a j98 lên chatgpt hỏi cho nó nhanh a ơi

`sqrt{10 + 4sqrt{6}}`

`=sqrt{10 + 2. 2sqrt{6}}`

`=sqrt{sqrt{6}^2 + 2. 2sqrt{6} + 2^2}`

`=sqrt{(sqrt{6}+ 2)^2}`

`= sqrt{6}+ 2`

\(\sqrt{10+4\sqrt{6}}=\sqrt{4.\dfrac{5}{2}+4\sqrt{6}}=2\sqrt{\dfrac{5\sqrt{6}}{2}}\)

em mới năm nay lên lớp 6 thôi nhưng vì em học tiếng anh sớm nên em có làm được một đoạn văn tiếng anh giống với đề bài,một số từ ngữ em không biết nên dùng sách tham khảo ạ:School days in the past for fifth graders were a mix of routine and fun. Classes started early, around 8 a.m., with subjects like math, reading, and science. Students used chalk and blackboards, and textbooks were often well-worn. During breaks, kids played outside, enjoying games like tag or jump rope. Summer holidays were a highlight, offering endless days for playing with friends, exploring nature, or visiting family. Getting to school might mean walking, biking, or taking a horse-drawn carriage. These simpler times were filled with laughter, learning, and the excitement of childhood adventures.

oh bạn học cả t anhh lớp 9 trc aF

\(\left(x+2\right)^2-\left(2x+1\right)\left(x+2\right)=0\\ < =>\left(x+2\right)\left[\left(x+2\right)-\left(2x+1\right)\right]=0\\ < =>\left(x+2\right)\left(x+2-2x-1\right)=0\\ < =>\left(x+2\right)\left(1-x\right)=0\\ < =>\left[{}\begin{matrix}x+2=0\\1-x=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

Vậy: ... 

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

=>(x+2)(x+2-2x-1)=0

=>(x+2)(-x+1)=0

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

Để hpt có nghiệm thì: 

\(\dfrac{m}{4}\ne\dfrac{1}{-m}\Leftrightarrow m^2\ne-4\Leftrightarrow m\in R\)

\(\left\{{}\begin{matrix}mx+y=5\\4x-my=1\end{matrix}\right.< =>\left\{{}\begin{matrix}m^2x+my=5m\\4x-my=1\end{matrix}\right.< =>\left\{{}\begin{matrix}\left(m^2+4\right)x=5m+1\\mx+y=5\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x=\dfrac{5m+1}{m^2+4}\\\dfrac{5m^2+m}{m^2+4}+y=5\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{5m+1}{m^2+4}\\y=5-\dfrac{5m^2+m}{m^2+4}\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x=\dfrac{5m+1}{m^2+4}\\y=\dfrac{5m^2+20-5m^2-m}{m^2+4}\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{5m+1}{m^2+4}\\y=\dfrac{20-m}{m^2+4}\end{matrix}\right.\)

Ta có: 

\(2y=1-x=>2\cdot\dfrac{20-m}{m^2+4}=1-\dfrac{5m+1}{m^2+4}\\ \Leftrightarrow\dfrac{40-2m}{m^2+4}=\dfrac{m^2+4-5m-1}{m^2+4}\\ \Leftrightarrow40-2m=m^2-5m+3\\ \Leftrightarrow m^2-5m+3+2m-40=0\\ \Leftrightarrow m^2-3m-37=0\)  

\(\Delta=\left(-3\right)^2-4\cdot1\cdot\left(-37\right)=157>0\\ m_1=\dfrac{3+\sqrt{157}}{2}\\ m_2=\dfrac{3-\sqrt{157}}{2}\)