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a: \(=a^2-b^4\)
b: \(=\left(a^2+2a\right)^2-9\)
c: \(=a^2-\left(2a+3\right)^2\)
d: \(=a^4-\left(2a-3\right)^2\)
e: \(=\left(-a^2-2a+3\right)^2\)
g: \(=4a^2-a^4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2 :
a ) \(A=\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(A=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\)
\(A=\left(a^2+2ab+b^2\right)+\left(a^2+2ac+c^2\right)+\left(b^2+2bc+c^2\right)\)
\(A=\left(a+b\right)^2+\left(a+c\right)^2+\left(b+c\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+2\left(x+1\right)^2+3\left(x-2\right)^2+4\left(x+3\right)^2\)
\(=x^2+2\left(x^2+2x+1\right)+3\left(x^2-4x+4\right)+4\left(x^2+6x+9\right)\)
\(=x^2+2x^2+4x+2+3x^2-12x+12+4x^2+24x+36\)
\(=10x^2+16x+50\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/\(=4a^2+4b^2+c^2+8ab-4bc-4ca+4b^2+4c^2+a^2+8bc-4ca-4ab+4a^2+4c^2+b^2+8ca-4bc-4ab=\)
\(=9a^2+9b^2+9c^2=9\left(a^2+b^2+c^2\right)\)
2/
Ta có
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge0\)
\(\Leftrightarrow a^2+b^2+c^2\ge-2\left(ab+bc+ca\right)=2\)
\(\Rightarrow P=9\left(a^2+b^2+c^2\right)\ge18\)
\(\Rightarrow P_{min}=18\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
\(A=4(a+b)^2+c^2-4c(a+b)+4(b+c)^2+a^2-4a(b+c)+4(c+a)^2+b^2-4b(a+c)\)
\(\Leftrightarrow A=4(a+b)^2+4(b+c)^2+4(c+a)^2-8(ab+bc+ac)\)
\(\Leftrightarrow A=4(a^2+b^2+2ab)+4(b^2+c^2+2bc)+4(c^2+a^2+2ac)-8(ab+bc+ac)\)
\(\Leftrightarrow A= 8(a^2+b^2+c^2)=8m\)
\(\left(a^2+2a\right)\left(2a-a^2\right)=4a^2-a^4\)
Bài làm:
Ta có: \(\left(a^2+2a\right)\left(2a-a^2\right)\)
\(=4a^2-a^4\)