a: \(a^4+b^4\ge2a^2b^2\)

\(\Leftrightarrow a^4-2a^2b^2+b^4>=0\)

hay \(\left(a^2-b^2\right)^2\ge0\)(luôn đúng)

d: \(\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)

Trả lời nhanh giúp mình ạ

1) FD n 

2) BK n

3) RT k

4) LT k

5) PU

6) Home

7) CS

8) HT

9) PD

10) Clean

Xin lỗi mik chỉ nhớ được có nhiêu đó à ;-; thông cảm nha

\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{\left(x-1\right)^2}{x^2-1}\right).\frac{x+2003}{x}\)ĐKXĐ: \(x\ne-1;0;1\)

\(A=\frac{\left(x+1\right)^2-\left(x-1\right)^2+\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}.\frac{x+2003}{x}\)

\(A=\frac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}.\frac{x+2003}{x}\)

\(A=\frac{x+1}{x-1}.\frac{x+2003}{x}\)

\(A=\frac{x^2+2004x+2003}{x^2-x}\)

Bài làm:

Ta có: \(a^2+2ab+b^2\)

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

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

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

\(=\left(a+b\right)^2\)

(a+b)²=(a+b)(a+b)=a²+ab+ba+b²=a²+2ab+b²

Trong SGK có ko nhỉ '-', cái này dễ mà:<

cái này t lm trc chx lấy đc sgk nên ko bt lm

a)\(\dfrac{-2}{17}+\dfrac{8}{17}=\dfrac{6}{17}\)

b) \(\dfrac{-9}{16}+\dfrac{7}{24}=\dfrac{-13}{48}\)

c)\(\dfrac{1}{8}+\dfrac{-5}{6}+\dfrac{7}{-36}=\dfrac{-17}{24}+\dfrac{7}{-36}=\dfrac{-65}{72}\)

a) =\(\dfrac{-2+8}{17}=\dfrac{6}{17}\)
b) =\(\dfrac{-27}{48}+\dfrac{14}{48}=\dfrac{-27+14}{48}=\dfrac{-13}{48}\)
c) \(=\dfrac{9}{72}+\dfrac{-60}{72}+\dfrac{-14}{72}=\dfrac{9+\left(-60\right)+\left(-14\right)}{72}=\dfrac{-65}{72}\)

b: Ta có: \(N=a^3+b^3+3ab\)

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

\(=1-3ab+3ab\)

=1