Question

1) Rút gọn: A= 7/x+4 + 8/x-4 + 14x/x^2-16

                  B= x^2-2x+1/x-1 + x^2-9/x+3

2) Tìm 2 số tự nhiên hơn kém nhau 5 đơn vị và có tổng bình phương bằng 73

Answer 1

https://www.youtube.com/watch?v=ADwFwrLD-_I

Answer 2

\(A=\frac{7}{x+4}+\frac{8}{x-4}+\frac{14x}{x^2-16}=\frac{7}{x+4}+\frac{8}{x-4}+\frac{14x}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{7\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}+\frac{8\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\frac{14x}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{7x-28+8x+32+14x}{\left(x-4\right)\left(x+4\right)}=\frac{29x+4}{\left(x-4\right)\left(x+4\right)}\)

\(B=\frac{x^2-2x+1}{x-1}+\frac{x^2-9}{x+3}=\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x-3\right)\left(x+3\right)}{x+3}\)

\(=x-1+x-3=2x-4\)

Answer 3

1) \(A=\frac{7}{x+4}+\frac{8}{x-4}+\frac{14x}{x^2-16}\)( ĐKXĐ : \(x\ne\pm4\))

\(A=\frac{7}{x+4}+\frac{8}{x-4}+\frac{14x}{\left(x-4\right)\left(x+4\right)}\)

\(A=\frac{7\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\frac{8\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\frac{14x}{\left(x-4\right)\left(x+4\right)}\)

\(A=\frac{7x-28+8x+32+14x}{\left(x-4\right)\left(x+4\right)}\)

\(A=\frac{29x+4}{\left(x-4\right)\left(x+4\right)}\)

\(B=\frac{x^2-2x+1}{x-1}+\frac{x^2-9}{x+3}\)( ĐKXĐ : \(\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\))

\(B=\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x-3\right)\left(x+3\right)}{x+3}\)

\(B=\left(x-1\right)+\left(x-3\right)\)

\(B=x-1+x-3=2x-4=2\left(x-2\right)\)

2) Gọi số bé là x ( x ∈ N )

=> Số lớn = x + 5

Theo đề bài ta có phương trình :

x² + ( x + 5 )² = 73

<=> x² + x² + 10x + 25 = 73

<=> 2x² + 10x + 25 - 73 = 0

<=> 2x² + 10x - 48 = 0

<=> 2x² - 6x + 16x - 48 = 0

<=> 2x( x - 3 ) + 16( x - 3 ) = 0

<=> ( x - 3 )( 2x + 16 ) = 0

<=> \(\orbr{\begin{cases}x-3=0\\2x+16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-8\end{cases}}\)

Đối chiếu với ĐKXĐ ta thấy x = 3 thỏa mãn

Vậy số bé là 3

       số lớn là 3 + 5 = 8