Question

cho \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)

a, rút gọn A

b, choA=-3. tính giá trị của biểu thức 9x²-42x+49

Answer 1

a ) Rút gọn : \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)

\(\Leftrightarrow A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}\)

\(\Leftrightarrow A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2}{x-5}\)

\(\Leftrightarrow A=\dfrac{1}{x+5}.\)

Khi \(A=-3\),thì :

\(\dfrac{1}{x+5}=-3\Leftrightarrow x=-\dfrac{16}{3}\)

Ta có : \(9x^2-42x+49\)

\(=\left(3x\right)^2-2.3x.7+49\)

\(=\left(3x-7\right)^2\)

Thay \(x=-\dfrac{16}{3},\) ta có :

\(\left(3.\dfrac{-16}{3}-7\right)^2=\left(-16-7\right)^2=\left(-23\right)^2=529\)