3.

\(\left|x-2\right|=2-x\Leftrightarrow\left|2-x\right|=2-x\)

\(\Leftrightarrow2-x\ge0\Rightarrow x\le2\) (quy tắc trị tuyệt đối: \(\left|A\right|=A\Leftrightarrow A\ge0\))

6. Đề bài sai (có lẽ do in nhầm)

Tập xác định của pt này là R

8.

Đặt \(\sqrt{x^2+3x+3}=t>0\Rightarrow x^2+3x+1=t^2-2\)

\(\Rightarrow t^2+t-2=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-2\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x^2+3x+3=1\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)

9.

\(\Leftrightarrow\left|\left(x+1\right)\left(x+4\right)\right|=x+4\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+4\ge0\\\left[{}\begin{matrix}x+4=0\\\left|x+1\right|=1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\\left[{}\begin{matrix}x=-4\\x=0\\x=-2\end{matrix}\right.\end{matrix}\right.\) (3 nghiệm đều thỏa mãn)

\(=\dfrac{3}{7}\left(\dfrac{15}{13}-1-1\right)=\dfrac{3}{7}\cdot\dfrac{-11}{13}=-\dfrac{33}{91}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}+\dfrac{2}{143}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{10}{39}=\dfrac{5}{39}\)

= 5/39

Thực hiện nhân tung ra ta có .

a.\(x^3+3x^2+3x+1-\left(x^3-3x+2\right)-3\left(x^2-9\right)=5\)

\(\Leftrightarrow6x+1-2+27=5\Leftrightarrow6x=-21\Leftrightarrow x=-\frac{7}{2}\)

b.\(x^3+3x^2-4+x^3-3x+2-\left(x^3+3x^2+3x+1\right)=4\)

\(\Rightarrow x^3=7\Leftrightarrow x=\sqrt[3]{7}\)

c.\(x^3+3x^2+3x+1+x^3-3x^2+3x-1=x^3+6x^2+12x+8+x^3-6x^2+12x-8\)

\(\Leftrightarrow2x^3+6x=2x^3+24x\Leftrightarrow18x=0\Leftrightarrow x=0\)

a) \(\left(x+1\right)^3-\left(x+2\right)\left(x-1\right)^2-3\left(x-3\right)\left(x+3\right)\)

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

\(=x^3+3x^2+3x+1-\left(x^3-x^2-x+1\right)-\left(3x^2-27\right)\)

\(=x^3+3x^2+3x+1-x^3+x^2+x+1-3x^2+27\)

\(=6x+26\)

\(\left(5x+3y\right)\left(25x^2-15xy+9y^2\right)\)

\(=\left(5x+3y\right)\left[\left(5x\right)^2-5x.3y+\left(3y\right)^2\right]\)

\(=\left(5x\right)^3+\left(3y\right)^3=125x^3-27y^3\)

\(4,=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{5-2\sqrt{6}-9}=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{-4-2\sqrt{6}}\\ =\dfrac{3\left(3-\sqrt{2}-\sqrt{3}\right)}{2+\sqrt{6}}=\dfrac{\left(9-3\sqrt{2}-3\sqrt{3}\right)\left(\sqrt{6}-2\right)}{2}\\ =\dfrac{9\sqrt{6}-18-6\sqrt{3}+6\sqrt{2}-9\sqrt{2}+6\sqrt{3}}{2}\\ =\dfrac{9\sqrt{6}-3\sqrt{2}-18}{2}\)

\(7,=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-2-\sqrt{3}\\ =\sqrt{3}+2+\sqrt{2}+1-2-\sqrt{3}=1+\sqrt{2}\)

\(10,\dfrac{1}{\sqrt{a}+\sqrt{a+2}}=\dfrac{\sqrt{a}-\sqrt{a+2}}{a-a-2}=\dfrac{\sqrt{a-2}-\sqrt{a}}{2}\)

Do đó \(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+...+\dfrac{1}{\sqrt{47}+\sqrt{49}}\)

\(=\dfrac{\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{49}-\sqrt{47}}{2}=\dfrac{-1+\sqrt{49}}{2}=\dfrac{7-1}{2}=3\)

10, \(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+...+\dfrac{1}{\sqrt{17}+\sqrt{19}}=\dfrac{\sqrt{1}-\sqrt{3}}{\left(\sqrt{1}+\sqrt{3}\right)\left(\sqrt{1}-\sqrt{3}\right)}+\dfrac{\sqrt{3}-\sqrt{5}}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}-\sqrt{5}\right)}+...+\dfrac{\sqrt{17}-\sqrt{19}}{\left(\sqrt{17}+\sqrt{19}\right)\left(\sqrt{17}-\sqrt{19}\right)}=\dfrac{1-\sqrt{3}+\sqrt{3}-\sqrt{5}+...+\sqrt{17}-\sqrt{19}}{-2}=-\dfrac{1-\sqrt{19}}{2}\)

=a³-b³