\(\Delta=\left(m-1\right)^2-4.\left(-m\right)\)

\(=\left(m^2-2m+1\right)+4m=\left(m+1\right)^2\)

Để pt có 2 nghiệm phân biệt => \(m\ne-1\)

\(\left[{}\begin{matrix}x_1=\dfrac{m-1+m+1}{2}=m\\x_2=\dfrac{m-1-m-1}{2}=-1\end{matrix}\right.\)

Để pt có 2 nghiệm phân biệt bé 1

\(\Rightarrow m< 1\)

Ta có:

\(x^2+\dfrac{1}{x^2}=4\)\(\left(x\ne0\right)\)

\(\left(x^2+\dfrac{1}{x^2}\right)^2=16\)

\(x^4+\dfrac{2.x^2}{x^2}+\dfrac{1}{x^4}=16\)

\(x^4+\dfrac{1}{x^4}=16-2=14\)

\(2024.3+2024.9-12.24\)

=\(2024\left(3+9\right)-12.24\)

\(=12.2024-12.24\)

\(=12\left(2024-24\right)\)

\(=12.2020=24240\)

a) Ta có:

\(CA\perp AB;DB\perp AB\)

\(\Rightarrow AC//BD\)

b) \(\widehat{ACD}=\widehat{BDE}=78^o\) (đồng vị)

\(\left(2x-3\right)^{2022}=\left(2x-3\right)^{2021}\)

\(\left(2x-3\right)^{2021}\left(2x-3-1\right)=0\)

\(\left(2x-3\right)^{2021}\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)

\(A=\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\)

\(=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\)

\(=a-b\)

Ta có:

\(\widehat{xOn}=180^o-\widehat{nOy}=180^o-30^o=150^o\)

\(\widehat{xOm}=\widehat{nOy}=30^o\)(đối đỉnh)

\(\widehat{mOy}=\widehat{xOn}=150^o\)(đối đỉnh)

a) \(\widehat{ABy}=\widehat{xAB}=120^o\)(so le trong)

b) \(\widehat{BCz}=\widehat{BAx}=120^o\) mà 2 góc có vị trí đồng vị

=> Cz//Ax

=> By//Ax//Cz

a) x=16

=> \(A=\dfrac{\sqrt{16}+2}{\sqrt{16}+5}=\dfrac{6}{9}=\dfrac{2}{3}\)

b) \(B=\dfrac{x+20+\sqrt{x}-2-6\left(\sqrt{x}+2\right)}{x-4}\)

\(=\dfrac{x-5\sqrt{x}+6}{x-4}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x+2}}\)

c) \(\sqrt{AB}=\sqrt{\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x+5}\right)\left(\sqrt{x}+2\right)}}=\sqrt{\dfrac{\sqrt{x}-3}{\sqrt{x}+5}}\left(đk:x\ge9\right)\)

\(\sqrt{AB}< \dfrac{1}{2}\Rightarrow\sqrt{\dfrac{\sqrt{x}-3}{\sqrt{x}+5}}< \dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{x}+5>4\left(\sqrt{x}-3\right)\)

\(\Leftrightarrow3\sqrt{x}< 17\)

\(\Leftrightarrow\sqrt{x}< \dfrac{17}{3}\) \(\Rightarrow0\le x< \dfrac{289}{3}\)

Gọi \(d=\left(n+2;2n+3\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n+2⋮d\\2n+3⋮d\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2n+4⋮d\\2n+3⋮d\end{matrix}\right.\)

\(\Rightarrow\left(2n+4\right)-\left(2n+3\right)⋮d\)

\(\Rightarrow\)\(1⋮d\Rightarrow d=1\)