ago là dấu hiệu thì quá khứ đơn, phía sau phần cần điền không có danh từ hay tân ngữ\(\Rightarrow\) câu bị động chia dạng quá khứ.

Chọn D: was built

a) \(MgSO_4\)

b) \(Al\left(OH\right)_3\)

Bài 2.

a) \(=3x\left(x^2-4x+4\right)=3x\left(x-2\right)^2\)

b) \(=\left(3x^2-3y^2\right)-\left(12x-12y\right)=3\left(x-y\right)\left(x+y\right)-12\left(x-y\right)=3\left(x-y\right)\left(x+y-4\right)\)

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

1. a) \(x\left(x+1\right)\left(7-x\right)=\left(x^2+x\right)\left(7-x\right)=7x^2-x^3+7x-x^2=-x^3+6x^2+7x\)

b) \(\left(2x+1\right)^2+\left(x-5\right)\left(x+5\right)-x\left(5x+7\right)\)

    \(=4x^2+4x+1+x^2-25-5x^2-7x\)

    = \(-3x-24\)

   \(\left(3x-1\right)\left(x+3\right)-\left(x+2\right)=1\)

\(\Rightarrow3x^2+9x-x-3-x-2=1\)

\(\Rightarrow3x^2-7x-6=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=3\end{matrix}\right.\)

C) Pt \(\Rightarrow m\cdot\dfrac{1-cos2x}{2}-\left(m-1\right)sin2x+\left(2m+1\right)\cdot\dfrac{1+cos2x}{2}=0\)

\(\Rightarrow\left(m+1\right)cos2x-\left(2m-2\right)sin2x=-1-3m\)

Pt có nghiệm:  \(\Leftrightarrow\) \(\left(m+1\right)^2+\left[-\left(2m-2\right)\right]^2\ge\left(1+3m\right)^2\)

                        \(\Rightarrow\dfrac{-3-\sqrt{13}}{2}\le m\le\dfrac{-3+\sqrt{13}}{2}\)

Pt vô nghiệm: \(\Rightarrow\left\{{}\begin{matrix}m>\dfrac{-3+\sqrt{13}}{2}\\m< \dfrac{-3-\sqrt{13}}{2}\end{matrix}\right.\)

11 B

13 D

b) \(Q=\left(\sqrt{75}-\dfrac{3}{2}:\sqrt{3}-\sqrt{48}\right)\cdot\sqrt{\dfrac{16}{3}}\)

        \(=\left(5\sqrt{3}-\dfrac{3}{2}\cdot\dfrac{1}{\sqrt{3}}-4\sqrt{3}\right)\cdot\dfrac{4}{\sqrt{3}}\)

        \(=\sqrt{3}\left(5-\dfrac{1}{2}-4\right)\cdot\dfrac{4}{\sqrt{3}}\)

        \(=\left(1-\dfrac{1}{2}\right)\cdot4=2\)

a) \(P=\dfrac{\sqrt{3}+\sqrt{6}}{1+\sqrt{2}}=\dfrac{\left(\sqrt{3}+\sqrt{6}\right)\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}\)

                             \(=\dfrac{\sqrt{3}-\sqrt{6}+\sqrt{6}-\sqrt{12}}{1-2}=\sqrt{12}-\sqrt{3}\)

d) \(\left(y+3\right)^3-\left(y+1\right)^3=56\)

   \(\Rightarrow y^3+9y^2+27y+27-\left(y^3+3y^2+3y+1\right)=56\)

\(\Rightarrow6y^2+24y+26=56\Rightarrow6y^2+24y-30=0\)

\(\Rightarrow\left[{}\begin{matrix}y=1\\y=-5\end{matrix}\right.\)