9=1/16-1/25-1/36-1/49)>0
=>x=0