Tọa độ giao điểm của `(d)` và `(d')` là:

`(m+1)x+3=2x+3`

`<=>mx+x+3-2x-3=0`

`<=>mx-x=0`

`<=>x(m-1)=0`

`<=>[(x=0),(m=1 (loại)):}`

`=>y=2.0+3=0+3=3`

`=>` Tọa độ giao điểm của `(d)` và `(d')` là `(0;3)`.

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^2-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)^3-3ab\left(a+b\right)-3\left(a+b\right).c\left(a+b+c\right)-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)^3-3ab\left(a+b+c\right)-3\left(a+b\right).c\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b+c\right)^2-3ab-3ab-3bc\right]=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow\dfrac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

Ta có:

\(a;b;c>0\)

\(\Rightarrow a+b+c>0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Rightarrow a=b=c\)

\(A=2020\left(1-\dfrac{a}{b}\right)\left(1-\dfrac{b}{c}\right)\left(1-\dfrac{c}{a}\right)-2021\left(\dfrac{a}{b}-\dfrac{b}{c}+\dfrac{c}{a}\right)^3\)

\(\Rightarrow A=2020.\left(1-1\right)\left(1-1\right)\left(1-1\right)-2021\left(1-1+1\right)^3\)

\(\Rightarrow A=-2021\).

`a(b+1)+b(a+1)=(a+1)(b+1)`

`VT=ab+a+ab+b`

`=2ab+a+b`

`=2.1+a+b`

`=2+a+b(1)`

`VP=(a+1)(b+1)`

`=a(b+1)+1(b+1)`

`=ab+a+b+1`

`=1+a+b+1`

`=2+a+b(2)`

`(1)(2)=>VT=VP`.

\(A=3.\sqrt{4x^6}-3x^2\)

\(=3.\sqrt{\left(2x^3\right)^2}-3x^2\)

\(=3.\left|2x^3\right|-3x^2\)

\(=3.\left(-2x^3\right)-3x^2\)

\(=-6x^3-3x^2\).

\(\dfrac{5}{8}+\dfrac{5}{24}+\dfrac{5}{48}+...+\dfrac{5}{528}\)

\(=\dfrac{5}{2.4}+\dfrac{5}{4.6}+\dfrac{5}{6.8}+...+\dfrac{5}{22.24}\)

\(=\dfrac{5}{2}.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{22.24}\right)\)

\(=\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-...-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{24}\right)\)

\(=\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{24}\right)\)

\(=\dfrac{5}{2}.\dfrac{11}{24}\)

\(=\dfrac{55}{48}\).

\(\sqrt{28-16\sqrt{3}}-\sqrt{\left(4-3\sqrt{3}\right)^2}\)

\(=\sqrt{16-16\sqrt{3}+12}-\left|4-3\sqrt{3}\right|\)

\(=\sqrt{4^2-2.4.2\sqrt{3}+\left(2\sqrt{3}\right)^2}-\left(3\sqrt{3}-4\right)\)

\(=\sqrt{\left(4-2\sqrt{3}\right)^2}-3\sqrt{3}+4\)

\(=\left|4-2\sqrt{3}\right|-3\sqrt{3}+4\)

\(=4-2\sqrt{3}-3\sqrt{3}+4\)

\(=8-5\sqrt{3}\).

\(a,\)

Biểu thức \(\sqrt{-5x-10}\) có nghĩa khi:
\(-5x-10\ge0\)

\(\Leftrightarrow5x\le-10\)

\(\Leftrightarrow x\le-2\)

Vậy \(x\le-2\).

\(b,\)

Biểu thức \(\sqrt{x^2-3x+2}\) có nghĩa khi:

\(x^2-3x+2\ge0\)

\(\Leftrightarrow x^2-x-2x+2\ge0\)

\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)\ge0\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2\ge0\\x-1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2\le0\\x-1\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2\\x\ge1\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2\\x\le1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le1\end{matrix}\right.\)

Vậy \(x\ge2\) hoặc \(x\le1\).

\(c,\)

Biểu thức \(\sqrt{\dfrac{x+3}{5-x}}\) có nghĩa khi:
\(\dfrac{x+3}{5-x}\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0\\5-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3\le0\\5-x< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-3\\x< 5\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-3\\x>5\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-3\le x< 5\\5< x\le-3\left(\text{vô lí}\right)\end{matrix}\right.\)

Vậy \(-3\le x< 5\).

\(d,\)

Biểu thức \(\sqrt{-x^2+4x-4}\) có nghĩa khi:

\(-x^2+4x-4\ge0\)

\(\Leftrightarrow x^2-4x+4\le0\)

\(\Leftrightarrow\left(x-2\right)^2\le0\left(1\right)\)

Ta thấy:

\(\left(x-2\right)^2\ge0\forall x\in R\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow\left(x-2\right)^2=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\).

\(A,\)

Với \(x=-1;y=0,5\)

\(\Rightarrow P=\left(-1\right)^3.0,5-14.\left(-1\right)^3-6.\left(-1\right).0,5^2+0,5+2\)

\(=-0,5+14+1,5+0,5+2\)

\(=17,5\)

\(B,\)

Với \(x=0,2;y=-1,2\)

\(\Rightarrow Q=15.0,2^2.\left(-1,2\right)-5.0,2.\left(-1,2\right)^2+7.0,2.\left(-1,2\right)-21\)

\(=-0,72-1,44-1,68-21\)

\(=-24,84\).

\(A,\)

\(9x^3y^6+4x^3y^6+7x^3y^6\)

\(=\left(9+4+7\right)x^3y^6\)

\(=20x^3y^6\)

\(B,\)

\(9x^5y^6-14x^5y^6+5x^5y^6\)

\(=\left(9-14+5\right)x^5y^6\)

\(=0\).