a: Thay x=-3 vào B, ta được:

\(B=\dfrac{2\cdot\left(-3\right)^2}{3\cdot\left(-3\right)+6}=\dfrac{2\cdot9}{-9+6}=\dfrac{18}{-3}=-6\)

b: \(A=\dfrac{2x^2+20+3x-6-7x-14}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x^2-4x}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x}{x+2}\)

\(\left(3\sqrt{7}\right)^2=63>28=\left(\sqrt{28}\right)^2\) hoặc \(3\sqrt{7}>2\sqrt{7}=\sqrt{28}\)

C1: $\sqrt{28}=\sqrt{4.7}=2\sqrt 7$

Ta có: $3>2$

$\Leftrightarrow 3\sqrt 7>3\sqrt 7$ hay $3\sqrt 7>\sqrt{28}$

C2: $3\sqrt{7}=\sqrt{63}$

Ta có: $63>28$

$\Leftrightarrow\sqrt{63}>\sqrt{28}$ hay $3\sqrt 7>\sqrt{28}$

b: Ta có: \(N=a^3+b^3+3ab\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\)

\(=1-3ab+3ab\)

=1

\(\dfrac{3}{4}\left(x^2y\right)^2:\dfrac{1}{8}xy^2\\ =\dfrac{3}{4}x^4y^2:\dfrac{1}{8}xy^2\\ =\left(\dfrac{3}{4}:\dfrac{1}{8}\right)\left(x^4:x\right)\left(y^2:y^2\right)\\ =6x^3\)

\(\dfrac{3}{4}\left(x^2y\right)^2\div\dfrac{1}{8}xy^2\)

\(=\dfrac{3}{4}x^4y^2\div\dfrac{1}{8}xy^2\)

\(=6x^3\)

\(a,\Leftrightarrow x^2-x+2021x-2021=0\\ \Leftrightarrow\left(x-1\right)\left(x+2021\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2021\end{matrix}\right.\\ b,\Leftrightarrow-5x^2+15x+x-3=0\\ \Leftrightarrow\left(x-3\right)\left(1-5x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

b: \(-5x^2+16x-3=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

câu 1:

a,x²+2x-4z²+1

=x²+2x.1+1²-(2z)²

=(x+1)²-(2z)²

=(x+1-2z)(x+1+2z)

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