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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{2ab}{4ab}+\frac{2ab}{a^2+4b^2}+\frac{1}{8ab}-\frac{1}{2}\)
áp dụng bđt AM-GM , a,b> 0
\(\Rightarrow A\ge2ab\left(\frac{4}{4ab+a^2+4b^2}\right)+\frac{1}{8ab}-\frac{1}{2}\)
\(\Rightarrow A\ge\frac{8ab}{1}+\frac{1}{8ab}-\frac{1}{2}\)
\(\Rightarrow A\ge2-\frac{1}{2}=\frac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
b: Ta có: \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x^3-4x-x^4+1\)
\(=-x^4+x^3-4x+1\)
c: Ta có: \(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2ab\)
\(=\left(a+b-c-a+c\right)\left(a+b-c+a-c\right)\)
\(=b\left(2a+b-2c\right)\)
\(=2ab+b^2-2bc\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(\left(a+b\right)^2\)
\(=\left(a+b\right)\left(a+b\right)\)
\(=a^2+ab+ab+b^2\)
\(=a^2+2ab+b^2\left(dpcm\right)\)
2) \(\left(a-b\right)^3\)
\(=\left(a-b\right)\left(a-b\right)\left(a-b\right)\)
\(=\left(a^2-ab-ab+b^2\right)\left(a-b\right)\)
\(=\left(a^2-2ab+b^2\right)\left(a-b\right)\)
\(=a^3-a^2b-2a^2+2ab^2+ab^2-b^3\)
\(=a^3-3a^2b+3ab^2-b^3\left(dpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)