\(\left(x+8\right)\left(x+6\right)-x^2=104\)

\(\Rightarrow x^2+14x+48-x^2=104\)

\(\Rightarrow14x=56\Rightarrow x=4\)

\(\Rightarrow x^2+14x+48-x^2=104\\ \Rightarrow14x=56\\ \Rightarrow x=4\)

\(\left(x+8\right)\left(x+6\right)-x^2=104\)

\(\Leftrightarrow14x=56\)

hay x=4

* ĐỀ như dưới phải ko ??? Ai lại ghi x2 bao giờ ??

a) (x + 8)(x + 6) - x² =104

\(\Leftrightarrow\) x² + 6x + 8x + 48 - x² = 104

\(\Leftrightarrow\) 14x + 48 = 104

\(\Leftrightarrow\) 14x = 56

\(\Leftrightarrow\) x = 4

Ta có:

(x+8)(x+6)-x^2=104

x^2+6x+8x+48-x^2=104

14x+48=104

14x=104-48

14x=56

x=4

(x+8)(x+6)-x²=104
<=> x²+6x+8x+48-x²=104
<=> 14x+48=104
<=> 14x=104-48
<=> 14x=56
<=> x= 56:14
<=> x=4

a x 7 + a x 8 + a = 104

a x ( 7 + 8 + 1 ) = 104

a x 16 = 104

a = 104 : 16

a = 6,5

Vậy a = 6,5

a x 7 + a x 8 + a = 104

a x 16 = 104

        a = 104 / 16 

         a = 6,5

a x 7 + a x 8 + a = 104

a x 16 = 104

        a = 104 / 16 

         a = 6,5

\(\frac{x+110}{100}+\frac{x+8}{102}+\frac{x+6}{104}=\frac{x+4}{106}+\frac{x+2}{108}\)

\(\Leftrightarrow\frac{x+110}{100}+\left(\frac{x+8}{102}+1\right)+\left(\frac{x+6}{104}+1\right)=\left(\frac{x+4}{106}+1\right)+\left(\frac{x+2}{108}+1\right)\)

\(\Leftrightarrow\frac{x+110}{100}+\frac{x+110}{102}+\frac{x+110}{104}=\frac{x+110}{106}+\frac{x+110}{108}\)

\(\Leftrightarrow\frac{x+110}{100}+\frac{x+110}{102}+\frac{x+110}{104}-\frac{x+110}{106}-\frac{x+110}{108}=0\)

\(\Leftrightarrow\left(x+110\right)\left(\frac{1}{100}+\frac{1}{102}+\frac{1}{104}+\frac{1}{106}+\frac{1}{108}\right)=0\)

\(\Leftrightarrow x+110=0\) (vì \(\frac{1}{100}+\frac{1}{102}+\frac{1}{104}-\frac{1}{106}-\frac{1}{108}>0\))

\(\Leftrightarrow x=-110\)

Vậy \(x=-110\)

mn giải hộ mik ik