ĐKXĐ : \(\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\)

\(\sqrt{x^2-1}+\sqrt{x^2-2x+1}=0\)\(\)

\(\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2-1=0\\x^2-2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=1\\\left(x-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\pm1\\x=1\end{matrix}\right.\)\(\)

\(\Leftrightarrow x=1\)

Vậy S = {1}

Bổ sung câu hệ

b, \(\left\{{}\begin{matrix}x^2+x-y^2-y=0\left(1\right)\\x^2+y^2-2\left(x+y\right)=0\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left(x-y\right)\left(x+y\right)+\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+y+1\right)=0\)\(\)

Th1 : \(x-y=0\Leftrightarrow x=y\), Thế vào (2)

\(\left(2\right)\Leftrightarrow x^2+x^2-2\left(x+x\right)=0\Leftrightarrow2x^2-4x=0\)

\(\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\Leftrightarrow y=0\\x=2\Leftrightarrow y=2\end{matrix}\right.\)

Th2: \(x+y+1=0\Leftrightarrow y=-\left(x+1\right)\), thế vào (2)

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

\(\Leftrightarrow2x^2+2x+3=0\)

Mà \(2x^2+2x+3=2\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{2}>0\)

-> Vô nghiệm

Vậy \(\left(x,y\right)\in\left\{\left(0;0\right);\left(2;2\right)\right\}\)

\(M=3\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}+4\right)^2+14\)

\(=3\left(x+2\sqrt{x}+1\right)-\left(x+8\sqrt{x}+16\right)+14\)

\(=3x+6\sqrt{x}+3-x-8\sqrt{x}-16+14\)

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

\(=2\left(x-4\sqrt{x}+4\right)+6\sqrt{x}-7\)

\(=2\left(\sqrt{x}-2\right)^2+6\sqrt{x}-7\ge2.0+6.\sqrt{4}-7=5\)

Dấu "=" \(x=4\)

Vậy GTNN của M là 4 <=> x = 4

Dó là \(\dfrac{-12}{-7}\)

Chọn C

\(9x^2-30x+225=0\)(1)

\(VT_{\left(1\right)}=9x^2-30x+25+200=\left(3x-5\right)^2+200\ge200>0\)

-> Phương trình vô nghiệm

\(3x^2+12x-66=0\)

\(\Leftrightarrow x^2+4x-22=0\)

\(\Leftrightarrow x^2+4x+4-26=0\)

\(\Leftrightarrow\left(x+2\right)^2=26\)

\(\Leftrightarrow x+2=\pm\sqrt{26}\)

\(\Leftrightarrow x=\pm\sqrt{26}-2\)

pt bậc 2 

\(ax^2+bx+c=0\)

Nếu : \(a+b+c=0\), thì có nghiệm là 1 và nghiệm còn lại là \(\dfrac{c}{a}\)

Nếu : \(a-b+c=0\), thì có nghiệm là -1 và nghiệm còn lại là \(-\dfrac{c}{a}\)

\(M=\dfrac{x+6\sqrt{x}+9+25}{\sqrt{x}+3}=\dfrac{\left(\sqrt{x}+3\right)^2+25}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}\)Áp dụng Cô si có

\(M\ge2\sqrt{\left(\sqrt{x}+3\right).\dfrac{25}{\sqrt{x}+3}}=10\)

Dấu "=" \(\sqrt{x}+3=\dfrac{25}{\sqrt{x}+3}\leftrightarrow x=4\)

Vậy GTNN của M = 10 <=> x = 4

\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{\left(2^4\right)^3.3^{10}+3.5.2^3.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)

\(=\dfrac{2^{12}.3^{10}+3.5.2^3.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}+5.2^{12}.3^{10}}{2^{12}.3^{12}+2^{11}.3^{11}}\)

\(=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)\(=\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.7}=\dfrac{2.2}{7}=\dfrac{4}{7}\)