gọi số đó là a

theo đề bài ta có

\(\frac{67-a}{112-a}=\frac{1}{6}\)

\(\Leftrightarrow\) \(6\) x   \(\left(67-a\right)\)= \(112-a\)

\(\Leftrightarrow\) \(402-6a=112-a\)

\(\Leftrightarrow\)\(402-112=-a+6a\)

\(\Leftrightarrow290=5a\)

\(\Leftrightarrow a=58\)

vậy số cần tìm là 58

biết thế sao còn hỏi người ta

1) ĐKXĐ: \(x\ne1\)

\(pt\Leftrightarrow\dfrac{2x-1+x-1-1}{x-1}=0\Leftrightarrow3x-3=0\Leftrightarrow3x=3\Leftrightarrow x=1\left(ktm\right)\)

Vậy \(S=\varnothing\)

2) ĐKXĐ: \(x\ne7\)

\(pt\Leftrightarrow\dfrac{x-8-8\left(x-7\right)+1}{x-7}=0\Leftrightarrow-7x+49=0\Leftrightarrow x=7\left(ktm\right)\)

Vậy \(S=\varnothing\)

3) ĐKXĐ: \(x\ne2\)

\(pt\Leftrightarrow\dfrac{1+3\left(x-2\right)-\left(3-x\right)}{x-2}=0\Leftrightarrow4x-8=0\Leftrightarrow x=2\left(ktm\right)\Rightarrow S=\varnothing\)

4) ĐKXĐ: \(x\ne2\)

\(pt\Leftrightarrow\dfrac{1+3\left(x-2\right)+x-3}{x-2}=0\Leftrightarrow4x-8=0\Leftrightarrow x=2\left(ktm\right)\Rightarrow S=\varnothing\)

5) ĐKXĐ: \(x\ne-1\)

\(pt\Leftrightarrow\dfrac{5x+2x+2+12}{2\left(x+1\right)}=0\Leftrightarrow7x+14=0\Leftrightarrow x=-2\left(tm\right)\)

6) ĐKXĐ: \(x\ne0,x\ne2\)

\(pt\Leftrightarrow\dfrac{x\left(x+2\right)-\left(x-2\right)-2}{x\left(x-2\right)}=0\Leftrightarrow x^2+x+2=0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(VLý\right)\Rightarrow S=\varnothing\)

1: =>2x-1+x-1=1

=>3x-2=1

=>3x=3

hay x=1(loại)

2: \(\Leftrightarrow\dfrac{-1}{x-7}=\dfrac{x-8}{x-7}-\dfrac{8\left(x-7\right)}{x-7}\)

=>x-8-8x+56=-1

=>-7x+48=-1

=>-7x=-49

hay x=7(loại)

3: \(\Leftrightarrow1+3\left(x-2\right)=3-x\)

=>3x-6+1-3+x=0

=>4x-8=0

hay x=2(loại)

8: \(\Leftrightarrow3\left(x+1\right)=-3x-2-4\left(x-1\right)\)

=>3x+3=-3x-2-4x+4

=>3x+3=-7x+2

=>10x=-1

hay x=-1/10(nhận)

35 nha bạn

\(\frac{28}{35}\)\(=\frac{4}{5}\)nhé

1, Gỉa sử m = 1 

Thay m = 1 vào hpt trên ta được 

\(\left\{{}\begin{matrix}x+y=1\\4x+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{2}{3}\end{matrix}\right.\)

2, Để hệ có nghiệm duy nhất \(\dfrac{m}{4}\ne\dfrac{1}{m}\Leftrightarrow m^2\ne4\Leftrightarrow m\ne\pm2\)

\(\left\{{}\begin{matrix}m^2x+my=m\\4x+my=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(m^2-4\right)x=m-2\\y=1-mx\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{m+2}\\y=1-\dfrac{m}{m+2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{m+2}\\y=\dfrac{2}{m+2}\end{matrix}\right.\)

Ta có : \(\dfrac{1}{m+2}-\dfrac{2}{m+2}=1\Rightarrow1-2=m+2\Leftrightarrow-1=m+2\Leftrightarrow m=-3\)(tmđk)

a, Với m = 1 

\(\left\{{}\begin{matrix}x+y=1_{\left(1\right)}\\4x+y=2_{\left(2\right)}\end{matrix}\right.\) 

Lấy (2) - (1) ta được

\(3x=1\Leftrightarrow x=\dfrac{1}{3};\Rightarrow y=1-x=1-\dfrac{1}{3}=\dfrac{2}{3}\) 

Vậy (x,y) = \(\left(\dfrac{1}{3};\dfrac{2}{3}\right)\) 

c, n^o của hệ là 

\(\left(\dfrac{-1}{m+2};\dfrac{2m+2}{m+2}\right)\\ Theo.bài:\\ x-y=1\\ \Leftrightarrow\dfrac{-1}{m+2}-\dfrac{2m+2}{m+2}=1\\ \Leftrightarrow-1-2m-2=m+2\\ \Leftrightarrow3m=-5\\ m=\dfrac{-5}{3}\)

\(\frac{24}{27}\)

ht

24/27=16/18 nha

~HT~

k cho mình nha mn

@@@@@@@@@@@@@@@@@