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a) \(a^2+b^2+c^2\ge ab+bc+ac\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)( luôn đúng )
Dấu "=" \(\Leftrightarrow a=b=c\)
b) \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
+) vế 1 bđt \(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ac\)( CMTT câu a )
+) vế 2 bđt \(\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ac\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ac\)( CMTT câu a )
Từ đây ta có đpcm
Dấu "=" \(\Leftrightarrow a=b=c\)
c) \(a^3+b^3\ge ab\left(a+b\right)\)
\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)\ge ab\left(a+b\right)\)
\(\Leftrightarrow a^2-ab+b^2\ge ab\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)( luôn đúng )
Dấu "=" \(\Leftrightarrow a=b\)
2. \(\left(a+b\right)^3-\left(a-b\right)^3\)
\(=\left(a+b-a+b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2b\left[\left(a^2+2ab+b^2\right)+\left(a^2-b^2\right)+\left(a^2-2ab+b^2\right)\right]\)
\(=2b\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)
\(=2b\left(3a^2+b^2\right)\)
Ta có
x + y = 2
=> (x+y)^2 = 4
=> x^2 + 2xy + y^2 = 4
=> 10 + 2xy= 4
=> 2xy = -6
=> xy= -3
x^3 + y^3 = ( x+Y) ( x^2 - xy + y^2) = 2 ( 10 -- 3) = 2( 10 + 3 ) = 2.13 = 26
\(ĐK:x\ne-1\\ \left|x\right|=2\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
Với \(x=2\Leftrightarrow A=\dfrac{3}{2+1}=1\)
Với \(x=-2\Leftrightarrow A=\dfrac{3}{-2+1}=-3\)
\(A=a^3+b^3+3ab\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\)
\(=a^2-ab+b^2+3ab\)
\(=a^2+2ab+b^2\)
\(=\left(a+b\right)^2=1^2=1\)
\(B=4\left(a^3+b^3\right)-6\left(a^2+b^2\right)\)
\(=4\left(a+b\right)\left(a^2-ab+b^2\right)-6a^2-6b^2\)
\(=4\left(a^2-ab+b^2\right)-6a^2-6b^2\)
\(=4a^2-4ab+4b^2-6a^2-6b^2\)
\(=-2a^2-4ab-2b^2\)
\(=-2\left(a^2+2ab+b^2\right)\)
\(=-2\left(a+b\right)^2=-2.1^2=-2\)