\(\text{Δ}=\left(-2m\right)^2-4\left(m^2-m-2\right)\)

\(=4m^2-4m^2+4m+8\)

=4m+8

Để phương trình có hai nghiệm thì 4m+8>=0

hay m>=-2

Theo đề, ta có: \(\left(x_1+x_2\right)^2-2x_1x_2=4\)

\(\Leftrightarrow\left(-2m\right)^2-2\left(m^2-m-2\right)=4\)

\(\Leftrightarrow4m^2-2m^2+2m=0\)

=>2m(m+1)=0

=>m=0 hoặc m=-1

Bài 2:

Theo đề, ta có hệ phương trình:

\(\left\{{}\begin{matrix}-2m-n+1=3\\4m-n+1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2m+n=-2\\4m-n=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6m=-4\\4m-n=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=-\dfrac{2}{3}\\n=4m+2=-\dfrac{8}{3}+2=-\dfrac{2}{3}\end{matrix}\right.\)

a, Theo tc 2 tt cắt nhau: \(AE=EC;BF=CF\)

Vậy \(AE+BF=EC+CF=EF\)

b, Vì \(\left\{{}\begin{matrix}AE=EC\\\widehat{EAO}=\widehat{ECO}=90^0\\OE.chung\end{matrix}\right.\) nên \(\Delta AOE=\Delta COE\)

\(\Rightarrow\widehat{AOE}=\widehat{EOC}\) hay OE là p/g \(\widehat{AOC}\)

Cmtt: \(\Delta BOF=\Delta COF\Rightarrow\widehat{BOF}=\widehat{COF}\) hay OF là p/g \(\widehat{BOC}\)

Vậy \(\widehat{EOF}=\widehat{COF}+\widehat{COE}=\dfrac{1}{2}\left(\widehat{AOC}+\widehat{BOC}\right)=90^0\) hay OE⊥OF

a, Vì \(\left\{{}\begin{matrix}OB=OC=R\\\widehat{OBM}=\widehat{OCN}=90^0\\\widehat{BOM}=\widehat{CON}\left(cùng.phụ.\widehat{MON}\right)\end{matrix}\right.\) nên \(\Delta BOM=\Delta CON\)

\(\Rightarrow OM=ON\)

b, Vì OM//AC(⊥OC) và ON//AB(⊥OB) nên AMON là hbh

Mà \(OM=ON\) nên AMON là hthoi

DKXD : \(x\ge-1;y\ne-1\)

Dat : \(\left\{{}\begin{matrix}\sqrt{x+1}=a\left(a\ge0\right)\\y+1=b\left(b\ne0\right)\end{matrix}\right.\)

hpt<=>\(\left\{{}\begin{matrix}a+2-\dfrac{2}{y+1}=2\\2a-\dfrac{1}{y+1}=\dfrac{3}{2}\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}a+2-\dfrac{2}{b}=2\\2a-\dfrac{1}{b}=\dfrac{3}{2}\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}a-\dfrac{2}{b}=0\\4a-\dfrac{2}{b}=3\end{matrix}\right.< =>\left\{{}\begin{matrix}3a=3\\a=\dfrac{2}{b}\end{matrix}\right.< =>\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)(tmdk)

\(=>\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)(tmdk)

Chọn A

\(=\dfrac{\sqrt{a}+2+\sqrt{a}-2}{a-4}:\dfrac{\sqrt{a}+2-2}{\sqrt{a}+2}\)

\(=\dfrac{2\sqrt{a}}{a-4}\cdot\dfrac{\sqrt{a}+2}{\sqrt{a}}=\dfrac{2}{\sqrt{a}-2}\)

a:

Bảng giá trị:

Vẽ đồ thị:

b; Tọa độ C là:

\(\begin{cases}2x+1=-2x+4\\ y=2x+1\end{cases}\Rightarrow\begin{cases}4x=3\\ y=2x+1\end{cases}\Rightarrow\begin{cases}x=\frac34\\ y=2\cdot\frac34+1=\frac32+1=\frac52\end{cases}\)

=>C(3/4;5/2)

Tọa độ A là:

\(\begin{cases}y=0\\ 2x+1=0\end{cases}\Rightarrow\begin{cases}y=0\\ 2x=-1\end{cases}\Rightarrow\begin{cases}y=0\\ x=-\frac12\end{cases}\)

=>A(-1/2;0)

Tọa độ B là:

\(\begin{cases}y=0\\ -2x+4=0\end{cases}\Rightarrow\begin{cases}y=0\\ -2x=-4\end{cases}\Rightarrow\begin{cases}y=0\\ x=2\end{cases}\)

=>B(2;0)

c: C(3/4;5/2); A(-1/2;0); B(2;0)

\(CA=\sqrt{\left(-\frac12-\frac34\right)^2+\left(0-\frac52\right)^2}=\sqrt{\left(-\frac54\right)^2+\left(-\frac52\right)^2}\)

\(=\sqrt{\frac{25}{16}+\frac{25}{4}}=\sqrt{\frac{25}{16}+\frac{100}{16}}=\sqrt{\frac{125}{16}}=\frac{5\sqrt5}{4}\)

\(CB=\sqrt{\left(2-\frac34\right)^2+\left(0-\frac52\right)^2}=\sqrt{\left(\frac54\right)^2+\left(\frac52\right)^2}=\frac{5\sqrt5}{4}\)

\(AB=\sqrt{\left(2+\frac12\right)^2+\left(0-0\right)^2}=\frac52\)

Chu vi tam giác ABC là:

\(C_{ABC}=AB+AC+BC\)

\(=\frac{5\sqrt5}{4}+\frac{5\sqrt5}{4}+\frac52=\frac{10\sqrt5}{4}+\frac52=\frac{5\sqrt5+5}{2}\)

Xét ΔABC có

\(cosC=\frac{CA^2+CB^2-AB^2}{2\cdot CA\cdot CB}\)

\(=\left(\frac{125}{16}+\frac{125}{16}-\frac{25}{4}\right):\left(2\cdot\frac{5\sqrt5}{4}\cdot\frac{5\sqrt5}{4}\right)=\left(\frac{125}{8}-\frac{25}{4}\right):\frac{2\cdot25\cdot5}{16}\)

\(=\frac{75}{8}\cdot\frac{16}{50\cdot5}=\frac{75}{50\cdot5}\cdot2=\frac{3}{2\cdot5}\cdot2=\frac35\)

=>\(\sin C=\sqrt{1-\left(\frac35\right)^2}=\frac45\)

Diện tích tam giác CAB là:

\(S_{CAB}=\frac12\cdot CA\cdot CB\cdot\sin C\)

\(=\frac12\cdot\frac{5\sqrt5}{4}\cdot\frac{5\sqrt5}{4}\cdot\frac45=\frac12\cdot\frac{25\cdot5}{16}\cdot\frac45=\frac{1}{32}\cdot25\cdot4=\frac{100}{32}=3,125\)