\(\rightarrow\) Nên suy nghĩ trước khi chép nha =)) . Này tui tấu ** á .... Khéo tẩu hỏa nhập viện à nhen ^^ 

One of the most unforgettable experiences I’ve had happened when I was a teenager, and it involved a rather silly incident where I, along with some friends, stole green mangoes from a neighbor’s tree. On a summer day, we were hanging out when we noticed the tree in the neighbor’s yard, covered in unripe mangoes. Without thinking twice, we decided to pick a few for ourselves.

We snuck over to the tree, climbing over the fence carefully and silently. It felt thrilling, like a little adventure, as we reached for the best mangoes. However, just as we were about to leave, the neighbor saw us from her window. She came out, furious, and demanded to know what we were doing. I was caught red-handed, and in that moment, I felt a wave of embarrassment wash over me.

The neighbor called my parents, and I had to apologize and return the mangoes. My parents were not happy, and I was grounded for a week, having to do extra chores around the house. Although it was a shameful experience, I learned an important lesson about respecting others’ property and thinking carefully before acting. Looking back, I realize it was a mistake, but also a valuable life lesson.

\(\rightarrow\) Duong Lam is one of the most famous  villages in Vietnam. 

Duong Lam village is one of the most famous villages in Vietnam

a: \(\text{Δ}=2^2-4\cdot1\cdot\left(-m^2-1\right)\)

\(=4+4m^2+4=4m^2+8>=8>0\forall m\)

=>Phương trình luôn có hai nghiệm phân biệt

b: Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-2\\x_1x_2=\dfrac{c}{a}=-m^2-1\end{matrix}\right.\)

a: Thay m=-1 vào phương trình, ta được:

\(x^2-2\left(-1-1\right)x-4\cdot\left(-1\right)=0\)

=>\(x^2+4x+4=0\)

=>\(\left(x+2\right)^2=0\)

=>x+2=0

=>x=-2

b: \(\text{Δ}=\left[-2\left(m-1\right)\right]^2-4\cdot1\cdot\left(-4m\right)\)

\(=4\left(m^2-2m+1\right)+16m\)

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

Để phương trình có hai nghiệm phân biệt thì Δ>0

=>\(4\left(m+1\right)^2>0\)

=>\(m+1\ne0\)

=>\(m\ne-1\)

a: \(\text{Δ}=\left(m+1\right)^2-4\cdot1\cdot\left(m-2\right)\)

\(=m^2+2m+1-4m+8=m^2-2m+9\)

\(=m^2-2m+1+8=\left(m-1\right)^2+8>=8>0\forall m\)

=>Phương trình luôn có hai nghiệm phân biệt

b: Theo vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=m+1\\x_2x_1=\dfrac{c}{a}=m-2\end{matrix}\right.\)

c: \(A=x_1^2+x_2^2-6x_1x_2\)

\(=\left(x_1+x_2\right)^2-8x_1x_2\)

\(=\left(m+1\right)^2-8\left(m-2\right)\)

\(=m^2+2m+1-8m+16\)

\(=m^2-6m+17=\left(m-3\right)^2+8>=8\forall m\)

Dấu '=' xảy ra khi m-3=0

=>m=3

Thay x=2 vào phương trình, ta được:

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

=>4+4m-2-2m=0

=>2m+2=0

=>2m=-2

=>m=-1

Theo Vi-et, ta có:

\(x_1+x_2=-\dfrac{b}{a}=-2m+1=-2\cdot\left(-1\right)+1=3\)

=>\(x_2=3-x_1=3-2=1\)

a: Thay m=2 vào (1), ta được:

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

=>\(x^2+2x+1=0\)

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

=>x+1=0

=>x=-1

b: \(\text{Δ}=m^2-4\left(m-1\right)=m^2-4m+4=\left(m-2\right)^2>=0\forall m\)

=>Phương trình luôn có nghiệm với mọi m

c: Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-m\\x_1x_2=\dfrac{c}{a}=m-1\end{matrix}\right.\)

\(A=\left(x_1+1\right)^2\cdot\left(x_2+1\right)^2+2016\)

\(=\left[\left(x_1+1\right)\left(x_2+1\right)\right]^2+2016\)

\(=\left[x_1x_2+x_1+x_2+1\right]^2+2016\)

\(=\left[m-1-m+1\right]^2+2016=2016\)