câu a: khi m= 2 => y=2x+2

y y=2x+2 x -1 2 0

với x=0=> y =2

với y=0 =>x -1

câu b : y = xm+2 cắt ox,oy lần lượt tại A,B mà tam giác OAB cân tại O nên OB=OA \(OA^2=OB^2\)

Y X 0 A B

Với x=0=>y=2 => A(0,2) => \(0A=\sqrt{0^2+2^2}=2\)

Với y=0=> x= \(x=\frac{-2}{m}\)nên \(B\left(\frac{-2}{m},0\right)\) ,\(OB=\sqrt{\frac{4}{m^2}+0^2}=\sqrt{\frac{4}{m^2}}\)

theo giả thiết OA=OB nên \(\sqrt{\frac{4}{m^2}}=\sqrt{4}\Leftrightarrow m^2=1\Leftrightarrow\orbr{\begin{cases}m=1\\m=-1\end{cases}}\)

ai giup mik vs!

có \(\Delta^'=\left(m+1\right)^2-2m-1=m^2\) phương trình có hai nghiệm khi \(\Leftrightarrow\Delta^'=m^2>0\Leftrightarrow m\ne0\)

TH1:

\(\orbr{\begin{cases}x_1=1\\x_2=2m+1\end{cases}}\) => \(\left(2m+1\right)=1\Leftrightarrow m=0\)loại

TH2

\(\orbr{\begin{cases}x_1=2m+1\\x_2=1\end{cases}}\)=>\(\left(2m+1\right)^2=1\Leftrightarrow4m^2+4m=0\Leftrightarrow4m+1=0\Leftrightarrow m=\frac{-1}{4}\)

\(\frac{x^2}{\sqrt{5}}-2\sqrt{5}=0\\ \frac{x^2}{\sqrt{5}}=2\sqrt{5}\\ \frac{x^2\sqrt{5}}{\sqrt{5}}=2\sqrt{5}.\sqrt{5}\\ x^2=10\\ x=+-\sqrt{10}\)

2)\(\sqrt{25\left(2x+1\right)^2}=0\\ 5\left(2x+1\right)=0\\ x=\frac{-1}{2}\)

\(\sqrt{9\left(x-1\right)}=21\)

\(\Leftrightarrow9\left(x-1\right)=441\)

\(\Leftrightarrow x-1=49\)

\(\Leftrightarrow x=50\)

Vậy x = 50

b) \(\sqrt{3}x+\sqrt{3}=\sqrt{12}+\sqrt{27}\)

\(\Leftrightarrow\left(x+1\right)\sqrt{3}=2\sqrt{3}+3\sqrt{3}\)

\(\Leftrightarrow\left(x+1\right)\sqrt{3}=\left(2+3\right)\sqrt{3}\)

\(\Leftrightarrow x+1=5\)

\(\Leftrightarrow x=4\)

Vậy x = 4

\(\sqrt{9\left(x-1\right)}=21\\9\left(x-1\right)=21^2\\x-1=49\\ x=48 \)\(\sqrt{3}x+\sqrt{3}=2\sqrt{3}+3\sqrt{3}\\ 0=\sqrt{3}\left(2+3-1-x\right)\\ 0=\sqrt{3}\left(4-x\right)\\ x=4\\ \)

a) \(\sqrt{\frac{\left(165-124\right)\left(165+124\right)}{164}}=\sqrt{\frac{41.289}{164}}=\sqrt{\frac{289}{4}}=\frac{17}{2}\)

b) tương tự ý a

c) \(\left(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\right)^2=7+4\sqrt{3}+7-4\sqrt{3}-2.\sqrt{7+4\sqrt{3}}.\sqrt{7-4\sqrt{3}}\)

\(=14-2\sqrt{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)

\(=14-2\sqrt{49-48}\)

\(=14-2.1=12\)

\(\Rightarrow\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}=\sqrt{12}=2\sqrt{3}\)

\(5xy\sqrt{\frac{x^2}{y^6}}=5\sqrt{\frac{x^4y^2}{y^6}}=5\sqrt{\frac{x^4}{y^4}}=5\left|\frac{x^2}{y^2}\right|=-5\)

\(5xy\sqrt{\frac{x^2}{y^6}}=5\sqrt{\frac{x^4y^2}{y^6}}=5\sqrt{\frac{x^4}{y^4}}=5\)

\(=\sqrt{2}\left(\sqrt{5}-1\right)\sqrt{3+\sqrt{5}}=\left(\sqrt{5}-1\right)\sqrt{6+2\sqrt{5}}\)

\(=\left(\sqrt{5}+1\right)\sqrt{\sqrt{5}^2+2\sqrt{5}+1}\)

\(=\left(\sqrt{5}+1\right)\sqrt{\left(\sqrt{5}+1\right)^2}=\left(\sqrt{5}+1\right)\left(\sqrt{5}+1\right)\)

\(=\left(\sqrt{5}+1\right)^2=5+2\sqrt{5}+1=6+\sqrt{5}\)