\(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)

\(=a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)

\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)

\(=a^4-2a^3b+2ab^3-b^4\)

\(=\left(a^4-b^4\right)-\left(2a^3b-2ab^3\right)\)

\(=\left(a^2-b^2\right)\left(a^2+b^2\right)-2ab\left(a^2-b^2\right)\)

\(=\left(a^2-b^2\right)\left(a^2+b^2-2ab\right)\)

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

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

       \(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\left(ĐK:x\ge3\right)\)

\(\Leftrightarrow\sqrt{x^2-9}=-\sqrt{x^2-6x+9}\)

\(\Leftrightarrow\sqrt{x^2-9}=-\sqrt{x^2-6x+9}\)

\(\Leftrightarrow\left(\sqrt{x^2-9}\right)^2=\left(-\sqrt{x^2-6x+9}\right)^2\)

\(\Leftrightarrow x^2-9=\left(\sqrt{x^2-6x+9}\right)^2\)

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

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

\(\Leftrightarrow6x=18\)

\(\Leftrightarrow x=3\)

\(ĐKXĐ:x\ge3\)

\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)

\(\Leftrightarrow\sqrt{\left(x+3\right)\left(x-3\right)}+\sqrt{\left(x-3\right)^2}=0\)

\(\Leftrightarrow\sqrt{x+3}\sqrt{x-3}+\sqrt{\left(x-3\right)}\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-3}=0\\\sqrt{x+3}+\sqrt{x-3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x\in\left\{\varnothing\right\}\end{cases}}\)

Vậy nghiệm duy nhất của pt là 3.

#)Góp ý :

Mấy câu như thế này thì bạn nên lên Discuss with Math You - MathYou hỏi nhé ^^

Vì web này chuyên cho học sinh giải toán tiếng anh nên lên đây hỏi sẽ tiện hơn

2 và 1 + √2
ta có :
1 + √2
= 1,5 + 1 
= 2,5
<=> 2 và 2,5
<=> 2 < 2,5
<=> 2 < 1 + √2

\(\sqrt{3}< \sqrt{4}\)

\(\Rightarrow\sqrt{3}-1< \sqrt{4}-1\)

\(\Rightarrow\sqrt{3}-1< 2-1=1\)

Pt a: Đk \(1< x\le6\)
\(\frac{\sqrt{6-x}-2x+3}{\sqrt{x-1}}=\sqrt{x-1}\Rightarrow\sqrt{6-x}-2x+3=x-1\)
\(\Leftrightarrow\sqrt{6-x}=3x-4\Rightarrow6-x=\left(3x-4\right)^2\)
\(\Leftrightarrow6-x=9x^2-24x+16\Leftrightarrow9x^2-23x+10=0\)
\(\Leftrightarrow9x^2-18x-5x+10=0\Leftrightarrow9x\left(x-2\right)-5\left(x-2\right)=0\Leftrightarrow\left(9x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}9x-5=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{9}\left(Lọai\right)\\x=2\left(Thoả\right)\end{cases}}\)
Vậy \(S=\left\{2\right\}\)
Pt b :
Đk: \(x^2-4\ge0\Leftrightarrow x^2\ge4\Leftrightarrow\left|x\right|\ge2\Leftrightarrow\orbr{\begin{cases}x\ge2\\x\le-2\end{cases}}\)
\(\left(x+1\right)\sqrt{x^2-4}=2x+2\Leftrightarrow\left(x+1\right)\left(\sqrt{x^2-4}-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\sqrt{x^2-4}-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\left(Lọai\right)\\\sqrt{x^2-4}=2\end{cases}}\)
\(\Leftrightarrow\sqrt{x^2-4}=2\Rightarrow x^2-4=4\Leftrightarrow x^2=8\Leftrightarrow x=2\sqrt{2}\left(Thoả\right)\)
Vậy \(S=\left\{2\sqrt{2}\right\}\)

\(a,\sqrt{4-4x+x^2}+\sqrt{\frac{2}{x^2+6x+9}}=\sqrt{\left(x-2\right)^2}+\sqrt{\frac{2}{\left(x+3\right)^2}}\)

\(đkxđ\Leftrightarrow\hept{\begin{cases}x+2\ge0\\x+3>0\end{cases}\Rightarrow\hept{\begin{cases}x\ge-2\\x>-3\end{cases}\Rightarrow}x\ge-2}\)

\(b,\frac{5\sqrt{x}}{\sqrt{x}-3}+\frac{2}{\sqrt{x}}\)

\(đkxđ\Leftrightarrow\hept{\begin{cases}x>0\\\sqrt{x}-3\ne0\end{cases}\Rightarrow\hept{\begin{cases}x>0\\\sqrt{x}\ne\sqrt{9}\end{cases}\Rightarrow}\hept{\begin{cases}x>0\\x\ne9\end{cases}}}\)

\(c,\sqrt{3-\sqrt{x}}\)

\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\3-\sqrt{x}\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}\le3\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x>0\\\sqrt{x}\le9\end{cases}\Rightarrow\hept{\begin{cases}x>0\\x\le3\end{cases}}}\)

\(\Rightarrow0< x\le3\)

c,  Để BT có nghĩa thì  \(x^2-4x+3\ge0\)

                                    \(\Leftrightarrow x^2-4x+4\ge1\)

                                    \(\Leftrightarrow\left(x-2\right)^2\ge1\)

                                    \(\Leftrightarrow\sqrt{\left(x-2\right)^2}\ge1\)

                                      \(\Leftrightarrow|x-2|\ge1\)

\(\Leftrightarrow x-2\ge1\) và   \(x-2\le-1\)

\(\Leftrightarrow x\ge3;x\le1\)

Bằng không!

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

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

\(=\sqrt{\left(1+\sqrt{3}\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)

\(=|1+\sqrt{3}|-|1-\sqrt{3}|\)

\(=1+\sqrt{3}-\left(\sqrt{3}-1\right)\)

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