\(\frac{x-4}{2015}-\frac{1}{2015}=\frac{10-2x}{2015}\)

\(\Rightarrow\frac{x-4}{2015}-\frac{10-2x}{2015}=\frac{1}{2015}\)

\(\Rightarrow\frac{x-4-\left(10-2x\right)}{2015}=\frac{1}{2015}\)

\(\Rightarrow\frac{\left(x+2x\right)-\left(4+10\right)}{2015}=\frac{1}{2015}\)

\(\Rightarrow\frac{3x-14}{2015}=\frac{1}{2015}\)

\(\Rightarrow\left(3x-14\right).2015=2015\)

\(\Rightarrow3x-14=1\) ( bớt cả 2 vế đi 2015 lần )

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)

Vậy \(x=5\)

X=5

x = 5

Thử lại :

5 - 4 / 2015 - 1/2015 = 0/2015

x-4/2015 - 1/2015=10-2x/2015 

ĐÁP SỐ : x=5

x-4-1 = 10-2x

x = 5

nhanh chưa bn

x=5

k mik nha

mik k lại cho bạn heng

bài dễ z mà k hỉu hả, mẫu số = nhau thi tử số = nhau,

mẫu số = nhau nên tử số = nhau

x-4 -1 = 10-2x

3x = 10+4+1

3x = 15

x = 5

Giá trị x thỏa mãn : \(\frac{x-4}{2015}\)-\(\frac{1}{2015}\)=\(\frac{10-2x}{2015}\)

mẫu số bằng nhau nên tử số cũng bằng  nhau

x-4 -1 = 10-2x

3x = 10+4+1

3x = 15

x = 5

= x-4-1 =10-2x

x= 5

tôi yêu violympic

<=> x - 4 - 1 = 10 - 2x

x = 5

(đúg chứ?)

=> x-4-1=10-2x hay x-4-1-10+2x=0

                            => 3x-15=0

                             => 3x=15

                             => x=5

 mình chắc chắn

\(2^x+2^{x+4}=272\)

\(< =>2^x.\left(1+2^4\right)=272\)

\(< =>2^x.17=272\)

\(< =>2^x=272:17\)

\(< =>2^x=16\)

\(< =>2^x=2^4\)

\(=>x=4\)

a, x-4/2-15-1/2015=10-2x

qua điều trên:

Ta thấy rằng :

x-5=10-2x

=>

x=15-2x

=>

0=15-3x

=>x=5

Ta có:

2^x+2^x+4=272

=> 2^x.(1+16)=272

=>2^x.17=272

=>2^x=16

=>x=4

\(\frac{x-4}{2015}-\frac{1}{2015}=\frac{10-2x}{2015}\)

\(< =>\frac{x-4-1}{2015}=\frac{10-2x}{2015}\)

\(< =>\frac{x-5}{2015}=\frac{10-2x}{2015}\)

\(< =>\frac{x-5}{2015}-\frac{10-2x}{2015}=0\)

\(< =>\frac{x-5-\left(10-2x\right)}{2015}=0\)

\(< =>\frac{x-5-10+2x}{2015}=0\)

\(< =>\frac{3x-15}{2015}=0\)

\(< =>3x-15=0\)

\(< =>3x=15\)

\(< =>x=5\)

\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\left(\frac{2016}{1}-1\right)+\left(\frac{2017}{2}-1\right)+...+\left(\frac{4030}{2015}-1\right)\)

\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\frac{2015}{1}+\frac{2015}{2}+...+\frac{2015}{2015}\)

\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=2015.\left(1+\frac{1}{2}+...+\frac{1}{2015}\right)\)

=> x = 2015

\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right).x+2015=\frac{2016}{1}+\frac{2017}{2}+\frac{2018}{3}+...+\frac{4030}{2015}\)

\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right).x=\left(\frac{2016}{1}-1\right)+\left(\frac{2017}{2}-1\right)+...+\left(\frac{4030}{2015}-1\right)\)

\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right).x=\frac{2015}{1}+\frac{2015}{2}+...+\frac{2015}{2015}=2015.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)\)\(\Rightarrow x=2015\)

giải rồi mà cũng hỏi