\(a,=x^2-1-\left(x^2+4x+4\right)=x^2-1-x^2-4x-4=11\)
\(\Leftrightarrow-5x=15\)
\(\Leftrightarrow x=-3\)
Vậy ...
\(b,=\left(x-3-2x+5\right)\left(x-3+2x-5\right)=0\)
\(\Leftrightarrow\left(-x+2\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{8}{3}\end{matrix}\right.\)
Vậy ...
a) \(\left(x-1\right)\left(x+1\right)-\left(x+2\right)^2=11\)
\(\Rightarrow x^2-1-x^2-4x-4-11=0\)
=> -4x - 16 = 0
=> -4x = 16
=> x = -4
b) \(\left(x-3\right)^2-\left(2x-5\right)^2=0\)
=> (x - 3 + 2x - 5).(x - 3 - 2x + 5) = 0
=> (3x - 8).(-x + 2) = 0
=> x = 8/3 hoặc x = 2
a) Ta có: \(\left(x-1\right)\left(x+1\right)-\left(x+2\right)^2=11\)
\(\Leftrightarrow x^2-1-x^2-4x-4=11\)
\(\Leftrightarrow-4x=16\)
hay x=-4
Vậy: S={-4}
b) Ta có: \(\left(x-3\right)^2-\left(2x-5\right)^2=0\)
\(\Leftrightarrow\left(x-3-2x+5\right)\left(x-3+2x-5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{8}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{2;\dfrac{8}{3}\right\}\)
