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Ta có:\(3\left(a^2+b^2+c^2\right)=\left(a+b+c\right)^2\)
\(\Leftrightarrow3a^2+3b^2+3c^2=a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\) với mọi \(a;b;c\inℝ\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) với mọi \(a;b;c\inℝ\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)
\(\Rightarrow a=b=c\)
\(\Rightarrow P=a^3+a^3+c^3-3.a.a.a\)
\(\Leftrightarrow P=3a^3-3a^3\)
\(\Leftrightarrow P=0\)
Vậy ...
`556^2 - 553 . 559 `
`= 556^2 - (556 - 3) . (556 + 3) `
`= 556^2 - (556^2 - 3^2)`
`= 556^2 - 556^2 + 9`
`= 0 + 9`
= 9
`456^2 + 456 . 88 + 44^2`
`= 456^2 + 456 . 88 + 44^2`
`= 456^2 + 2 .456 . 4 + 44^2`
`= (456 + 44)^2`
`= 500^2`
`= 250000`
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Áp dụng các HDT sau nhé:
`(a+b)^2 = a^2 + 2ab + b^2`
`a^2 - b^2 = (a+b)(a-b)`
(\(x+1\)) + (\(x-1\))2
= \(x\) + 1 + \(x^2\) - 2\(x\) + 1
= \(x^2\) - (2\(x\) - \(x\)) + (1 + 1)
= \(x^2\) - \(x\) + 2
\(\left(x+1\right)+\left(x-1\right)^2\\ =\left(x+1\right)+\left(x^2-2x+1\right)\\ =x+1+x^2-2x+1\\ =x^2+\left(x-2x\right)+\left(1+1\right)\\ =x^2-x+2\)
Đặt \(x^2-x+1=a;x+1=b\)
Phương trình sẽ trở thành: \(3a^2-2b^2=5ab\)
=>\(3a^2-5ab-2b^2=0\)
=>\(3a^2-6ab+ab-2b^2=0\)
=>3a(a-2b)+b(a-2b)=0
=>(a-2b)(3a+b)=0
=>\(\left[{}\begin{matrix}a-2b=0\\3a+b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2-x+1-2\left(x+1\right)=0\\3\left(x^2-x+1\right)+x+1=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x^2-x+1-2x-2=0\\3x^2-3x+3+x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2-3x-1=0\\3x^2-2x+4=0\end{matrix}\right.\)
=>\(x^2-3x-1=0\)
=>\(x=\dfrac{3\pm\sqrt{13}}{2}\)
`(x+1)^2 + (x-1)^2`
`= x^2 + 2x + 1 + x^2 - 2x + 1`
`= 2x^2 + 2`
`= 2(x^2 +1)`
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Áp dụng hằng đẳng thức:
\(\left(a\pm b\right)^2=a^2\pm2ab+b^2\)
\(\left(x+1\right)^2-\left(x-1\right)^2\\ =\left[\left(x+1\right)-\left(x-1\right)\right]\left[\left(x+1\right)+\left(x-1\right)\right]\\ =\left(x+1-x+1\right)\left(x+1+x-1\right)\\ =2\cdot2x\\ =4x\)
\(a.\left(x+y+4\right)\left(x+y-4\right)\\ =\left[\left(x+y\right)+4\right]\left[\left(x+y\right)-4\right]\\ =\left(x+y\right)^2-4^2\\ b.\left(x-y+6\right)\left(x+y-6\right)\\ =\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\\ =x^2-\left(y-6\right)^2\\ c.\left(y+2z-3\right)\left(y-2z-3\right)\\ =\left[\left(y-3\right)+2z\right]\left[\left(y-3\right)-2z\right]\\ =\left(y-3\right)^2-\left(2z\right)^2\\ d.\left(x+2y+3z\right)\left(2y+3z-x\right)\\ =\left[\left(2y+3z\right)+x\right]\left[\left(2y+3z\right)-x\right]\\ =\left(2y+3z\right)^2-x^2\)