\(\Leftrightarrow x^2-6x+9-x^2+4=1\)

=>-6x=-12

hay x=2

\(a,\dfrac{1}{2}x=3+2\)

\(\dfrac{1}{2}x=5\)

\(x=5\div\dfrac{1}{2}\)

\(x=10\)

\(b,\dfrac{1}{4}x^2-\sqrt{36}=10\)

\(\dfrac{1}{4}x^2-6=10\)

\(\dfrac{1}{4}x^2=10+6\)

\(\dfrac{1}{4}x^2=16\)

\(x^2=16\div\dfrac{1}{4}\)

\(x^2=64\)

\(x^2=\left(8\right)^2\)

\(\Rightarrow x=8\)

\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x-y\right)\left(x+y\right)-4x^2\\ P=\left(x-y-x-y\right)^2-4x^2\\ P=4y^2-4x^2=4\left(y-x\right)\left(x+y\right)\)

4Al + 3O₂ ---> 2Al₂O₃

Na₂O + H₂O ---> 2NaOH

2KMnO₄ ---> K₂MnO₄ + MnO₂ + O₂

Fe + 2HCl ---> FeCl₂ + H₂

H₂ + CuO ---> Cu + H₂O

2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

A = \(\dfrac{x+13}{x+2}\)  (đk \(x\) ≠ -2)

Em cần làm gì với biểu thức này?

Ta có: \(\int\dfrac{xdx}{x^2+3}\)

Đặt \(u=x^2+3\left(u>0\right)\) 

Có \(du=2xdx\)

\(\Rightarrow\int\dfrac{xdx}{x^2+3}=\)\(\int\dfrac{du}{2u}=\dfrac{1}{2}ln\left(u\right)=\dfrac{1}{2}ln\left(x^2+3\right)\)