Question

tìm x

(x+4)²- (x+1)(x-1)= 16

(2x-1)²+ (x+3)²- 5(x+7)(x-7)=0

 

Answer 1

a) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

(x + 4)² - (x + 1) (x - 1) = 16

<=> (x² + 8x + 16) - (x² - 1) = 16

<=> x² + 8x + 16 - x² + 1 = 16

<=> 8x + 17 = 16

<=> 8x = -1

<=> x = 

Answer 2

\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(4x^2+1-4x+\left(x^2+9+6x\right)-5\left(x^2-7^2\right)=0\)

\(4x^2+1-4x+x^2+9+6x-5x^2+245=0\)

\(\left(4x^2+x^2-5x^2\right)-\left(4x+6x\right)+\left(1+9+245\right)=0\)

\(2x+255=0\)

\(2x=-255\)

\(x=\dfrac{-255}{2}\)

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Answer 3

a) Ta có: \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Leftrightarrow x^2+8x+16-x^2+1=16\)

\(\Leftrightarrow8x+17=16\)

\(\Leftrightarrow8x=-1\)

hay \(x=-\dfrac{1}{8}\)

b) Ta có: \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(\Leftrightarrow5x^2+2x+10-5x^2+245=0\)

\(\Leftrightarrow2x=-255\)

hay \(x=-\dfrac{255}{2}\)