a: \(x\left(x+7\right)-\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow x^2+7x-x^2-x+6=0\)
hay x=-1
b: Ta có: \(\left(x+2\right)^2-\left(x^2-4\right)=0\)
\(\Leftrightarrow x+2=0\)
hay x=-2
b. (x + 2)2 - x2 + 4 = 0
<=> (x + 2 - x)(x + 2 + x) + 4 = 0
<=> 2(2 + 2x) + 4 = 0
<=> 4(1 + x) + 4 = 0
<=> 4(1 + x) = -4
<=> 1 + x = -1
<=> x = -1 - 1
<=> x = -2
\(a,\) \(x\left(x+7\right)-\left(x-2\right)\left(x+3\right)\)
\(\Leftrightarrow x^2+7x-x^2-3x+2x+6\\ \Leftrightarrow6x=0\\ \Leftrightarrow x=0\)
\(Vậy...\)
\(b,\) \(\left(x+2\right)^2-\left(x^2-4\right)=0\)
\(\Leftrightarrow x^2+4x+4-x^2+4=0\\ \Leftrightarrow4x+8=0\\ \Leftrightarrow x=-2\)
a. \(x\left(x+7\right)-\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow x^2+7x-x^2-x+6=0\)
\(\Leftrightarrow6\left(x+1\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
b. \(\left(x+2\right)^2-\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+2-x+2\right)=0\)
\(\Leftrightarrow4\left(x+2\right)=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
a. x(x + 7) - (x - 2)(x + 3) = 0
<=> x2 + 7x - (x2 + 3x - 2x - 6) = 0
<=> x2 + 7x - x2 - 3x + 2x + 6 = 0
<=> x2 - x2 + 7x - 3x + 2x + 6 = 0
<=> 6x + 6 = 0
<=> 6x = -6
<=> x = -1
