7. A = (x + y)^2 - 4y^2

= (x + y - 2y)(x + y + 2y)

= (x - y)(x + 3y)

2. x^4 + 4

= x^4 + 4x^2 + 4 - 4x^2

= (x^2 + 2)^2 - (2x)^2

= (x^2 + 2x + 2)(x^2 - 2x + 2)

3. 4x^4 + 16

= 4(x^4 + 4)

= 4(x^2 + 2x + 2)(x^2 - 2x + 2)

\(x^6+x^4-3x^2-4x+6\)

\(=\left(x^6+2x^5+4x^4+6x^3+5x^2\right)-\left(2x^5+4x^4+8x^3+12x^2+10x\right)+\left(x^4+2x^3+4x^2+6x+5\right)+1\)

\(=x^2\left(x^4+2x^3+4x^2+6x+5\right)-2x\left(x^4+2x^3+4x^2+6x+5\right)+\left(x^4+2x^3+4x^2+6x+5\right)+1\)

\(=\left(x^4+2x^3+4x^2+6x+5\right)\left(x^2-2x+1\right)+1\)

\(=\left[\left(x^4+2x^3+x^2\right)+3\left(x^2+2x+1\right)+2\right]\left(x-1\right)^2+1\)

\(=\left[\left(x^2+x\right)^2+3\left(x+1\right)^2+2\right]\left(x-1\right)^2+1\ge1\)

Dấu "=" xảy ra khi \(x=1\)

\(=ab\left(a-b\right)\left(a+b\right)+c^3\left(a-b\right)-c\left(a^3-b^3\right)\)

\(=\left(a-b\right)\left(a^2b+ab^2\right)+c^3\left(a-b\right)-\left(a-b\right)\left(a^2c+abc+b^2c\right)\)

\(=\left(a-b\right)\left(a^2b+ab^2+c^3-a^2c-abc-b^2c\right)\)

\(=\left(a-b\right)\left[ab\left(a-c\right)+b^2\left(a-c\right)-c\left(a^2-c^2\right)\right]\)

\(=\left(a-b\right)\left[ab\left(a-c\right)+b^2\left(a-c\right)-\left(a-c\right)\left(ac+c^2\right)\right]\)

\(=\left(a-b\right)\left(a-c\right)\left(ab+b^2-ac-c^2\right)\)

\(=\left(a-b\right)\left(a-c\right)\left[a\left(b-c\right)+\left(b-c\right)\left(b+c\right)\right]\)

\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)

1: Ta có: \(A=25x^4-24x^2-1\)

\(=25x^4-25x^2+x^2-1\)

\(=\left(x^2-1\right)\left(25x^2+1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(25x^2+1\right)\)

2: Ta có: \(A=64x^4+63x^2-1\)

\(=64x^4+64x^2-x^2-1\)

\(=\left(x^2+1\right)\left(64x^2-1\right)\)

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

3: Ta có: \(A=x^4-15x^2+50\)

\(=x^4-5x^2-10x^2+50\)

\(=\left(x^2-5\right)\left(x^2-10\right)\)

4: Ta có: \(A=-10x^4+9x^2+1\)

\(=-10x^4+10x^2-x^2+1\)

\(=\left(x^2-1\right)\left(-10x^2-1\right)\)

\(=-\left(10x^2+1\right)\left(x-1\right)\left(x+1\right)\)

8a.

\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\left(3x^2-5x+1\right)=3-5+1=-1\)

\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\left(-3x+2\right)=-3+2=-1\)

\(\Rightarrow\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)\Rightarrow\) hàm có giới hạn tại \(x=1\)

Đồng thời \(\lim\limits_{x\rightarrow1}f\left(x\right)=-1\)

b.

\(\lim\limits_{x\rightarrow2^+}f\left(x\right)=\lim\limits_{x\rightarrow2^+}\dfrac{x^3-8}{x-2}=\lim\limits_{x\rightarrow2^+}\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{x-2}\)

\(=\lim\limits_{x\rightarrow2^+}\left(x^2+2x+4\right)=12\)

\(\lim\limits_{x\rightarrow2^-}f\left(x\right)=\lim\limits_{x\rightarrow2^-}\left(2x+1\right)=5\)

\(\Rightarrow\lim\limits_{x\rightarrow2^+}f\left(x\right)\ne\lim\limits_{x\rightarrow2^-}f\left(x\right)\Rightarrow\) hàm ko có giới hạn tại x=2

9.

\(\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^+}\dfrac{x^2+mx+2m+1}{x+1}=\dfrac{0+0+2m+1}{0+1}=2m+1\)

\(\lim\limits_{x\rightarrow0^-}f\left(x\right)=\lim\limits_{x\rightarrow0^-}\dfrac{2x+3m-1}{\sqrt{1-x}+2}=\dfrac{0+3m-1}{1+2}=\dfrac{3m-1}{3}\)

Hàm có giới hạn khi \(x\rightarrow0\) khi:

\(\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^-}f\left(x\right)\Rightarrow2m+1=\dfrac{3m-1}{3}\)

\(\Rightarrow m=-\dfrac{4}{3}\)

?????

cho mình xin đề bài với ạ

1: does not/doesn't
2:did not/didn't
3:are not/aren't
4:are not/aren't
5:does not/doesn't
mình cũng ko chắc vài câu vì ko có từ ngữ cụ thể ạ

mik nghĩ là:

Cho em gặt hái mỗi ngày.